Give an example of an entire function of order \(1/3\).
Solution
An explicit example is \[ \boxed{ f(z)=\prod_{n=1}^{\infty} \left(1-\frac{z}{n^3}\right).} \]
1. The displayed infinite product defines an entire function.
The proposed zero sequence is \[ a_n=n^3. \] Since \[ \sum_{n=1}^{\infty}\frac1{|a_n|} =\sum_{n=1}^{\infty}\frac1{n^3}<\infty, \] the genus-zero canonical product \[ \prod_{n=1}^{\infty}\left(1-\frac z{a_n}\right) \] converges uniformly on compact subsets of \(\mathbb C\). Hence it defines an entire function whose zeros are exactly the points \(n^3\), all simple.
2. The critical exponent of the zero sequence \(a_n=n^3\) is \(1/3\).
For \(s>0\), \[ \sum_{n=1}^{\infty}\frac1{|a_n|^s} =\sum_{n=1}^{\infty}\frac1{n^{3s}}. \] This series converges exactly when \(3s>1\), i.e. when \(s>1/3\), and diverges when \(s\le1/3\). Therefore \[ \alpha(a_n)=\frac13. \]
3. The entire function \(f\) has order exactly \(1/3\).
By Borel’s theorem, the canonical product associated to a zero sequence has order equal to the critical exponent of that sequence. Applying it to \(a_n=n^3\) and using 2 gives \[ \rho(f)=\alpha(a_n)=\frac13. \] Thus the displayed product is an entire function of the required order.