Challenge Math Challenge - March 2021

[Not anonymous]

Spoiler: Question 15

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]

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.

 

[archaic]

Question 13:

Spoiler

$$\begin{align*}
Q(n)&=\prod_{i=4}^n\frac{i^2-9}{i^2-4}\\
&=\prod_{i=4}^n\frac{(i+3)(i-3)}{(i+2)(i-2)}\\
&=\prod_{i=4}^n\frac{i+3}{i+2}\prod_{i=4}^n\frac{i-3}{i-2}\\
&=\frac{\frac{(n+3)!}{6!}}{\frac{(n+2)!}{5!}}\frac{(n-3)!}{(n-2)!}\\
&=\frac16\frac{n+3}{n-2}
\end{align*}$$
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 $Q(25003)<Q(25002)$.

 

[archaic]

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 $y_1$.

 

[fresh_42]

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.

 

[Not anonymous]

Spoiler: Question 13

(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]

Re #15; Assume f:R-->R is continuous and for every x, there exists n ≥ 1 such that the nth iteration of f equals x at x, i.e. f(f(f(...(x)))))))), n iterates, = x. In particular, n can depend on x. Prove, or disprove, for all x, f(x) =x.Re #2: I can think of three,1) the quotient group Z/pZ; 2) Z localized at the multiplicative set of integers not divisible by p; and 3) the p-adic integers. 4) ? I guess I could cheat and say also the quotient ring Z/pZ!