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 of 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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0Shouldn't it be $|z|<1$ according to your notation (according to http://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... – 2012-12-05
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0Convergence 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. – 2012-12-05
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0Evaluating 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 ? – 2012-12-05
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0Correct. It's just the original statement, written out in polar coordinates, if you will. – 2012-12-05