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 u_{n} = ∫_{0}^{∞} x^{n}f(x) dx for n ≥ 0. Express u_{n} in terms of u_{0}, prove that the sequence u_{n}3^{n}/n! converges, and show that the limit is 0 iff u_{0} = 0.

**Solution**

Integrating by parts and using the relation given, we get u_{n} = 3/(n+1) u_{n+1} - 3/(n+1) (1/2)^{n+1} u_{n+1}. So u_{n} = u_{0} n!/3^{n} ∏_{1}^{n} 1/(1 - 1/2^{r}).

The product ∏ (1 - 1/2^{r}) converges to a positive value since ∑ 1/2^{r} converges. Hence 3^{n}/n! u_{n} converges to a positive multiple of u_{0}. Hence the limit is zero iff u_{0} is zero.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001