Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
Why do you want these in terms of Legendres? The radial polynomials are defined over $[0,1]$ while the Legendres are defined over $[-1,1]$. –  Ron Gordon Apr 10 '13 at 14:18
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
Strictly speaking, Zernikes are special cases of Jacobi polynomials and the radial functions are defined over $[0,1]$. (Zernike polynomials are used to represent wavefront aberrations over a circular exit pupil.) So while I applaud your effort (+1), keep in mind that I think the OP may be a bit misguided here. –  Ron Gordon Apr 10 '13 at 14:16
Ron, I'm well aware of the optical applications and the different domains of interest, but it was a nice exercise in basis changes... :) –  J. M. Apr 10 '13 at 14:18
Nice to know someone else knows this stuff! –  Ron Gordon Apr 10 '13 at 14:18
Also, I got lucky here; the general formula for expressing $P_n^{(\alpha,\beta)}(x)$ in terms of $P_n^{(\mu,\nu)}(x)$ (i.e., express one Jacobi polynomial in terms of another) is something I won't wish to write out in full. –  J. M. Apr 10 '13 at 14:20
That's why I applaud your effort: I played around with this stuff many moons ago and did not wish to do so again. –  Ron Gordon Apr 10 '13 at 14:23
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.