Math Challenge - March 2021

  • Challenge
  • Thread starter fresh_42
  • Start date
  • Featured
  • #51
125
48
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##
 
  • #52
15,542
13,639
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.
 
  • #53
688
210
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:
  • #54
688
210
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##.
 
  • #55
15,542
13,639
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.
 
  • #56
125
48
(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##
 
  • #57
mathwonk
Science Advisor
Homework Helper
2020 Award
11,242
1,447
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 coud cheat and say also the quotient ring Z/pZ!
 
Last edited:

Related Threads on Math Challenge - March 2021

Replies
93
Views
2K
Replies
102
Views
3K
Replies
67
Views
4K
  • Last Post
5
Replies
114
Views
3K
Replies
86
Views
6K
Replies
100
Views
2K
  • Sticky
  • Last Post
2
Replies
35
Views
724
Replies
97
Views
15K
Replies
77
Views
9K
Replies
61
Views
3K
Top