Bounds for Weierstrass Elementary Factors

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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:

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


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 Infinite 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:

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

$\blacksquare$


Proof of another bound


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Also see


Sources

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