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Raabe's test says Legendre functions always converge?

  1. May 13, 2015 #1
    The Legendre functions are the solutions to the Legendre differential equation. They are given as a power series by the recursive formula (link [1] given below):

    ##\begin{align}y(x)=\sum_{n=0}^\infty a_n x^n\end{align}##

    ##\begin{align}a_{n+2}=-\frac{(l+n+1)(l-n)}{(n+1)(n+2)}a_n\end{align}##

    If ##l## is a non-negative integer, The Legendre functions reduce to the Legendre polynomials. Otherwise, they are divergent (for ##x\in[-1,1]##).

    However, I did Raabe's test and found that they are convergent for all values of ##l##, a wrong result.

    My workings:

    ##\begin{align}y(1)&=\sum_{n=0}^\infty a_n\,1^n\\
    &=\sum_{n=0,2,4...}^\infty a_n+\sum_{n=1,3,5...}^\infty a_n\end{align}##

    Let ##\,y_1=\sum_{n=0,2,4...}^\infty a_n\,\,\,\,\,\,\,\,## and ##\,\,\,\,\,\,\,\,y_2=\sum_{n=1,3,5...}^\infty a_n##

    Applying Raabe's test to ##\,y_1## (note that the ##\,a_{n+2}## term in (2) becomes the ##\,a_{n+1}## term in ##y_1##),

    ##\begin{align}R&=\displaystyle\lim_{n\rightarrow +\infty}n\big(\big|{\frac{a_n}{a_{n+1}}}\big|-1\big)\\
    &=\displaystyle\lim_{n\rightarrow +\infty}n\big[\frac{(n+1)(n+2)}{(l+n+1)(n-l)}-1\big]\\
    &=\frac{2n^2+\big(l(l+1)+2\big)n}{n^2+n-l(l+1)}\\
    &=2\end{align}##

    ##R>1##. Thus, ##\,y_1## is absolutely convergent.

    Since both ##\,y_1## and ##\,y_2## follows the same recursive formula (2), ##\,y_2## is also absolutely convergent. So ##\,y(1)## is absolutely convergent. Since ##\,y(1)## is absolutely convergent, ##\,y(-1)## is convergent.

    But ##\,y(1)## or ##\,y(-1)## must diverge. Where's my mistake?

    Background information:

    http://mathworld.wolfram.com/LegendreDifferentialEquation.html

    http://en.wikipedia.org/wiki/Ratio_test
     
  2. jcsd
  3. May 18, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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