Bounds for Weierstrass Elementary Factors
Until this has been finished, please leave {{refactor}} in the code.
New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.
Contents
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:
| \(\displaystyle \left\vert{\log \left({1 - z}\right) + \sum_{k \mathop = 1}^p \frac {z^k} k}\right\vert\) | \(=\) | \(\displaystyle \left\vert{- \sum_{k \mathop = p + 1}^\infty \frac{z^k} k}\right\vert\) | Series Expansion of Complex Logarithm | ||||||||||
| \(\displaystyle \) | \(\le\) | \(\displaystyle \sum_{k \mathop = p + 1}^\infty \frac{\left\lvert{z}\right\rvert^k} k\) | Triangle Inequality for Series | ||||||||||
| \(\displaystyle \) | \(\le\) | \(\displaystyle \sum_{k \mathop = p + 1}^\infty \left\lvert{z}\right\rvert^k\) | because $k \ge 1$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac{\left\lvert{z}\right\rvert^{p + 1} } {1 - \left\lvert{z}\right\rvert}\) | Sum of Geometric Progression | ||||||||||
| \(\displaystyle \) | \(\le\) | \(\displaystyle 2 \left\lvert{z}\right\rvert^{p + 1}\) | because $\left\lvert{z}\right\rvert \le \dfrac 1 2$ |
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
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
When this page/section has been completed, {{ProofWanted}} should be removed from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page (see the proofread template for usage).
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 $5.11$
