# Problem 6 - IMO 1985

For every real number $x_1$ construct the sequence $x_1,x_2,x_3,\ldots$ by setting $x_{n+1}=x_n(x_n+\frac{1}{n})$ for each $n \ge 1$. Prove that there exists exactly one value of $x_1$ for which $0 < x_n < x_{n+1} < 1$ for every $n$.

• Solution here: artofproblemsolving.com/Forum/… – Sandeep Silwal Apr 18 '14 at 19:34
• I think this proof is very artificial... I want to work in a more simple solution... – lele Apr 18 '14 at 19:55
• Nice question! Any chance you can tell us what that value of $x_1$ is? – barak manos Jul 5 '14 at 7:26
• @boywholived: Thanks. – barak manos Jul 5 '14 at 7:40
• @barakmanos: See my too long comment ("answer") below for a numerical approximation of $x_1$. – Han de Bruijn Jul 7 '14 at 11:10

Assume that there is some $x_1$ such that $0<x_n<x_{n+1}<1$ then $\{x_n\}$ is monotonically increasing and bounded. Hence converges by monotone convergence theorem.

Let $\lim\limits_{n\rightarrow \infty}{x_n}=l$, then obtain $l^2=l \Rightarrow l=0,1$, we can readily omit $0$ and hence $x_n\rightarrow 1$.

Proof for uniqueness of $x_1$.

Assume for the first term $x_1$ and $a_1$, the sequence $\{x_n\}$ and $\{a_n\}$ converges. Without loss of generality assume $x_1>a_1$(Recall that the definition states that all terms of the sequence are positive). Using induction we can prove that $x_n>a_n \quad\forall n\in \mathbb{N}$.

As $x_n$ and $a_n$ both converges to $1$, there exits an $N$ such that for all $n\ge N\implies \frac{3}{4}<a_n<x_n<1$ (Take $\varepsilon=\frac{1}{4}$)

Observe that $$x_{n+1}-a_{n+1}=\left(x_n-a_n\right)\left(x_n+a_n\right)+\frac{1}{n}\left(x_n-a_n\right)>\frac{3}{2}(x_n-a_n)$$

Hence $$\begin{array}{rcl}\frac{x_{n+1}-a_{n+1}}{x_n-a_n}&>&\frac{3}{2}\implies\\ x_{n+1}-a_{n+1}&>&\left(\frac{3}{2}\right)^{n+1-N}\cdot(x_N-a_N) \\ \end{array}$$

The divergence of $x_n-a_n$ is clear. A contradiction.

To prove the existence, denote $f_{1}(x)=x$ and $f_{n+1}(x)=f_{n}(x)\left(f_{n}(x)+\frac{1}{n}\right)$. Using induction prove that $f_{n}(x)$ is a strictly increasing function in $x$.

Let $b_n$, $c_n$ be such that $f_{n}(b_n)=1-\frac{1}{n}$ and $f_{n}(c_n)=1$, the existence and uniqueness of $b_n$, $c_n$ follows from monotonicity of $f_n$, the observations $f_{n}(0)=0$, $f_{n}(1)>1$ and the continuity of $f_{n}$. Also observe that $b_n<c_n$ and using induction that $b_n$ is strictly increasing and $c_n$ is strictly decreasing sequences.

The set $A=\{b_n\}$ is bounded above. Hence by least upper bound property of $\mathbb{R}$ there exists a $\gamma =\sup \{b_n\}$. We shall show that $\gamma$ is indeed our $x_1$!

$$\color{blue}{f_{n+1}{(\gamma)}}=f_{n}(\gamma)\left(f_n{(\gamma)}+\frac{1}{n}\right)\color{blue}{>}f_{n}(\gamma)\left(f_{n}(b_n)+\frac{1}{n}\right)=\color{blue}{f_{n}(\gamma)>0}$$ $$\color{blue}{1}=f_{n}(c_{n})\color{blue}{>f_{n}(\gamma)} \quad \forall n$$ $$\implies 1>f_{n+1}{(\gamma)}>f_{n}(\gamma)>0 \qquad \square$$

Proofs of interest

$f_n(x)$ has non negative integer coefficients hence $f_n(x)$ is strictly increasing on $[0,1]$.

Notice that $\color{blue}{f_{n+1}(b_n)}=f_{n}(b_n)\left(f_{n}(b_n)+\frac{1}{n}\right)=1-\frac{1}{n}\color{blue}{<}1-\frac{1}{n+1}=\color{blue}{f_{n+1}(b_{n+1})} \implies b_n< b_{n+1}$ or $b_n$ is a strictly increasing sequence.

Notice that $\color{blue}{f_{n+1}(c_n)}=f_{n}(c_n)\left(f_{n}(c_n)+\frac{1}{n}\right)=1+\frac{1}{n}\color{blue}{>}1=\color{blue}{f_{n+1}(c_{n+1})} \implies c_{n+1}<c_n$ or $c_n$ is a strictly decreasing sequence.

The fact that $b_n$ is bounded is clear as $b_n<c_1 \quad \forall n$.

• How is the sequence bounded? That depends on $x_1$, which is yet to be determined! For example $x_1>\frac{\sqrt 5-1}2$ leads to $x_2>1$ and from then on the sequnce diverges exponentially. - And monotonic depends on the choiice of $x_1$ as well. - Finally, I seem to have missed your definitions of $a_n, b_n$. – Hagen von Eitzen Jul 7 '14 at 17:22
• @boywholived: I've withdrawn my downvote, but still find your proof as a whole not very convincing. Which may be due to my own dyslexia. – Han de Bruijn Jul 7 '14 at 18:08
• Check, check, double check .. Upon repeatedly re-reading your purported proof, I couldn't detect any serious flaw in it. Therefore I've removed the question mark in my answer and the accompanying comment. I always have an uneasy feeling, though, with such abstract non-constructive proofs establishing existence and uniqueness, without actually calculating the number in question. Therefore, in my own too long comment, I've "replaced" existence by explicit calculation and uniqueness by error analysis. Thus trying to fill that gap. – Han de Bruijn Jul 9 '14 at 9:09

This is not a separate answer, but rather a (too long) comment in response to the proper answer by boywholived. Sure, there exists exactly one such value of $x_1$, but the comment by barak manos still stands: any chance you can tell us what that value of $x_1$ is? Here is a little computer program that does the job:

program IMO_1985;
procedure find; const LARGE : integer = 50; var n,i : integer; x,x_1,bit : double; H,L : boolean; begin { Initialize } x_1 := 1/2; bit := 1/2; for i := 0 to LARGE do begin n := 1; H := true; L := true; { Writeln(n:2,' : ',x_1); } x := x_1; while true do begin x := x*(x+1/n); { Writeln((n+1):2,' : ',x); } { Decreasing towards zero: } L := (x < 1-1/n); { Exploding towards infinity: } H := (x > 1); if H or L then Break; n := n+1; end; { Readln; } { Find x_1 bit by bit: } bit := bit/2; if H then x_1 := x_1 - bit; if L then x_1 := x_1 + bit; end; Writeln(x_1:0:16); end;
begin find; end.
The program uses a few easy to prove statements about the sequence:

• If the sequence $x_n$ is decreasing for some value of $n$ ( $x_{n+1} < x_n$ ) then it will remain that way until it reaches a limit, which is zero.
• The sequence is decreasing if and only if $x_n < 1-1/n$ . Due to the previous statement, we can put a Break in the calculations then. And conclude that our estimate of the initial value $x_1$ must have been greater (by one bit).
• If some value of $x_n$ becomes greater than $1$ then the sequence will be increasing forever (i.e. it explodes). We can put a Break in the calculations then. And conclude that our estimate of the initial value $x_1$ must have been smaller (by one bit).
It turns out that the convergence of the sequence towards the desired limit $\;x_n \to 1\;$ is extremely unstable numerically. So what we actually have is a switch between $\,0\,$ and $\,\infty\,$ that generates the outcome for $x_1$ bit by bit, until double precision is reached.
Now I almost forgot to tell you what the outcome is: $$x_1 \approx 0.4465349172642295$$ Reverse method. If a method is (numerically) unstable, then try the reverse. Because it's not uncommon that the latter will be stable then. So let's solve $x_n \ge 0$ from $x_{n+1} = x_n(x_n+1/n)$: $$x_n^2 + \frac{1}{n} x_n - x_{n+1} = 0 \quad \Longrightarrow \quad x_n = -\frac{1}{2n}+\sqrt{\left(\frac{1}{2n}\right)^2+x_{n+1}} = \frac{x_{n+1}}{\frac{1}{2n}+\sqrt{\left(\frac{1}{2n}\right)^2+x_{n+1}}}$$ The latter trick is for reasons of numerical stability as well. Now start with some LARGE value of $n$ and put $x_{n+1}=1-1/(n+1)$. Then iterate backwards until $x_1$ is reached. The resulting (Pascal) program has been an almost one-liner (i.e. without the Error Analysis):

program IMO_1985;
procedure find; const LARGE : integer = 50; var n : integer; x,E : Extended; begin { Initialize } x := 1-1/(LARGE+1); E := 1/(LARGE+1); for n := LARGE downto 1 do begin x := x/(sqrt(1/sqr(2*n)+x)+1/(2*n)); { Error Analysis: } E := E/(2*x+1/n); end; Writeln('Outcome x_1 = ',x:0:16); Writeln(' # decimals = ',-Trunc(ln(E)/ln(10))); Writeln('# good bits = ',-Trunc(ln(E)/ln(2))); end;
begin find; end.
Needless to say that the outcome is the same as with the previous method.

Error Analysis. Estimate with infinitesimals: $$x_{n+1} = x_n(x_n+1/n) \quad \Longrightarrow \quad dx_{n+1} = (2x_n+1/n)dx_n$$ Initialize with $\;dx_{n+1} = 1/(n+1)$ . It follows that: $$dx_1 = \frac{dx_{n+1}}{(2x_1+1)(2x_2+1/2)(2x_3+1/3)\cdots(2x_n+1/n)}$$ The Reverse method program has been slightly modified to take this into account. The variable E corresponds with the errors $dx_n$ . Minus the 10-logarithm of the error $dx_1$ in $x_1$ is the number of significant decimals, $16$ in our case, which indeed seems to be alright. Minus the 2-logarithm of the error $dx_1$ in $x_1$ is the number of significant bits: $53$ . This number, not at all by coincidence (why?), is close to the number LARGE $=50$ in both programs.

Update. Actually, our value $\,x_1\,$ as well as the error $\,dx_1\,$ are functions of $\,n$ , so let's replace $\,x_1\,$ by $\,X(n)\,$ for all finite values of $\,n\,$ and likewise $\,dx_1\,$ by $\,\epsilon(n)$ : $$\epsilon(n) = \frac{1/(n+1)}{(2x_1+1)(2x_2+1/2)(2x_3+1/3)\cdots(2x_n+1/n)} \quad \Longrightarrow \\ \epsilon(n) = \epsilon(n-1)\frac{n/(n+1)}{2x_n+1/n}$$ With $\,1-1/n < x_n < 1\,$ we have $\,2-1/n < 2x_n+1/n < 2+1/n\,$ and: $$\epsilon(n) < \epsilon(n-1)\frac{1}{(1+1/n)(2-1/n)} = \frac{1}{2+1/n-1/n^2} < \frac{1}{2} \quad \Longrightarrow \quad \epsilon(n) < \epsilon(1)\left(\frac{1}{2}\right)^{n-1}$$ For $\,n=1\,$ we have $\,1 < 2x_1+1 < 3\,$ hence $\,\epsilon(1) < 1/2$ . This proves a
Lemma. (see above: so that's why!) : $$\epsilon(n) < \left(\frac{1}{2}\right)^n$$ Note. I'm not quite sure of this, because $\,\epsilon(1)=dx_1=1/2\,$ is not an infinitesimal ( i.e sufficiently small ) error; perhaps $\,\epsilon(n)\,$ must be multiplied with a proper constant?
Theorem. As expected: $$\lim_{n\to\infty} X(n) = x_1$$ Proof. For all real $\epsilon > 0$ there exists an integer $N > 0$ such that if $n > N$ then $\left|X(n) - x_1\right| < \epsilon$ .
Indeed, define $N$ by $\,N = \ln(1/\epsilon)/\ln(2)$ , then $\,\epsilon = (1/2)^N\,$ and for $n > N$ we have with the above Lemma: $\,\left|X(n)-x_1\right| < (1/2)^n < \epsilon$ . This completes the proof.
Disclaimer: though apart from a few technicalities perhaps, see the above Note.