Past exam of the topics in applied mathematics course of the University of Oxford / 2008 / A / 1 / b / ii / Solution

First establish a uniform upper bound. Direct subtraction givesbecause the quadratic polynomial has positive leading coefficient and discriminant $49-56=-7$. The assumed positivity of $F(N_0)$ therefore places $N_1$ in $(0,3/2)$.

This interval is forward invariant. On $[0,1/2]$ and on $[1,3/2]$, the three factors in $P(N)$ have nonnegative product, hence $F(N)\ge 1/2$. On $[1/2,1]$ the product is nonpositive, butso $F(N)\ge 1/8>0$. Together with the upper bound this proves $F((0,3/2))\subset (0,3/2)$. Induction now givesThe hypothesis on the first image is essential for initial values outside that invariant interval.