In my studies for an upcoming analysis exam, I came across the following problem:
A sequence of functions is defined as $$f _0(x) = 0 \qquad f_{n+1}(x) = f_n(x) + \frac{x^2 - f_n(x)^2}{2} $$
Use the above recurrence relation to show that $$|x| - f_{n+1}(x) = (|x| - f_n(x)) \bigg(1 - \frac{ |x| + f_n(x) }{2} \bigg) $$ Hence, deduce that $0 \leq f_n(x) \leq f_{n+1}(x) \leq |x|$ for $|x| \leq 1$.
The first part is easy enough, just by performing a bit of algebra: \begin{align*} |x| - f_{n+1}(x) &= |x| - f_n(x) - \frac{x^2 + f_n(x)^2}{2} \\ &= |x| - f_n(x) - \frac{(|x| - f_n(x))(|x| + f_n(x))}{2} \\ &= (|x| - f_n(x)) \bigg(1 - \frac{ |x| + f_n(x) }{2} \bigg) \\ \end{align*}
But I'm struggling with the 'hence' portion. The wording seems to suggest that this new result makes it easier to prove this, but I just don't see it. Is there something painfully obvious that I'm missing? Or do we need to work with the original relationship a bit to get a jumping off point?
hint
Let us prove by induction that if $|x|\le1$ then $$0\le f_n (x)\le |x|.$$
it is true for $n=0$. assume it is true for some $n\ge 0$. then $$0\le\frac{|x|+f_n (x)}{2}\le |x|\le 1$$
thus $$0\le 1-\frac {|x|+f_n (x)}{2}\le 1$$ and $$0\le |x|-f_{n+1}(x)\le |x|-f_n (x) $$
or $$0\le f_n(x)\le f_{n+1}(x) \le |x|.$$