We have $\frac{1}{\sqrt{1+x^2}} = \sum^{\infty}_{n=0} P_n(0)x^n$ where $P_n(x)$ is a Legendre polynomial of degree $n$. Is there something similar for two dimensions i.e. $\frac{1}{\sqrt{1+x^2+y^2}}$ ?
Bivariate Legendre polynomials?
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ordinary-differential-equations
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0There is another way to generalize the [Legendre polynomials](http://en.wikipedia.org/wiki/Legendre_polynomials). The Legendre polynomials can be got by an expansion o$f$ the Newtonian potential in three dimensions. This expansion can be generalized to higher dimensions and the resulting polynomials are the [Gegenbauer polynomials](http://en.wikipedia.org/wiki/Gegenbauer_polynomials). – 2012-12-05
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The formula given in your question generalizes to $ \frac{1}{\sqrt{1-2xz+ x^2}} = \sum^{\infty}_{n=0} P_n(z)x^n $ for $|x| < 1$. Hence by setting $y^2 = -2xz$ $ \frac{1}{\sqrt{1+ x^2 + y^2}} = \sum^{\infty}_{n=0} P_n\left(\frac{-y^2}{2x} \right)x^n \, . $
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0Correct. It's just the original statement, written out in polar coordinates, if you will. – 2012-12-05