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The power series of $\exp (z_{0} + z_{1})$ on every bidisc

Example

For $f (z) = \exp (z_{0} + z_{1})$ on $C^{2}$ ,

$\exp (z_{0} + z_{1}) = \sum_{α \in N^{2}} \frac{z^{α}}{α!} = \sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} \frac{z_{0}^{m} z_{1}^{n}}{m! n!} .$

The series converges absolutely on every bidisc, and its coefficient at $z^{α}$ is exactly $1 / α!$ .

Facts & Assumptions

Given: The complex exponential and the multi-index notation on $C^{2}$ .

[L1]

$\exp (z_{0} + z_{1}) = \exp (z_{0}) \exp (z_{1})$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[L2]

The exponential series $\sum_{n \geq 0} z^{n} / n!$ converges for every complex $z$ and equals $\exp (z)$ (The complex exponential series converges absolutely for every complex argument, The complex exponential by its power series).

[L3]

The Cauchy product of two absolutely convergent complex series converges absolutely and has the product sum (The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums).

[L4]

An absolutely summable complex double array has the same sum along a bijective enumeration and in either iterated order: real double-series Fubini applies separately to its real and imaginary parts, whose absolute values are bounded by the complex moduli (Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value, Real and imaginary parts, complex conjugation, and modulus).

Verification

technique · direct

1.1

By [L1] and [L2], $\exp (z_{0} + z_{1}) = (\sum_{m \geq 0} z_{0}^{m} / m!) (\sum_{n \geq 0} z_{1}^{n} / n!)$ .

L1L2

2.1

Each one-variable exponential series is absolutely convergent. Moreover, for the double array $a_{m n} = z_{0}^{m} z_{1}^{n} / (m! n!)$ , [L2] gives $\sum_{n} ∣ a_{m n} ∣ = (∣ z_{0} ∣^{m} / m!) \exp (∣ z_{1} ∣)$ and then $\sum_{m} \sum_{n} ∣ a_{m n} ∣ = \exp (∣ z_{0} ∣) \exp (∣ z_{1} ∣) < \infty$ . Hence [L3] identifies the product in step 1.1 with the absolutely convergent diagonal enumeration $\sum_{N \geq 0} \sum_{m = 0}^{N} z_{0}^{m} z_{1}^{N - m} / (m! (N - m)!)$ . The substitution $(m, n) = (m, N - m)$ enumerates every pair in $N^{2}$ exactly once, and [L4] now permits regrouping the array in iterated order. Therefore $\exp (z_{0} + z_{1}) = \sum_{m \geq 0} \sum_{n \geq 0} \frac{z_{0}^{m} z_{1}^{n}}{m! n!} = \sum_{α \in N^{2}} \frac{z^{α}}{α!} .$

step 1.1L2L3L4

3.1

On every bidisc the series is absolutely convergent because it is the product of two absolutely convergent one-variable exponential series there, and step 2.1 already identifies the coefficient of $z^{α}$ as exactly $1 / α!$ .

step 2.1∎

Depends on

Used by

Cauchy estimates on a bidisc, computed and compared with the exact derivatives Example

Dependency tree · two levels

78 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

J. Lebl, Tasty Bits of Several Complex Variables, v4.4, §1.2 (standard reference, not scraped)

The power series of exp⁡(z0+z1)exp(z0​+z1​) on every bidisc