Convergence of expansion of Legendre generating function.by scorpion990 Tags: convergence, expansion, function, generating, legendre 

#1
Jan1909, 08:39 PM

P: 86

The Legendre functions may be defined in terms of a generating function: [tex]g(x,t) = \frac{1}{\sqrt{12xt+t^2}} [/tex]
Of course, [tex]\frac{1}{\sqrt{1+x}} =\sum^{\infty}_{n=0} (\stackrel{.5}{n})x^n [/tex]. However, this series doesn't converge for all x. It only converges if x < 1. In our case, [tex]t^2  2xt[/tex] would have to be less than 1. In the derivation of many recursion formulas, powers of t are set equal to each other. However, this isn't valid for all values of t and x... How come this method of derivation is still valid? Any help/insight would be appreciated. 



#2
Jan2409, 04:45 PM

P: 151

The interval of convergence can depend on the point at which you are taking the
Taylor expansion  and, for that matter, which variable you are doing the expansion in. The one you choose above is just one of many ways to expand it, but not the way the Legendre functions show up. The Legendre functions (http://en.wikipedia.org/wiki/Legendre_polynomials) would come from an expansion in your variable t. The assumption for certain applications would be that x < or =1 , t < 1 , which does not check for your series (eg. x=1, t=.999999999). [ I don't recall offhand how the series extends for t> or + 1 .] As an expansion in t (t<1) , you can quickly verify the region of convergence (going to the complex plane) by using Cauchy's integral formula for upper bounds on the Taylor coefficients followed by a series test. 


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