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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. $\frac{1}{\sqrt{1+x^2+y^2}}$ ?

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There is another way to generalize the Legendre polynomials. The Legendre polynomials can be got by an expansion of the Newtonian potential in three dimensions. This expansion can be generalized to higher dimensions and the resulting polynomials are the Gegenbauer polynomials. –  user26872 Dec 5 '12 at 22:44

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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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Shouldn't it be $|z|<1$ according to your notation (according to en.wikipedia.org/wiki/Legendre_polynomials) ? Besides I have $x,y \in \mathbb{R}$ and I would expect the solution to work for higher dimensions too... –  user49546 Dec 5 '12 at 18:01
    
Convergence requires $|x| < 1$, not $|z| < 1$, according to the NIST handbook (which may well have been the source of the Wikipedia article). What do you mean by higher dimensions? Higher-dim Legendre polynomials? It's easy to see that your formula implies $$\frac{1}{\sqrt{1 + x^2 + y^2}} = \sum_{n \ge 0} P_{2n}(0) (x^2 + y^2)^n$$ but I don't think that's what you have in mind. –  Hans Engler Dec 5 '12 at 18:35
    
Evaluating at dimension n would be $\frac{1}{\sqrt{1+\sum_{k=0}^n x_k^2}}$ In which case I would have $$ \frac{1}{\sqrt{1+\sum_{k=0}^n x_k^2}} = \sum_j P_{2j}(0)\left(\sum_{k=0}^n x_k^2\right)^j$$ is that correct ? –  user49546 Dec 5 '12 at 20:38
    
Correct. It's just the original statement, written out in polar coordinates, if you will. –  Hans Engler Dec 5 '12 at 21:07

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