# Bounds for Weierstrass Elementary Factors

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## Theorem

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

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

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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Let $\cmod z \le 1$.

We may assume $p \ge 1$.

For a fixed $p$, let:

- $\map {E_p} z = \paren {1 - z} \map \exp {z + \dfrac {z^2} 2 + \cdots + \dfrac {z^p} p} = 1 + \ds \sum_{k \mathop = 1}^\infty a_k z^k$

be its power series expansion about $z = 0$.

By Derivative of Complex Power Series we obtain:

- $\map {E_p'} z = \ds \sum_{k \mathop = 1}^\infty k a_k z^{k - 1} = -z^p \map \exp {z + \cdots + \dfrac {z^p} p}$

Comparing the two expressions gives two pieces of information about the coefficients $a_k$.

First:

- $a_1 = a_2 = \cdots = a_p = 0$

Second, since the coefficients of the expansion of $\map \exp {z + \cdots + \dfrac {z^p} p}$ are all positive:

- $a_k \le 0$ for $k \ge p + 1$

Thus:

- $\cmod {a_k} = -a_k$ for $k \ge p + 1$

This gives:

- $0 = \map {E_p} 1 = 1 + \ds \sum_{k\mathop = p + 1}^\infty a_k$

or:

- $\ds \sum_{k \mathop = p + 1}^\infty \cmod {a_k} = -\sum_{k \mathop = p + 1}^\infty a_k = 1$

Hence:

\(\ds \cmod {\map{E_p} z - 1}\) | \(=\) | \(\ds \cmod {\sum_{k \mathop = p + 1}^\infty a_k z^k}\) | ||||||||||||

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

\(\ds \) | \(\le\) | \(\ds \cmod z^{p + 1} \sum_{k \mathop = p + 1}^\infty \cmod {a_k}\) | Triangle Inequality for Series and $\cmod z\le1$ | |||||||||||

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

$\blacksquare$

## 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*$\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: Lemma $5.11$