Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26
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The power series of z0z1 on a bidisc centred away from the origin

Example

Let f(z)=z0z1 on C2, and expand about the centre a=(1,1). Then

f(z)=1+(z0−1)+(z1−1)+(z0−1)(z1−1).

So the multi-indexed power series at a has coefficients c(0,0)=c(1,0)=c(0,1)=c(1,1)=1 and cα=0 for every other α∈N2. Being finite, the series converges absolutely on every bidisc centred at (1,1).

Facts & Assumptions

Given: The function f(z)=z0z1 on C2, the centre a=(1,1), and the bidisc notation of Balls, polydiscs and the distinguished boundary in Cm.

Verification

technique · direct
1.1givenalgebra

Writing zk=1+(zk−1) for k=0,1 and expanding gives z0z1=(1+(z0−1))(1+(z1−1))=1+(z0−1)+(z1−1)+(z0−1)(z1−1).

2.1step 1.1

The right-hand side is a finite multi-indexed power series about (1,1), with exactly the four nonzero coefficients stated above, so it converges absolutely on every bidisc centred at (1,1).

3.1step 2.1algebra∎

The displayed finite series already equals f everywhere, so it is in particular the power-series expansion of f about (1,1). Also ∂(1,0)f(1,1)=1, ∂(0,1)f(1,1)=1, and ∂(1,1)f(1,1)=1 by direct differentiation, while every derivative of order at least 2 in one coordinate is 0; these values match the displayed coefficients.

Depends on

Used by

Dependency tree · two levels

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Sources