# Problem 3.

In-1 There is a function

$$
f(x)=\frac{1}{\sqrt[5]{1-x^{5}}}
$$

Calculate

$$
f(f(f(f(f(\ldots f(2022)))) \ldots))
$$

where the function $f$ is applied 1303 times.
Answer: $1 / \sqrt[5]{1-2022^{5}}$

Solution. For odd $n$

$$
\begin{gathered}
f(x)=\frac{1}{\sqrt[n]{1-x^{n}}} \\
f(f(x))=\frac{1}{\sqrt[n]{1-\frac{1}{1-x^{n}}}}=\sqrt[n]{1-\frac{1}{x^{n}}} \\
f(f(f(x)))=\sqrt[n]{1-\left(1-x^{n}\right)}=x \\
f(f(f(f(x))))=f(x)
\end{gathered}
$$

We see periodicity, period $=3$. The remainder of 1303 divided by 3 is 1, therefore

$$
f(f(f(f(f(\ldots f(2022)))) \ldots))=f(2022)
$$

V-2 There is a function

$$
f(x)=\frac{1}{\sqrt[7]{1-x^{7}}}
$$

Calculate

$$
f(f(f(f(f(\ldots f(2022)))) \ldots))
$$

where the function $f$ is applied 1304 times.

Answer: $\sqrt[7]{1-\frac{1}{2022^{7}}}$

V-3 There is a function

$$
f(x)=\frac{1}{\sqrt[9]{1-x^{9}}}
$$

Calculate

$$
f(f(f(f(f(\ldots f(2022)))) \ldots))
$$

where the function $f$ is applied 1305 times.

Answer: 2022

V-4 There is a function

$$
f(x)=\frac{1}{\sqrt[11]{1-x^{11}}}
$$

Calculate

$$
f(f(f(f(f(\ldots f(2022)))) \ldots))
$$

where the function $f$ is applied 1306 times.

Answer: $1 / \sqrt[11]{1-2022^{11}}$

## Lomonosov Olympiad for Schoolchildren in Mathematics

Final stage $2021 / 22$ academic year for 11th grade

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