where
| |εμ| < | g | Σ ∞n=m+1 | 1 | < | g | . |
| ρμ | 2n | ρμ 2m |
Now draw barriers as before, directed from the origin, joining the singular point of φ(z) to z = ∞, take a finite region excluding all these barriers, let ρ be a quantity less than the radii of convergence of all the power series developments of φ(z) about interior points of this region, so chosen moreover that no circle of radius ρ with centre at an interior point of the region includes any singular point of φ(z), let g be such that |φ(z)| < g for all circles of radius ρ whose centres are interior points of the region, and, x being any interior point of the region, choose the positive integer n so that 1/n |x| < 1⁄3ρ; then take the points a1 = x/n, a2 = 2x/n, a3 = 3x/n, ... an = x; it is supposed that the region is so taken that, whatever x may be, all these are interior points of the region. Then by what has been said, replacing x, z respectively by 0 and x/n, we have
| φ(μ) (a1) = Σ m1λ1=0 | φ(μ + λ1) (0) | ( | x | ) λ1 + αμ |
| λ1! | n |
with
αμ < g/ρμ 2m1,
provided (μ + m1 + 1)μ < (2⁄3)m1 + 1; in fact for μ ⋜ 2n2n−2 it is sufficient to take m1 = n2n; by another application of the same inequality, replacing x, z respectively by a1 and x/n, we have
| φ(μ) (a2) = Σ m2λ2=0 | φ(μ + λ2) (a1) | ( | x | ) λ2 + β′μ , |
| λ2! | n |
where