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The product $\prod_{n \geq 0} (1 + z / 2^{n})$ defines an entire zero-free tail limit after the first factor

Example

The product

$\prod_{n \geq 1} (1 + \frac{z}{2^{n}})$

converges normally on $C$ , hence defines an entire function. On every compact set, all but finitely many factors are zero-free, so the tail limit is zero-free there.

Facts & Assumptions

Given: The factors $f_{n} (z) = 1 + z / 2^{n}$ for $n \geq 1$ .

[F1]

A normally convergent holomorphic product defines a holomorphic function, and after finitely many factors the tail contributes no zeros on a fixed compact set (Normally convergent products define holomorphic functions with the expected zeros).

Verification

1.1

On a compact disc $∣ z ∣ \leq R$ one has $\sum_{n \geq 1} \sup ∣ f_{n} (z) - 1 ∣ = \sum_{n \geq 1} R / 2^{n} < \infty$ , so the product is normally convergent.

F1givenalgebra

2.1

Therefore [F1] makes the product entire, and on each compact disc only finitely many factors can vanish because $- 2^{n}$ eventually lies outside the disc.

F1step 1.1algebra∎

Depends on

Normally convergent products define holomorphic functions with the expected zeros

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Matthias Weber, Complex Analysis, Ch. 3 §3.2 (standard reference, not scraped)

The product ∏n≥0(1+z/2n)∏n≥0​(1+z/2n) defines an entire zero-free tail limit after the first factor