Let R be the reals and R* the non-negative reals. f: R* → R satisfies the following conditions: (1) it is differentiable and f '(x) = - 3 f(x) + 6 f(2x) for x > 0; (2) |f(x)| ≤ e-√x for x ≥ 0. Define un = ∫0∞ xnf(x) dx for n ≥ 0. Express un in terms of u0, prove that the sequence un3n/n! converges, and show that the limit is 0 iff u0 = 0.
Solution
Integrating by parts and using the relation given, we get un = 3/(n+1) un+1 - 3/(n+1) (1/2)n+1 un+1. So un = u0 n!/3n ∏1n 1/(1 - 1/2r).
The product ∏ (1 - 1/2r) converges to a positive value since ∑ 1/2r converges. Hence 3n/n! un converges to a positive multiple of u0. Hence the limit is zero iff u0 is zero.
© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2001