= Solution
{c}
The <hyperbolic cosine> is entire and $\cosh0=1$. Continuity makes it nonzero in a neighborhood of zero, so its reciprocal is holomorphic there and has a convergent <Taylor series>. Since $\cosh$ and its reciprocal are even, all odd coefficients vanish.
Multiply the proposed expansion by $\cosh z$:
$$
(1+c_2z^2+c_4z^4+\cdots)(1+\tfrac12z^2+\tfrac1{24}z^4+\cdots)=1.
$$
Coefficient comparison gives $c_2+1/2=0$ and $c_4+c_2/2+1/24=0$, hence
$$
\boxed{c_2=-\tfrac12,\qquad c_4=\tfrac5{24}.}
$$
The zeros of $\cosh z$ solve $e^{2z}=-1$, so they are $z=i(\pi/2+k\pi)$, $k\in\mathbb Z$. The nearest ones are $\pm i\pi/2$, and they are simple because $\sinh$ is nonzero there. The <hyperbolic secant> has genuine poles at these points and is holomorphic in the disk of radius $\pi/2$. The <radius of convergence> of its Taylor series is therefore
$$
\boxed{R=\pi/2.}
$$
Back to article page
