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Generating function proof

  1. Apr 17, 2012 #1
    Hey

    I've been trying to show that [itex]\frac{1}{\sqrt{1+u^2 -2xu}}[/itex] is a generating function of the polynomials,

    in other words that
    [itex]\frac{1}{\sqrt{1+u^2 -2xu}}=\sum\limits_{n=0}^{\infty }{{{P}_{n}}(x){{u}^{n}}}[/itex]

    My class was told to do this by first finding the binomial series of
    [itex]\frac{1}{\sqrt{1-s}}[/itex]

    And then insert s = -u2 + 2xu

    Then to expand out sn and group together all the un terms,

    this is what i've been doing for a while and i haven't been able to see the end,

    [itex]\frac{1}{\sqrt{1-x}}=\sum\limits_{n=0}^{\infty }{\frac{\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)....\left( \frac{1}{2}-n \right)}{n!}}{{(-x)}^{n?}}\,\,\,\,\,let\,\,x=2xu-{{u}^{2}}[/itex]

    [itex]
    (u(u-2x))^{n}=u^{n}\left[ {{u}^{n}}-2x\frac{n!}{(n-1)!}{{u}^{n-1}}+4{{x}^{2}}\frac{n!}{(n-2)!}{{u}^{n-2}}+...+\frac{n!}{r!(n-r)!}{{(-1)}^{r}}{{u}^{n-r}}{{(2x)}^{r}}+...+{{(-1)}^{n}}{{(2x)}^{n}} \right]
    [/itex]

    but I'm getting stuck at this point,

    every term here has a factor of un however there are still u's within the bracket so I can't simplify it to Pn(x),

    Have i miss interpreted the instructions or is straight forward way to do this?

    I have found online other methods of this proof but they don't follow the instruction we were given

    any help is verryy appreciated,

    thanks
     
  2. jcsd
  3. Apr 17, 2012 #2
    you are looking at the generating function of the Legendre polynomials
    http://en.wikipedia.org/wiki/Legendre_polynomials
    which does not assume a simply closed form, therefore, you don't expect the very tedious series to simplify. I'm not sure what your prof want to see though, but Pn(x) can only be given recursively or formally as the formal Taylor expansion of the generating function.
     
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