False statementConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-29

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FALSE: the order of an entire function always equals its canonical genus

Statement

The order of an entire function always equals its canonical genus.

Facts & Assumptions

Given: The zero-free entire function $f (z) = e^{z}$ .

[F1]

Weierstrass factorization represents a zero-free entire function as a pure exponential factor, so its canonical product part has genus $0$ (Weierstrass factorization for entire functions).

[F2]

The order of an entire function is defined from the growth of $M_{f} (r)$ (The order of an entire function).

Refutation

1.1

The function $e^{z}$ has no zeros, so [F1] makes its canonical-product part genus $0$ .

F1given

2.1

On the positive real axis one has $M_{e^{z}} (r) \geq e^{r}$ , while everywhere on $∣ z ∣ = r$ one has $∣ e^{z} ∣ = e^{Re z} \leq e^{r}$ ; therefore $M_{e^{z}} (r) = e^{r}$ , and [F2] gives $ρ (e^{z}) = {lim sup}_{r \to \infty} \log r / \log r = 1$ . Hence the order is $1$ while the canonical genus is $0$ .

F2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 Functions of finite order (standard reference, not scraped)