The even and odd Zernike polynomials are defined as follows: $$Z^{m}_n(\rho,\varphi) = R^m_n(\rho)\,\cos(m\,\varphi) \!$$ and: $$Z^{-m}_n(\rho,\varphi) = R^m_n(\rho)\,\sin(m\,\varphi), \!$$ with: $$R^m_n(\rho) = \! \sum_{k=0}^{(n-m)/2} \!\!\! \frac{(-1)^k\,(n-k)!}{k!\,((n+m)/2-k)!\,((n-m)/2-k)!} \;\rho^{n-2\,k}$$ My question: is there a way to express the Zernike polynomials in terms of Legendre polynomials? Thanks in advance
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First we rewrite your definition for the radial Zernike polynomial in a more convenient form: $$\mathcal R_n^m(\rho)=\sum_{k=0}^{(n-m)/2}(-1)^k \binom{n-k}{k} \binom{n-2k}{(n-m)/2-k} \rho^{n-2k}$$ Now, you are asking about how to expand the radial Zernike polynomial as a Legendre series. We first note that $\mathcal R_n^m(\rho)$ can be expressed solely in terms of odd-order Legendre polynomials for odd $n,m$, and in terms of even-order Legendre polynomials for even $n,m$. (Recall also that the radial Zernike polynomials are identically zero if $n,m$ are not of the same parity.) Now, for the Legendre expansion $$\mathcal R_n^m(\rho)=\sum_{k=0}^n c_k P_k(\rho)$$ where the coefficients are given by $$c_k=\left(k+\frac12\right)\int_{-1}^1 \mathcal R_n^m(t)P_k(t)\,\mathrm dt$$ we can derive an expression for $c_k$ by inserting the series definition of $\mathcal R_n^m(\rho)$ into the integral expression of $c_k$, and then using the identity $$\int_{-1}^1 u^{n-2j}P_k(u)\,\mathrm du=\frac2{k+1}\frac{\tbinom{(n-2j-1)/2}{(k-1)/2}}{\tbinom{(n-2j+k+1)/2}{(k+1)/2}}$$ to yield the expression $$c_k=\frac{2k+1}{k+1}\sum_{j=0}^{(n-m)/2}(-1)^j \frac{\tbinom{n-j}{j}\tbinom{n-2j}{(n-m)/2-j}\tbinom{(n-2j-1)/2}{(k-1)/2}}{\tbinom{(n-2j+k+1)/2}{(k+1)/2}}$$ $c_k$ can be expressed in terms of a ${}_4 F_3$ hypergeometric function, but I'll omit that expression for now. |
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