# Math Challenge - March 2021

• Challenge
• Featured
Given ##f(f(f(x))) =x##, suppose there exists ##x_1## such that ##f(x_1) \neq x_1##. Say ##f(x_1) = x_2##, where ##x_2 \neq x_1##. We see that ##f(x_2)## cannot be ##x_2## since that will imply ##f(f(f(x_1))) = x_2## and the given condition won't be met. So we must have ##f(x_2) =x_3## for where ##x_3## is different from ##x_2, x_1## and we must have ##f(x_2) = x_1## for the given condition to be met.

Without loss of generality we can assume ##x_1## to be the smallest of the 3 values. Two possibilities arise.
1. ##x_1 \lt x_2 \lt x_3##. This means ##f(x_1) = x_2 \lt f(x_2) = x_3## and ##f(x_2) = x_3 \gt f(x_2) = x_1## and ##f(x_3) \lt f(x_1)##. So the function ##f## must be increasing at least for a sub-interval of ##(x_1, x_2)## and decreasing for a sub-interval of ##(x_2, x_3)## and since it is continuous, there must be some ##x_4 \in (x_2, x_3)## such that ##f(x_4) = x_2##. This is because the continuous method should take every value between ##f(x_2) and f(x_3)## for ##x \in (x_2, x_3)## and ##f(x_2) = x_3 \gt x_2 \gt x_1 = f(x_3)##. But this introduces a contradiction because ##f(f(f(x_4))) = f(f(x_2)) = f(x_3) = x_1 \neq x_4## i.e. given condition is not met some real-valued ##x##. Hence the assumption that ##f(x) \neq x## for some ##x## must be wrong.

2. ##x_1 \lt x_3 \lt x_2##. Here too, we can prove like in case (1) that assuming ##f(x) \neq x## for some ##x## must be wrong

Since both cases lead to violation of given condition, it follows that ##f(x) =x## for all ##x## is a necessary condition for ##f(f(f(x)))=x## for all ##x##

• fresh_42
Mentor
So the function ##f## must be increasing at least for a sub-interval of ##(x_1, x_2)## and decreasing for a sub-interval of ##(x_2, x_3)## and since it is continuous, there must be some ##x_4 \in (x_2, x_3)## such that ##f(x_4) = x_2##.
You can drop this "at least". From ##f(a)=f(b)## follows ##a=f(f(f(a)))=f(f(f(b)))=b##, i.e. ##f## in injective (= into, =one to one). So it can only be either monotone decreasing or monotone increasing on the entire real number line.

Question 13:
\begin{align*} Q(n)&=\prod_{i=4}^n\frac{i^2-9}{i^2-4}\\ &=\prod_{i=4}^n\frac{(i+3)(i-3)}{(i+2)(i-2)}\\ &=\prod_{i=4}^n\frac{i+3}{i+2}\prod_{i=4}^n\frac{i-3}{i-2}\\ &=\frac{\frac{(n+3)!}{6!}}{\frac{(n+2)!}{5!}}\frac{(n-3)!}{(n-2)!}\\ &=\frac16\frac{n+3}{n-2} \end{align*}
a) ##Q(4)=\frac{7}{12}## and ##\lim_{n\to\infty}Q(n)=\frac16##.
$$\left(\frac{x+3}{x-2}\right)'=\frac{(x-2)-(x+3)}{(x-2)^2}=-\frac{5}{(x-2)^2}<0$$
So ##Q(n)## starts at ##\frac7{12}## for ##n=4## and decreases to ##\frac16## as ##n## gets bigger:
$$\frac16<Q(n)\leq\frac7{12}$$
b) No. Since ##0.1667>\frac16##, there should be an ##N>4## such that ##Q(n)\leq0.1667## for every ##n\geq N##.
Another way to do it is by solving ##Q(n)=\frac16\frac{n+3}{n-2}=0.1667## for ##n##. This gives us ##n=25002\geq4##, which means that the inequality is false, because ##Q(25003)<Q(25002)##.

Last edited:
I have tried 4), but got ##y_1(t)=y_2(t)=0## when the initial conditions are as stated. Is that wrong?
I started by solving for ##y_2## in the first equation, differentiating, then substituting by ##\dot y_2## from the second equation and ##y_2## from the first equation equation. This gave me a second order linear ode in ##\ddot y_1,\,\dot y_1## and ##y_1##.

Mentor
I have tried 4), but got ##y_1(t)=y_2(t)=0## when the initial conditions are as stated. Is that wrong?
I started by solving for ##y_2## in the first equation, differentiating, then substituting by ##\dot y_2## from the second equation and ##y_2## from the first equation equation. This gave me a second order linear ode in ##\ddot y_1,\,\dot y_1## and ##y_1##.
No, as far as part (a) is concerned. But this stable solution is only the warm up.

(a) The inequality can be proven by using method of induction. We first notice that the LHS is equivalent to ##P_{n} \equiv \prod_{i=4}^n \frac {(i-3)(i+3)} {(i-2)(i+2)}##. For induction, we claim that this product is equal to ##\left(\frac {1} {6}\right) \left(\frac {n+3} {n-2}\right)##. The base case is for ##n=4## and we see that that claim holds true for this base case since ##\prod_{i=4}^4 \dfrac {(4-3)(4+3)} {(4-2)(4+2)} = \left(\frac {1} {6}\right) \left(\frac {4+3} {4-2}\right)##.

Now assume that the claim is true for all ##n \in {4, 5, ..., k}## for some positive integer ##k >= 4##. Now consider the product up to ##k+1##, i.e. ##P_{k+1} = P_{k} \times \left(\frac {(k+1-3)(k+1+3)} {(k+1-2)(k+1+2)}\right)##. Substituting for ##P_{k}## based on inductive claim, this product becomes $$P_{k+1} = \left(\frac {1} {6}\right) \left(\frac {k+3} {k-2}\right) \left(\frac {(k-2)(k+4)} {(k-1)(k+3)}\right) \Rightarrow P_{k+1} = \left(\frac {1} {6}\right) \left(\frac {k+4} {k-1}\right) \equiv \left(\frac {1} {6}\right) \left(\frac {(k+1)+3} {(k+1)-2}\right)$$, so the inductive claim holds true for ##k+1## too, proving the original claim for all values of ##n >= 4##.
Clearly ##P_{n} = \left(\frac {1} {6}\right) \left(\frac {n+3} {n-2}\right) \gt \frac {1} {6}## since the ##\left(\frac {n+3} {n-2}\right) \gt 1## for any ##n \gt 2##.

(b) No, the statement isn't valid if RHS of inequality is replaced by ##0.1667## because ##\left(\frac {n+3} {n-2}\right)## tends to 1 as ##n## tends to infinity. In other words, ##P_{n} \rightarrow \frac {1} {6}## as ##n \rightarrow \infty## (but tending from right side, so always greater than ##\frac {1} {6}##), meaning ##P_{n}## can go arbitrarily close to ##\frac {1} {6}## from right. A simple counterexample is ##n=100002##. ##P_{100002} = \frac {1} {6} \frac {100005} {100000} = 0.166675 \lt 0.1667##

mathwonk