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Thread: radius of convergence

  1. #1
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    radius of convergence

    im learning about radius of convergence now but i still have yet to grasp its concept.

    let "R hat" = (lim sup la_nl^1/n )^-1
    assume that a_n is not 0 and the seq of positive numbers la_nl/la_n+1l converges to a limit. how do i prove that the lim la_nl/la_n+1l = 'R hat" where R hat is the radius of convergence of the series( sum from n = 1 to infinitiy a_n x^n?


    thanks!
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    im learning about radius of convergence now but i still have yet to grasp its concept.

    let "R hat" = (lim sup la_nl^1/n )^-1
    assume that a_n is not 0 and the seq of positive numbers la_nl/la_n+1l converges to a limit. how do i prove that the lim la_nl/la_n+1l = 'R hat" where R hat is the radius of convergence of the series( sum from n = 1 to infinitiy a_n x^n?


    thanks!

    u have Cauchy - Hadamard Th. saying that :

    let the $\displaystyle R$ be radius of convergence of the power series $\displaystyle \displaystyle \sum _{n=1} ^{\infty} a_n x^n $, then it is worth ::


    $\displaystyle \displaystyle R = \frac {1}{\lim_{n\to \infty} \sqrt[n] {|a_n|}}$


    then if , $\displaystyle \displaystyle \lim_{n\to \infty} \sqrt[n] {|a_n|}=+\infty$ then radius is $\displaystyle R=0$ which means that power series are converging just in point $\displaystyle x=0$ (all power series are converging at least at point x=0)


    or if $\displaystyle \displaystyle \lim_{n\to \infty} \sqrt[n] {|a_n|}=0$ then radius is $\displaystyle R=+\infty$ which means that series converge for all $\displaystyle x\in \mathbb{R}$.


    P.S. you can depending on the situations use another formula for calculating radius of convergence (less work sometimes)


    $\displaystyle \displaystyle R = \lim_{n\to \infty} |\frac {a_n}{a_{n+1}}|$




    now if i understand your question ... you are referring to the Th and need the proof of it which says :

    Th::

    let the power series $\displaystyle \displaystyle \sum _{n=1} ^{\infty} a_n x^n $ is converging for some value $\displaystyle x_0 \neq 0$. Than series are absolutely convergent for all $\displaystyle x$ for which is $\displaystyle |x|< |x_0| $ and that series uniformly convergent for all $\displaystyle x$ for which $\displaystyle |x|\leq r$ where we have that $\displaystyle r<|x_0|$.



    proof ::

    let's assume that for $\displaystyle x_0\neq 0 $ series $\displaystyle \displaystyle \sum _{n=1} ^{\infty} a_n x^n $ converge. That means that common member of the series must strive for zero, and thus he as a convergent sequence is limited. So there is constant $\displaystyle M>0$ such that for all $\displaystyle n\in \mathbb{N}$ this is true $\displaystyle |a_n x_0 ^n | \leq M$

    now let the $\displaystyle |x| < |x_0|$ than is true that

    $\displaystyle \displaystyle |a_n x ^n | = |a_n x_0 ^n \frac {x^n}{x_0 ^n} |= |a_n x_0 ^n | \cdot |\frac {x^n}{x_0 ^n} | \leq M |\frac {x^n}{x_0 ^n} |$


    because $\displaystyle |\frac {x^n}{x_0 ^n} | <1$ than , series $\displaystyle \displaystyle \sum _{n=1} ^{\infty} |\frac {x^n}{x_0 ^n} |^n $ as the sum of the geometrical series is converging, which now means that and series $\displaystyle \displaystyle \sum _{n=1} ^{\infty} |a_n x^n |$ is converging, and that $\displaystyle \displaystyle \sum _{n=1} ^{\infty} a_n x^n $ is absolutely converge.


    now we show second part

    let $\displaystyle |x|\leq r $ where $\displaystyle r< |x_0| $ Based on the proof above we know that series $\displaystyle \displaystyle \sum _{n=1} ^{\infty} a_n r^n $ absolutely converge. As for the selected $\displaystyle x$ is valid ::

    $\displaystyle |a_n x^n | = |a_n||x^n|\leq |a_n| r^n$

    using Weierstrass criteria we have uniform convergence of the series $\displaystyle \displaystyle \sum _{n=1} ^{\infty} a_n x^n $


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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    im learning about radius of convergence now but i still have yet to grasp its concept.

    let "R hat" = (lim sup la_nl^1/n )^-1
    assume that a_n is not 0 and the seq of positive numbers la_nl/la_n+1l converges to a limit. how do i prove that the lim la_nl/la_n+1l = 'R hat" where R hat is the radius of convergence of the series( sum from n = 1 to infinitiy a_n x^n?


    thanks!

    A hint for your question:

    Recall the proof of d'Alebert test of convergence and...

    Or:

    Look at:

    $\displaystyle \frac{a_2}{a_1}\frac{a_3}{a_2}\frac{a_4}{a_3}...\f rac{a_{n+1}}{a_n}$
    Last edited by Also sprach Zarathustra; Oct 24th 2010 at 05:55 AM.
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