Knowlpedia

Theorem

Smooth nonanalytic endpoint of a Gamma average

All one-sided derivatives of the Gamma average exist at zero, although its Taylor series has zero radius of convergence.

Statement

For a,b>0a,b>0, the F=Fa,bF=F_{a,b} is smooth up to z=0z=0 from the right, with

F(m)(0)=(−1)m(a)m(b)m.F^{(m)}(0)=(-1)^m(a)_m(b)_m.

Here (a)m(a)_m is the rising factorial. Its Taylor series at zero has radius of convergence zero, so this endpoint smoothness is not real analyticity.

Differentiation and finite remainders

Differentiation under the integral gives

F(m)(z)=(−1)m(b)mΓ(a)∫0∞e−vva+m−1(1+zv)−b−m dv.F^{(m)}(z)=\frac{(-1)^m(b)_m}{\Gamma(a)} \int_0^\infty e^{-v}v^{a+m-1}(1+zv)^{-b-m}\,dv.

Deleting the last factor gives an integrable bound valid for z≥0z\ge0. Thus ∣F(m)(z)∣≤(a)m(b)m|F^{(m)}(z)|\le(a)_m(b)_m, and finite Taylor expansions have the usual remainder bounds. For example, ∣F(z)−1+abz∣≤(a)2(b)2z2/2|F(z)-1+abz|\le(a)_2(b)_2z^2/2.

The absolute Taylor coefficients are (a)m(b)m/m!(a)_m(b)_m/m!; the ratio of consecutive coefficients is (a+m)(b+m)/(m+1)(a+m)(b+m)/(m+1), which tends to infinity. This proves the zero convergence radius. Fixed-order differentiated remainder estimates remain valid and do not require summing that divergent series.