Bounds for Weierstrass Elementary Factors

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Theorem

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

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

Let $z \in \C$.


Some bound

Let $\left\lvert{z}\right\rvert \le \dfrac 1 2$.

Then:

$\left\lvert{E_p \left({z}\right) - 1}\right\rvert \le 3 \left\lvert{z}\right\rvert^{p + 1}$


Another bound

Let $\left\lvert{z}\right\rvert \le 1$.

Then:

$\left\lvert{E_p \left({z}\right) - 1}\right\rvert \le \left\lvert{z}\right\rvert^{p + 1}$


Proof

Proof of some bound

Let $\left\lvert{z}\right\rvert \le \dfrac 1 2$.

We may assume $p \ge 1$.

We have:

$E_p \left({z}\right) = \exp \left({\log \left({1 - z}\right) + \displaystyle \sum_{k \mathop = 1}^p \frac{z^k} k}\right)$


Then:

\(\displaystyle \left\vert{\log \left({1 - z}\right) + \sum_{k \mathop = 1}^p \frac {z^k} k}\right\vert\) \(=\) \(\displaystyle \left\vert{- \sum_{k \mathop = p + 1}^\infty \frac{z^k} k}\right\vert\) $\quad$ Series Expansion of Complex Logarithm $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \sum_{k \mathop = p + 1}^\infty \frac{\left\lvert{z}\right\rvert^k} k\) $\quad$ Triangle Inequality for Series $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \sum_{k \mathop = p + 1}^\infty \left\lvert{z}\right\rvert^k\) $\quad$ because $k \ge 1$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac{\left\lvert{z}\right\rvert^{p + 1} } {1 - \left\lvert{z}\right\rvert}\) $\quad$ Sum of Geometric Progression $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle 2 \left\lvert{z}\right\rvert^{p + 1}\) $\quad$ because $\left\lvert{z}\right\rvert \le \dfrac 1 2$ $\quad$

Because $p \ge 1$:

$2 \left\lvert{z}\right\rvert^{p + 1} \le \dfrac 1 2$

By Bounds for Complex Exponential:

$\left\lvert{E_p \left({z}\right) - 1}\right \rvert \le 3 \left\lvert{z}\right\rvert^{p + 1}$

$\blacksquare$


Proof of another bound


Also see


Sources