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Radius of convergence derivation

  1. Nov 3, 2014 #1
    Hi,

    I am likely just missing something fundamental here, but I recently just revisited series and am looking over some notes.

    In my notes, I have written that if

    ## \lim_{x \to +\infty} \frac{a_{n+1}}{a_n} = L ##

    Then ## | x - x_o | = 1/L ##

    But shouldn't the correct expression be $$ | x - x_o | = L $$ ?

    Why is the radius of convergence 1/L instead of L? I was under the assumption since

    ## \frac{(x - x_o)^{n+1}} {(x-x_o)^n} = x - x_o ##

    that this term (absolute value) would be less than and/or (possibly) equal to L? Are my notes inaccurate or is there a reason for R being 1/L?
     
    Last edited: Nov 3, 2014
  2. jcsd
  3. Nov 3, 2014 #2

    Stephen Tashi

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    What does [itex] x [/itex] have to do with [itex] a_n [/itex]. Is [itex] a_n [/itex] some function of [itex] x [/itex] ?

    Did you mean [itex]lim_{n\to\infty} [/itex] ?
     
  4. Nov 3, 2014 #3
    Yes. Sorry, that's a typo. It is [itex]lim_{n\to\infty} [/itex] as you noted.
     
  5. Nov 3, 2014 #4

    mathman

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    It would be cleare if you started from the beginning. It is hard to guess the original statement of the situation.
     
  6. Nov 3, 2014 #5

    Mark44

    Staff: Mentor

    This doesn't make much sense. For starters, the limit should be as n changes, not x. More importantly, if L > 1, the series diverges, so it doesn't make any sense to talk about the radius of convergence.

    The limit above appears in the Ratio Test. If L > 1, the series diverges. If L < 1, the series converges. If L = 1, the test isn't conclusive.
     
  7. Nov 4, 2014 #6

    Stephen Tashi

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    Apparently the series in question has the n-th term [itex] T_n = a_n (x - x0)^n [/itex]. The ratio relevant to radius of convergence is [itex] \frac{T_{n+1}}{T_n} = \frac{a_{n+1}}{a_n} \frac{(x-x_0)^{n+1}} {(x-x_0)^n} [/itex]. If the ratio is to be 1 in the limit and the limit of [itex] \frac{ a_{n+1} } {a_n} [/itex] is [itex] L [/itex] as [itex] n [/itex] approaches infinity then you need the ratio [itex] \frac {(x-x_0)^{n+1}} {(x-x0)^n} [/itex] to be [itex] \frac{1}{L} [/itex].
     
  8. Nov 4, 2014 #7
    The power series ##\sum\limits_{n=0}^\infty a_n(x-x_0)^n## converges, by the ratio test, when ##\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}\right|<1## and diverges when ##\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}\right|>1##.

    After a bit of fiddling, you get that the power series converges when ##|x-x_0|<1/L## and diverges when ##|x-x_0|>1/L##, where ##L=\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|##.
     
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