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China Mathematical Competition

Algebra Difficulty 4.0 AMC 10/12 Find the answer China

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

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.