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The unit-disc estimate for Weierstrass elementary factors

Statement

For every integer $p \geq 0$ and every complex number $w$ with $∣ w ∣ \leq 1$ ,

$∣ 1 - E_{p} (w) ∣ \leq ∣ w ∣^{p + 1} .$

In particular, there is a universal constant $C = 1$ such that

$∣ 1 - E_{p} (w) ∣ \leq C ∣ w ∣^{p + 1} (∣ w ∣ \leq 1) .$

Facts & Assumptions

Given: An integer $p \geq 0$ .

[F1]

The elementary factor is $E_{p} (w) = (1 - w) \exp ⁣ (w + \frac{w^{2}}{2} + \dots + \frac{w^{p}}{p})$ (Weierstrass elementary factors).

[F2]

The complex exponential is entire (The complex exponential is entire and its complex derivative is itself).

[F3]

Complex derivatives satisfy the linearity and product rules (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[F4]

Complex derivatives satisfy the chain rule (The chain rule for complex derivatives).

Proof

technique · direct

1.1

Write $S_{p} (u) = u + u^{2} / 2 + \dots + u^{p} / p$ , with the empty sum $0$ when $p = 0$ . Using [F1], [F2], [F3], and [F4], differentiate $E_{p} (u) = (1 - u) e^{S_{p} (u)}$ to obtain $E_{p}^{'} (u) = - e^{S_{p} (u)} + (1 - u) e^{S_{p} (u)} (1 + u + \dots + u^{p - 1}) = - u^{p} e^{S_{p} (u)} .$

F1F2F3F4givenalgebra

2.1

Fix $w$ with $∣ w ∣ \leq 1$ . By step 1.1, $\frac{d}{d t} E_{p} (t w) = w E_{p}^{'} (t w) = - w^{p + 1} t^{p} e^{S_{p} (t w)} (0 \leq t \leq 1) .$ Integrating from $0$ to $1$ and using $E_{p} (0) = 1$ gives $1 - E_{p} (w) = w^{p + 1} \int_{0}^{1} t^{p} e^{S_{p} (t w)} d t .$

F1step 1.1algebra

3.1

For $0 \leq t \leq 1$ and $1 \leq k \leq p$ , one has $Re ⁣ (\frac{(t w)^{k}}{k}) \leq \frac{∣ t w ∣^{k}}{k} \leq \frac{t^{k}}{k},$ so $∣ e^{S_{p} (t w)} ∣ = e^{Re S_{p} (t w)} \leq e^{S_{p} (t)} .$ Taking absolute values in step 2.1 therefore yields $∣ 1 - E_{p} (w) ∣ \leq ∣ w ∣^{p + 1} \int_{0}^{1} t^{p} e^{S_{p} (t)} d t .$

step 2.1algebra

4.1

For real $t \in [0, 1]$ , step 1.1 gives $\frac{d}{d t} E_{p} (t) = - t^{p} e^{S_{p} (t)} .$ Hence $\int_{0}^{1} t^{p} e^{S_{p} (t)} d t = E_{p} (0) - E_{p} (1) = 1,$ because $E_{p} (0) = 1$ and $E_{p} (1) = 0$ by [F1]. Substituting this into step 3.1 proves $∣ 1 - E_{p} (w) ∣ \leq ∣ w ∣^{p + 1},$ and the displayed bound with $C = 1$ follows.

F1step 1.1step 3.1algebra∎

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Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 Infinite products (standard reference, not scraped)
Matthias Weber, Complex Analysis, Ch. 3 §3.2 (standard reference, not scraped)