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Radius of convergence

  1. Aug 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the radius of convergence for [tex]\Sigma[/tex] [tex]\frac{nx^{2n}}{2^{n}}[/tex]

    2. Relevant equations
    Ratio test


    3. The attempt at a solution

    I apply the ratio test to get [tex]\frac{(n+1)(x^{2})}{2n}[/tex]. I let n approach infinity, to get [tex]\frac{1}{2}[/tex]. So, this series converges when |x2|<1. So, I have -1<x2<1. Now I'm lost, where can I go from here?
     
  2. jcsd
  3. Aug 10, 2008 #2
    I think that should be x^2 < 2 actually. Since it will converge if x^2 / 2 is less than 1.Also, I don't believe you need to make it absolute value of x^2 since x^2 will always be positive.
     
    Last edited: Aug 10, 2008
  4. Aug 10, 2008 #3
    So it converges on this interval (-[tex]\infty[/tex],2)?
     
  5. Aug 10, 2008 #4
    Remember you have x^2 < 2. Just solve it for x. And about that absolute value thing, disregard what I said.
     
  6. Aug 10, 2008 #5
    So, I get x<[tex]\pm[/tex][tex]\sqrt{2}[/tex] (I know that doesn't make sense).
     
  7. Aug 10, 2008 #6
    Ok, well my instruction in my first post to remove the absolute value signs was dumb. If you take the root of both sides you get root of x^2 < root of 2. Remember that the root of x^2 is the absolute value of x anyway. So you see why what I told you was such a waste of time?

    To clarify, you have:

    [tex]\left|x^{2}\right|<2[/tex] then

    [tex]\left|x\right|<\sqrt{2}[/tex]
     
    Last edited: Aug 10, 2008
  8. Aug 10, 2008 #7
    Ok I'm feeling pretty dense right now, so I'm going to say the radius is -[tex]\sqrt{2}[/tex]<x<[tex]\sqrt{2}[/tex]?
     
  9. Aug 10, 2008 #8
    Haha, ok thank you! I see your an east coaster, lol. For me it's only 12.
     
  10. Aug 10, 2008 #9
    Just to clarify, I have no idea what I am talking about and should stop clarifying. I was right originally! It SHOULD be absolute x is less than root 2. You DO need need the radius to be in terms of x to the power of one. Live and learn. My apologies, I am watching a debate with creationists I could send you if you require me to make ammends.
     
  11. Aug 10, 2008 #10

    HallsofIvy

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    The instruction to take the root originally was correct. That is because the "2" in the exponent of x.

    By the ratio test we must have
    [tex]\frac{(n+1)|x|^{2(n+1)}}{2^{n+1}}\frac{2^n}{n|x|^n}= \frac{n+1}{n}\cdot\frac{1}{2}|x|^n< 1[/tex]

    The limit of that is not "1/2", it is (1/2)|x|2 which must be less than 1: (1/2)|x|2< 1 so |x|2< 2 and [itex]|x|< \sqrt{2}[/itex].
     
  12. Aug 10, 2008 #11
    Ok thank you. So because it is a power series, it will converge when [tex]\frac{1}{2}[/tex]|x2|<1, and diverge when [tex]\frac{1}{2}[/tex]|x2|>1.
     
    Last edited: Aug 10, 2008
  13. Aug 10, 2008 #12

    HallsofIvy

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    More correctly, it will converge absolutely for for (1/2)x2|< 1, diverge for (1/2)x2|> 1 and may converge conditionally or diverge when (1/2)x2|= 1.

    That is, of course, because the ratio test says that a sum, [itex]\sum_{n=0}^\infty a_n[/itex] of positive numbers will converge if [itex]lim a_{n+1}/a_n< 1[/itex] converge for [itex]lim a_{n+1}/a_n >1[/itex] and may converge or diverge if that limit is 1.
     
  14. Aug 10, 2008 #13
    Ok great! Thank you for the clarification.
     
  15. Aug 11, 2008 #14

    HallsofIvy

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    And, by the way, "the radius is [tex]-\sqrt{2}< x< \sqrt{2}[/tex]" is wrong. That is the interval of convergence. The radius of convergence is [itex]\sqrt{2}[/itex]
     
  16. Aug 11, 2008 #15
    That makes sense. Thanks.
     
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