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Power series

  1. Dec 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the radius of convergence and the interval of convergence


    2. Relevant equations

    A_n = Ʃ sum n =1 to infinity [((-1)^n) x^(2n+1)]/(2n+1)!

    3. The attempt at a solution
    All I thought was to use the ratio test so I did A_(n+1) /A_n

    = ((x^(2n+1))/(2n+1)!) ( (2n+1)!)/ [x^(2n+1)]
    I got |x^2| limit n ---> |x^2 / 4n^2 +10n +6 | when I simplified and such. So the limit is equal to 0 so my interval of convergence is 0 ? and R = 0?
     
  2. jcsd
  3. Dec 2, 2013 #2

    mfb

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    Staff: Mentor

    If the limit of A_(n+1)/A_n is zero, the radius of convergence is not 0. Check how the ratio test works.
     
  4. Dec 2, 2013 #3
    It is true that the limit is 0. However, this does not mean that the radius of convergence is 0. You did the more difficult part correctly, take a other look at the definition of ratio test and how to find radius of convergence from that.
     
  5. Dec 2, 2013 #4
    Less then one. Converges right? Radius of convergence umm |x^2| < 1 so x < +-1?
    No?
     
  6. Dec 2, 2013 #5
    Not quite. Are you familiar with a simple equation relating limit and radius of convergence?
     
  7. Dec 2, 2013 #6
    No. It converges for all values of x so radius of convergence is infinity for all x? What is the equation?
     
  8. Dec 2, 2013 #7
    That's correct. It does indeed converge for all values of x. And the small equation I was referring to is R (radius of convergence) = 1/L (the limit). You assume that 1/infinity = 0 and 1/0 = infinity. So in this case, your answer for the limit is 0, so 1/0 = Infinity which implies an infinite radius of convergence.
     
  9. Dec 2, 2013 #8
    Question when using the ratio test for power series and your limit is equal to 1 then you just...like this

    |x| limit n --> (n^2 +1)/(n^2 +2) = |x| (1) then you say

    You need |x| <1 and go from there? Even though your limit is one you can still use it although with the ratio test if you get a limit of 1 it is inclusive?
     
  10. Dec 2, 2013 #9
    Yes. What you can say is if the limit is 1, the radius is also one, so you can find the interval by adding and subtracting 1 from the center, even though the ratio test for convergence is I conclusive. That is because in this case, you aren't looking for the limit (which would be an inconclusive convergence test). Rather, you want 1/L, so it doesn't matter in this case that the ratio test for convergence is inconclusive.
     
  11. Dec 2, 2013 #10
    Where did you get your 1/L radius of convergence eq. ? I want to read this
     
  12. Dec 2, 2013 #11
    I am currently in the same class (i assume integral calculus) and that is what we were taught.
     
  13. Dec 2, 2013 #12

    mfb

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    It is true, and you can prove it. For a series around the origin (everything else is just more to type, it does not change anything):

    Assume the ratio of ##\displaystyle \frac{|A_{n+1}|}{|A_n|}## approaches a limit of L.

    Consider an arbitrary value of x with |x|<1/L.
    Then the ratio ##\displaystyle \frac{|x^{n+1}A_{n+1}|}{|x^n A_n|} = |x| \frac{|A_{n+1}|}{|A_n|}## approaches |x|*L < 1. Therefore, the series converges at that point via the regular ratio tests for series.


    Consider an arbitrary value of x with |x|>1/L.
    Then the ratio ##\displaystyle \frac{|x^{n+1}A_{n+1}|}{|x^n A_n|} = |x| \frac{|A_{n+1}|}{|A_n|}## approaches |x|*L > 1. Therefore, the series does not converge at that point via the regular ratio tests for series.


    Therefore, the radius of convergence is R=1/L.

    With the definitions 1/0=infinity, 1/infinity=0 this proof works for the two special cases as well.
     
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