= Solution
For the unnormalized <Zernike polynomial> with $n=4,m=0$, the three radial terms are
$$
\frac{4!}{2!2!}\rho^4-\frac{3!}{1!1!1!}\rho^2+\frac{2!}{2!0!0!}=6\rho^4-6\rho^2+1.
$$
Since $\cos0\phi=1$,
$$
\boxed{Z_4^0(\rho,\phi)=6\rho^4-6\rho^2+1.}
$$
Along a central chord with signed coordinate $s\in[-1,1]$, replace $\rho$ by $|s|$. The profile is even, equals one at the center and both edges, and has minima $-1/2$ at $s=\pm1/\sqrt2$. This <Zernike spherical mode> corresponds to \b[primary <spherical aberration>], balanced by defocus and piston terms rather than simply the raw quartic Seidel term.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-338-zernike-spherical-mode.png]
{title=The unnormalized Zernike spherical mode on a central chord}
{description=The even quartic has central and edge values one and two minima of minus one half. Its lower-order terms remove its projections onto piston and defocus.}
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