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



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


Sources