Infinite Products and the Weierstrass Factorisation Theorem — Examples

On a compact disc $\mathrm{\mid }z\mathrm{\mid }\le R$ one has ${\sum }_{n\ge 1}\sup \mathrm{\mid }{f}_n(z)-1\mathrm{\mid }={\sum }_{n\ge 1}R\mathrm{/}{2}^n<\mathrm{\infty }$, so the product is normally convergent.

Therefore [F1] makes the product entire, and on each compact disc only finitely many factors can vanish because $-2^n$ eventually lies outside the disc.

For each $N\ge 1$, the partial product $P_N(z):={\prod }_{n=1}^N(1-z^2\mathrm{/}{n}^2)$ has quadratic expansion $P_N(z)=1-\left({\sum }_{n=1}^{N}1\mathrm{/}{n}^2\right)z^2+z^4{R}_N(z)$, because every term beyond the linear choice from a single factor contains at least two copies of $z^2$.

The series $\sum 1\mathrm{/}{n}^2$ converges, so on a fixed neighbourhood of $0$ the functions $R_N$ stay bounded and the quadratic coefficients converge. Passing to the locally uniform limit supplied by [F1] gives ${\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)=1-\left({\sum }_{n\ge 1}1\mathrm{/}{n}^2\right)z^2+O(z^4)$.

Comparing the quadratic terms in step 2.1 with the expansion from [F2] yields ${\sum }_{n\ge 1}1\mathrm{/}{n}^2={\pi }^2\mathrm{/}6$.

The function satisfies $f(0)=1\mathrm{\ne }0$, and its unique zero is $a_1=1$, which lies in $\mathrm{\mid }z\mathrm{\mid }<R$ because $R>1$.

Applying [F1] with this single zero gives $0=\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }1-R{e}^{it}\mathrm{\mid }dt-\log R$, equivalently $\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }1-R{e}^{it}\mathrm{\mid }dt=\log R$.

Here ${\sum }_{n\ge 1}\mathrm{\mid }{a}_n{\mathrm{\mid }}^{-1}={\sum }_{n\ge 1}1\mathrm{/}{n}^2$ converges, and $E_0(w)=1-w$, so [F1] applies with $p=0$.

Therefore ${\prod }_{n\ge 1}(1-z\mathrm{/}{n}^2)$ converges normally on $\mathbb{C}$ and has zeros exactly at the points $n^2$.

The paired partial sums satisfy $a_{2m-1}+a_{2m}=1\mathrm{/}{m}^2$, so the series $\sum a_n$ converges because ${\sum }_{m\ge 1}1\mathrm{/}{m}^2$ converges; it is not absolutely convergent because $\sum a_{2m-1}=\sum 1\mathrm{/}\sqrt{m}$ diverges.

The paired product is $(1+a_{2m-1})(1+a_{2m})=(1+\frac{1}{\sqrt{m}})(1-\frac{1}{\sqrt{m}}+\frac{1}{m^2})=1-\frac{1}{m}+\frac{1}{m^2}+\frac{1}{m^{5\mathrm{/}2}}$. For all large $m$ this is at most $1-\frac{1}{2m}$.

Since ${\sum }_{m\ge 1}1\mathrm{/}(2m)$ diverges, [F1] makes ${\prod }_{m\ge M}(1-\frac{1}{2m})$ tend to $0$ for every large $M$; step 2.1 therefore forces the product of the paired factors, and hence $\prod (1+a_n)$ itself, to tend to $0$ rather than to a nonzero limit. This refutes the statement.

The function $1$ has no zeros, so [F1] allows the trivial product part and gives the factorization $1=e^0\cdot 1$.

But also $1=e^{2\pi i}\cdot 1$, and the exponential factors $e^0$ and $e^{2\pi i}$ come from different entire logarithms. Therefore the factorization is not unique.

The sequence $a_n=n$ has no finite accumulation point, so it is a legitimate zero sequence.

For this sequence the genus-zero canonical product is ${\prod }_{n\ge 1}{E}_0(z\mathrm{/}n)={\prod }_{n\ge 1}(1-\frac{z}{n})$ by [F1]. Evaluating at $z=-1$ gives the positive-factor product ${\prod }_{n\ge 1}(1+\frac{1}{n})\mathrm{.}$

The series ${\sum }_{n\ge 1}1\mathrm{/}n$ diverges by [F2], so [F3] implies that the product in step 2.1 does not converge. Therefore the genus-zero canonical product for $a_n=n$ fails even pointwise at $z=-1$, and the statement is false.

The function $e^z$ has no zeros, so [F1] makes its canonical-product part genus $0$.

On the positive real axis one has $M_{e^z}(r)\ge e^r$, while everywhere on $\mathrm{\mid }z\mathrm{\mid }=r$ one has $\mathrm{\mid }{e}^z\mathrm{\mid }=e^{\mathrm{Re}z}\le e^r$; therefore $M_{e^z}(r)=e^r$, and [F2] gives $\rho (e^z)={\mathrm{lim\hspace{0.17em}sup}}_{r\to \mathrm{\infty }}\log r\mathrm{/}\log r=1$. Hence the order is $1$ while the canonical genus is $0$.