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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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 $5.11$