Maths Olympiad Prep

Library / /16 of 100

Algebra Difficulty 4.0 AMC 10/12 Find the answer China

Suppose that f(x)=x1+x2f(x) = \frac{x}{\sqrt{1+x^2}} and f(n)(x)=f[f[f…f(x)]]f^{(n)}(x) = f[f[f\dots f(x)]]. Then f(99)(1)=‾f^{(99)}(1) = \underline{\hspace{2cm}}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

We have
f(1)(x)=f(x)=x1+x2,f(2)(x)=f[f(x)]=x1+2x2,⋮f(99)(x)=x1+99x2. \begin{aligned} f^{(1)}(x) &= f(x) = \frac{x}{\sqrt{1+x^2}}, \\ f^{(2)}(x) &= f[f(x)] = \frac{x}{\sqrt{1+2x^2}}, \\ &\vdots \\ f^{(99)}(x) &= \frac{x}{\sqrt{1+99x^2}}. \end{aligned}
Therefore, f(99)(1)=110f^{(99)}(1) = \frac{1}{10}.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.