- Find the Hadamard products for:
\(e ^ { z } - 1\)
cos πz.
[Hint: The answers are \(\scriptstyle e ^ { z / 2 } z \prod _ { n = 1 } ^ { \infty } ( 1 + z ^ { 2 } / 4 n ^ { 2 } \pi ^ { 2 } )\) and \(\scriptstyle \prod _ { n = 0 } ^ { \infty } ( 1 - 4 z ^ { 2 } / ( 2 n + 1 ) ^ { 2 } )\) , respectively.]
Solution
We use the canonical product \[ \frac{\sin \pi w}{\pi w} =\prod_{n=1}^{\infty}\left(1-\frac{w^2}{n^2}\right). \tag{1} \]
For (a), put \(w=z/(2\pi i)\) in (1). Since \[ \sin\!\left(\frac z{2i}\right)=-i\sinh\frac z2, \] we obtain \[ \frac{\sinh(z/2)}{z/2} =\prod_{n=1}^{\infty} \left(1+\frac{z^2}{4\pi^2n^2}\right). \] Using \[ e^z-1=2e^{z/2}\sinh(z/2), \] we get \[ \boxed{ e^z-1 =e^{z/2}z\prod_{n=1}^{\infty} \left(1+\frac{z^2}{4\pi^2n^2}\right)}. \]
For (b), use \[ \cos\pi z=\frac{\sin 2\pi z}{2\sin\pi z}. \] Applying (1) to numerator and denominator gives \[ \cos\pi z =\frac{2\pi z\prod_{n\ge1}(1-4z^2/n^2)} {2\pi z\prod_{n\ge1}(1-z^2/n^2)}. \] The factors indexed by even \(n\) in the numerator cancel the denominator factors, leaving exactly the odd indices: \[ \boxed{ \cos\pi z =\prod_{n=0}^{\infty} \left(1-\frac{4z^2}{(2n+1)^2}\right)}. \] These are the required Hadamard products.