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The value phi_1(0) is obtained directly from its series

Statement

For the zero $1 \times 1$ matrix, verify directly that $φ_{1} (0) = 1$ and that the augmented exponential returns this value without evaluating $(e^{z} - 1) / z$ .

Facts & Assumptions

Given: $X = [0] \in M_{1} (C)$ , $w = [1] \in C$ , and $k = 1$ .

[F1]

The phi-function family by everywhere-convergent series: $φ_{1} (X) = \sum_{j \geq 0} X^{j} / (j + 1)!$ , with $X^{0} = I_{n}$ .

[F2]

Phi recurrence and augmented-matrix action identities: the first coordinate of $e^{B_{1} (X, w)} (0, 1)^{T}$ is $φ_{1} (X) w$ , with no division by $X$ or a scalar.

Proof

1.1givenF1algebra

Since $X^{0} = [1]$ and $X^{j} = [0]$ for every $j \geq 1$ , the series in [F1] has only the term $1 / 1! = 1$ nonzero. Hence $φ_{1} (X) = [1]$ .

2.1givenstep 1.1F2algebra∎

Here $J_{1} = [0]$ and $B_{1} (X, w) = [\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}] = : N$ , so $N^{2} = 0$ and $e^{N} = I + N$ . Therefore $e^{N} (0, 1)^{T} = (1, 1)^{T}$ ; [F2] identifies its first coordinate with $φ_{1} (X) w = 1$ . The quotient $(e^{z} - 1) / z$ is undefined at $z = 0$ , while the series and augmented action above are defined there.

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Sources

Awad H. Al-Mohy and Nicholas J. Higham, Computing the action of the matrix exponential, with an application to exponential integrators (standard reference, not scraped)
Jitse Niesen and Will M. Wright, A Krylov subspace algorithm for evaluating the phi-functions appearing in exponential integrators (standard reference, not scraped)