x is a positive real such that 1 + x + x2 + ... xn-1 = xn. Show that 2 - 1/2n-1 ≤ x < 2 - 1/2n.
Solution
For n = 1, the unique root is x = 1, which satisfies 1 ≤ x ≤ 3/2. So assume n > 1. By Descartes' rule of signs, the polynomial xn - xn-1 - ... - x - 1 = 0 has one positive root (see below).
(1 - 1/2n)n ≥ (1 - 1/2n)2n-2 = (1 - 1/2n-1 + 1/22n)n-1 > (1 - 1/2n-1)n-1. So iterating, (1 - 1/2n)n > (1 - 1/2)1 = 1/2 (*).
Put f(x) = xn - xn-1 - ... - x - 1. Put g(x) = (x - 1) f(x) = xn+1 - 2xn + 1. We have g(2 - 1/2n) = -(1/2n) (2 - 1/2n)n + 1 = - (1 - 1/2n+1)n + 1 > 0 (since 0 < 1 - 1/2n+1 < 1). Hence f(2 - 1/2n) > 0.
Similarly, g(2 - 1/2n-1) = -(1/2n-1) (2 - 1/2n-1)n + 1 = -2(1 - 1/2n)n + 1 < 0 (by (*) above). Hence f(2 - 1/2n-1) < 0.
Descartes' rule of signs
Suppose anxn + ... + a0 has real coefficients. Ignoring zero coefficients, count the number of sign changes from an to a0. For example: x3 - 5x2 + 5x - 1 has three sign changes; x2 + 1 has none; x2 - 1 has one sign change. The result is that the number of positive roots is either the number of sign changes or an even number less. So x3 - 5x2 + 5x - 1 has 1 or 3 positive roots, x2 + 1 has none, x2 - 1 has 1.
© John Scholes
jscholes@kalva.demon.co.uk
8 Oct 2002