Bounds for Weierstrass Elementary Factors

Theorem

Let $E_p: \C \to \C$ denote the $p$th Weierstrass elementary factor:

$\map {E_p} z = \begin{cases} 1 - z & : p = 0 \\ \paren {1 - z} \map \exp {z + \dfrac {z^2} 2 + \cdots + \dfrac {z^p} p} & : \text{otherwise}\end{cases}$

Let $z \in \C$.

Some bound

Let $\cmod z \le \dfrac 1 2$.

Then:

$\cmod {\map {E_p} z - 1} \le 3 \cmod z^{p + 1}$

Another bound

Let $\cmod z \le 1$.

Then:

$\cmod {\map {E_p} z - 1} \le \cmod z^{p + 1}$

Proof

Proof of some bound

Let $\cmod z \le \dfrac 1 2$.

We may assume $p \ge 1$.

We have:

$\map {E_p} z = \map \exp {\map \log {1 - z} + \displaystyle \sum_{k \mathop = 1}^p \frac {z^k} k}$

Then:

\(\ds \cmod {\map \log {1 - z} + \sum_{k \mathop = 1}^p \frac {z^k} k}\)

\(=\)

\(\ds \cmod {-\sum_{k \mathop = p + 1}^\infty \frac{z^k} k}\)

Series Expansion of Complex Logarithm

\(\ds \)

\(\le\)

\(\ds \sum_{k \mathop = p + 1}^\infty \frac {\cmod z^k} k\)

Triangle Inequality for Series

\(\ds \)

\(\le\)

\(\ds \sum_{k \mathop = p + 1}^\infty \cmod z^k\)

because $k \ge 1$

\(\ds \)

\(=\)

\(\ds \frac {\cmod z^{p + 1} } {1 - \cmod z}\)

Sum of Geometric Sequence

\(\ds \)

\(\le\)

\(\ds 2 \cmod z^{p + 1}\)

because $\cmod z \le \dfrac 1 2$

Because $p \ge 1$:

$2 \cmod z^{p + 1} \le \dfrac 1 2$

By Bounds for Complex Exponential:

$\cmod {\map {E_p} z - 1} \le 3 \cmod z^{p + 1}$

$\blacksquare$

Also see

• Weierstrass Factorization Theorem, what this is made for
• Bounds for Complex Exponential
• Bounds for Complex Logarithm

Sources

• 1973: John B. Conway: Functions of One Complex Variable $VII$: Compact and Convergence in the Space of Analytic Functions: $\S5$: Weierstrass Factorization Theorem: Lemma $5.11$