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 \(n \gt 1\)).
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 \(n \gt 1\)).
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