What is the generating function proof for Legendre polynomials?

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SUMMARY

The generating function for Legendre polynomials is expressed as \(\frac{1}{\sqrt{1+u^2 -2xu}} = \sum\limits_{n=0}^{\infty }{{{P}_{n}}(x){{u}^{n}}}\). To derive this, one must start with the binomial series of \(\frac{1}{\sqrt{1-s}}\) and substitute \(s = -u^2 + 2xu\). The challenge lies in expanding the series and grouping terms correctly to isolate \(P_n(x)\), which can be complex due to the presence of multiple \(u\) factors. The discussion highlights that while alternative methods exist, they may not align with the specific instructional approach given in class.

PREREQUISITES
  • Understanding of generating functions in combinatorics
  • Familiarity with Legendre polynomials and their properties
  • Knowledge of binomial series and Taylor expansions
  • Basic proficiency in algebraic manipulation of series
NEXT STEPS
  • Study the derivation of the binomial series for \(\frac{1}{\sqrt{1-s}}\)
  • Explore the recursive relationships of Legendre polynomials
  • Investigate alternative proofs for generating functions of orthogonal polynomials
  • Learn about formal power series and their applications in combinatorial proofs
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Mathematicians, students studying advanced calculus or combinatorial theory, and anyone interested in the properties and applications of Legendre polynomials.

physicsjock
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Hey

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

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

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

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,

\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}}

<br /> (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]<br />

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
 
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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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