Tensor's test solutions

The problems were:

1) $f:\mathbb{R}\to\mathbb{R}$ is a continuous function such that $f(f(x)) = x,\ \forall x \in \mathbb{R}$. Show that, for some $c \in \mathbb{R}, f(c) = c$.

2) Find all natural numbers $x, y, x \gt y$, such that $x^y = y^x$. Hint (given as part of the test): What is the maximum value of $f(x) = x^{1/x}$?

3) $f_n$ is a sequence such that $f_1 \gt 0$ and $$3f_{n+1} = 2f_n + \frac{A}{f_n^2}$$ for some constant $A \gt 0$ and all $n \ge 1$. Show that $f_{n+1} \le f_n \forall n \gt 1$


Solution to Problem 1)

Consider $g(x) = f(x) - x$. If there was no such $c$ such that $f(c) = c$, then we must have that $g(x)$ is always positive or always negative, as we can show that a continuous function that is never zero is only always positive or negative, using the intermediate value theorem.

If $g(x) \gt 0, \forall x$, then $f(f(x)) \gt f(x) \gt x$, a contradiction. The other case is similar.


Solution to Problem 2)

The equation is basically $x^{1/x} = y^{1/y}$.

$f(x) = x^{1/x}$ has a maximum value at $x = e$. The function is increasing in $(1, e)$ and decreasing in $(e, \infty)$. Since $x \gt y$, we must have $y = 1$ or $y = 2$ (as $e = 2.718...$) and there is a unique $x \gt e$ corresponding to that $y$.

For $y = 2$, $x = 4$ is easily seen to be a solution.

Solution to Problem 3)

We can rewrite the sequence as

$$3f_{n+1} = f_n + f_n + \frac{A}{f_n^2}$$

Using the Arithmetic and Geometric means inequality (for another usage, see here: nth root of n), we see that

$$f_{n+1} \ge \sqrt[3]{A} \implies f_{n+1}^3 \ge A$$

Now $3f_{n+1} - 3f_n = \frac{A}{f_n^2} - f_n = \frac{A - f_n^3}{f_n} \le 0$ for $n \gt 1$.

We have shown that the sequence is both monotonically decreasing, and bounded below (for $n \gt 1$).