Worked problem 2 ($✓$) 1999 Paper I

The $n$ positive numbers ${x}_{1},{x}_{2},\dots ,{x}_{n}$, where $n\ge 3$, satisfy

${x}_{1}=1+\frac{1}{{x}_{2}}\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{2}=1+\frac{1}{{x}_{3}}\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{n-1}=1+\frac{1}{{x}_{n}}\phantom{\rule{0.3em}{0ex}},$

and also

$\phantom{\rule{1em}{0ex}}{x}_{n}=1+\frac{1}{{x}_{1}}\phantom{\rule{0.3em}{0ex}}.$

Show that

(i)
${x}_{1},{x}_{2},\dots ,{x}_{n}>1$,
(ii)
${x}_{1}-{x}_{2}=-\frac{{x}_{2}-{x}_{3}}{{x}_{2}{x}_{3}}\phantom{\rule{0.3em}{0ex}}$,
(iii)
${x}_{1}={x}_{2}=\cdots ={x}_{n}\phantom{\rule{0.3em}{0ex}}$.

Hence ﬁnd the value of ${x}_{1}$.

First thoughts

My ﬁrst thought is that this question has an unknown number of variables: ${x}_{1}$, … , ${x}_{n}$. That makes it seem rather complicated. I might, if necessary, try to understand the result by choosing an easy value for $n$ (maybe $n=3$). If I manage to prove some of the results in this special case, I will certainly go back to the general case: doing the special case might help me tackle the general case, but I don’t expect to get many marks in an exam if I just prove the result in one special case.

Next, I see that the question has three sub-parts, then a ﬁnal one. The ﬁnal one begins ‘Hence ...’. This means that I must use at least one of the previous parts in my working for the ﬁnal part. It is not clear from the structure of the question whether the three sub-parts are independent; the proof of (ii) and (iii) may require the previous result(s), or it may not.

Actually, I think I can see how to do the very last part. If I assume that (iii) holds, so that ${x}_{1}={x}_{2}=\cdots ={x}_{n}=x$ (say), then each of the equations given in the question is identical and each gives a simple equation for $x$.

It is surprising, isn’t it, that the ﬁnal result doesn’t depend on the value of $n$? That makes the idea of choosing $n=3$, just to see what is going on, quite attractive, but I’m not going to resort to that idea unless a get very stuck.

The notation in part (i) is a bit odd. I’m not sure that I have seen anything like it before.13 But it can only mean that each of the variables ${x}_{1}$, ${x}_{2}$, $\dots \phantom{\rule{0.3em}{0ex}}$ ${x}_{n}$ is greater than 1.

In fact, now I think about it, I am puzzled about part (i). How can an inequality help to derive the equalities in the later parts? I can think of a couple of ways in which the result ${x}_{i}>1$ could be used. One is that I may need to cancel, say, ${x}_{1}$ from both sides of an equation in which case I would need to know that ${x}_{1}\ne 0$. But looking back at the question, I see I already know that ${x}_{1}>0$ (it is given right at the start of the question), so this cannot be the right answer. Maybe I need to cancel some other factor, such as $\left({x}_{1}-1\right)$. Another possibility is that I get two or more solutions by putting ${x}_{1}={x}_{2}=\cdots ={x}_{n}=x$ and I need the one with $x>1$. This may be the answer: looking back at the question again, I see that it asks for the value of ${x}_{1}$ — so I am looking for a single value. I’m still puzzled, but I will remember to keep a sharp look out for ways of using part (i).

One more thing strikes me about the question. The equations satisﬁed by the ${x}_{i}$ are given on two lines (‘and also’). This could be for typographic reasons (the equations would not all ﬁt on one line) but more likely it is to make sure that I have noticed that the last equation is a bit different: all the other equations relate ${x}_{i}$ to ${x}_{i+1}$, whereas the last equation relates ${x}_{n}$ to ${x}_{1}$. It goes back to the beginning, completing the cycle. I’m pleased that I thought of this, because this circularity must be important.

Doing the question

I think I will do the very last part ﬁrst, and see what happens.

Suppose, assuming the result of part (iii), that ${x}_{1}={x}_{2}=\cdots ={x}_{n}=x$. Then substituting into any of the equations given in the question gives

$x=1+\frac{1}{x}$

i.e. ${x}^{2}-x-1=0$. Using the quadratic formula gives

$x=\frac{1±\surd 5}{2}\phantom{\rule{0.3em}{0ex}},$

which does indeed give two answers (despite the fact that the question asks for just one). However, I see that one is negative and can therefore be eliminated by the condition ${x}_{i}>0$ which was given in the question (not, I note, the condition ${x}_{1}>1$ from part (i); I still have to ﬁnd a use for this).

I needed only part (iii) to ﬁnd ${x}_{1}$, so I expect that either I need both (i) and (ii) directly to prove (iii), or I need (i) to prove (ii), and (ii) to prove (iii).

Now that I have remembered that ${x}_{i}>0$ for each $i$, I see that part (i) is obvious. Since ${x}_{2}>0$ then $1∕{x}_{2}>0$ and the ﬁrst equation given in the question, ${x}_{1}=1+1∕{x}_{2}\phantom{\rule{0.3em}{0ex}}$, shows immediately that ${x}_{1}>1$ and the same applies to ${x}_{2}$, ${x}_{3}$, etc.

Now what about part (ii)? The given equation involves ${x}_{1}$, ${x}_{2}$ and ${x}_{3}$, so clearly I must use the ﬁrst two equations given in the question:

${x}_{1}=1+\frac{1}{{x}_{2}}\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{2}=1+\frac{1}{{x}_{3}}\phantom{\rule{0.3em}{0ex}}.$

Since I want ${x}_{1}-{x}_{2}$, I will see what happens if I subtract the two equations:

 ${x}_{1}-{x}_{2}=\left(1+\frac{1}{{x}_{2}}\right)-\left(1+\frac{1}{{x}_{3}}\right)=\frac{1}{{x}_{2}}-\frac{1}{{x}_{3}}i=\frac{{x}_{3}-{x}_{2}}{{x}_{2}{x}_{3}}=-\frac{{x}_{2}-{x}_{3}}{{x}_{2}{x}_{3}}\phantom{\rule{2.77695pt}{0ex}}.$ ($\ast$)

That seems to work!

One idea that I haven’t used so far is what I earlier called the circularity of the equations: the way that ${x}_{n}$ links back to ${x}_{1}$. I’ll see what happens if I extend the above result. Since there is nothing special about ${x}_{1}$ and ${x}_{2}$, the same result must hold if I add 1 to each of the suffices:

${x}_{2}-{x}_{3}=\frac{{x}_{3}-{x}_{4}}{{x}_{3}{x}_{4}}\phantom{\rule{2.77695pt}{0ex}}.$

I see that I can combine this with the previous result:

${x}_{1}-{x}_{2}=-\frac{{x}_{2}-{x}_{3}}{{x}_{2}{x}_{3}}=\frac{{x}_{3}-{x}_{4}}{{x}_{2}{x}_{3}^{2}{x}_{4}}\phantom{\rule{2.77695pt}{0ex}}.$

I now see where this is going. The above step can be repeated to give

${x}_{1}-{x}_{2}=\frac{{x}_{3}-{x}_{4}}{{x}_{2}{x}_{3}^{2}{x}_{4}}=-\frac{{x}_{4}-{x}_{5}}{{x}_{2}{x}_{3}^{2}{x}_{4}^{2}{x}_{5}}=\cdots \phantom{\rule{2.77695pt}{0ex}}$

and eventually I will get back to ${x}_{1}-{x}_{2}$:

${x}_{1}-{x}_{2}=-\frac{{x}_{4}-{x}_{5}}{{x}_{2}{x}_{3}^{2}{x}_{4}^{2}{x}_{5}}=\cdots =±\frac{{x}_{1}-{x}_{2}}{{x}_{2}{x}_{3}^{2}{x}_{4}^{2}{x}_{5}^{2}\dots {x}_{n}^{2}{x}_{1}^{2}{x}_{2}}\phantom{\rule{2.77695pt}{0ex}}$

i.e.

$\left({x}_{1}-{x}_{2}\right)\left(1\mp \frac{1}{{x}_{1}^{2}{x}_{2}^{2}{x}_{3}^{2}{x}_{4}^{2}{x}_{5}^{2}\dots {x}_{n}^{2}}\right)=0\phantom{\rule{2.77695pt}{0ex}}.$

I have put in a $±$ because each step introduces a minus sign and I’m not sure yet whether the ﬁnal sign should be ${\left(-1\right)}^{n}$ or ${\left(-1\right)}^{n-1}$. I can check this later (for example, by working out one simple case such as $n=3$); but I may not need to.

I deduce from this last equation that either

${x}_{1}^{2}{x}_{2}^{2}{x}_{3}^{2}{x}_{4}^{2}{x}_{5}^{2}\dots {x}_{n}^{2}=±1$

or

${x}_{1}={x}_{2}$

(which is what I want). At last I see where to use part (i): I know that

${x}_{1}^{2}{x}_{2}^{2}{x}_{3}^{2}{x}_{4}^{2}{x}_{5}^{2}\dots {x}_{n}^{2}\ne ±1$

because ${x}_{1}>1$, ${x}_{2}>1$, etc. Thus the only possibility is ${x}_{1}={x}_{2}$. Since there was nothing special about ${x}_{1}$ and ${x}_{2}$, I deduce further that ${x}_{2}={x}_{3}$, and so on, as required.

Post-mortem

There were a number of useful points in this question.

1.
The ﬁrst point concerns using the information given in the question. The process of teasing information from what is given is fundamental to the whole of mathematics. It is very important to study what is given (especially seemingly unimportant conditions, such as ${x}_{i}>0$) to see why they have been given. If you ﬁnd you reach the end of a question without apparently using some given information, then you should look back over your work: it is very unlikely that a condition has been given that is not used in some way. It may not be a necessary condition — and we will see that the condition ${x}_{i}>0$ is not, in a sense, necessary in this question — but it should be sufficient. The other piece of information in the question which you might easily have overlooked is the use of the singular rather than the plural in referring to the solution (‘ ... ﬁnd the value of ...’), implying that there is just one value, despite the fact that the ﬁnal equation is quadratic.
2.
The second point concerns the structure of the question. Here, the position of the word ‘Hence’ suggested strongly that none of the separate parts were stand-alone results; each had to be used for a later proof. Understanding this point made the question much easier, because I was always on the look out for an opportunity to use the earlier parts. Of course, in some problems (without that ‘hence’) some parts may be stand-alone; though this is rare in STEP questions.

You may think that this is like playing a game according to hidden (STEP) rules, but that is not the case. Precision writing and precision reading is vital in mathematics and in many professions (law, for example). Mathematicians have to be good at it, which is the reason why so many employers want to recruit people with mathematical training.

3.
The third point was the rather inconclusive speculation about the way inequalities might help to derive an equality. It turned out that what was actually required was ${x}_{1}{x}_{2}\dots {x}_{n}\ne ±1$. I was a bit puzzled by this possibility in my ﬁrst thoughts, because it seemed that the result ought to hold under conditions different from those given; for example, ${x}_{i}<0$ for all $i$ (does this condition work??). Come to think of it, why are conditions given on all ${x}_{i}$ when they are all related by the given equations? This makes me think that there ought to be a better way of proving the result which would reveal exactly the conditions under which it holds.
4.
Then there was the idea (which I didn’t actually use) that I might try to prove the result for, say, $n=3$ to help me understand what was going on. This would not have counted as a proof of the result (or anything like it), but it might have given me ideas for tackling the question.
5.
A key observation was that the equations given in the question are ‘circular’. It was clear that the circularity was essential to the question and it turned out to be the key to the most difficult part. Having identiﬁed it early on, I was ready to use it when the opportunity arose.
6.
Finally, I was pleased that I read the whole question carefully before plunging in. This allowed me to see that I could easily do the last part before the preceding parts, which I found very helpful in getting into the question.

Final thoughts

It occurs to me only now, after my post-mortem, that there is another way of obtaining the ﬁnal result.

Suppose I start with the idea of circularity (as indeed I might have, had I not been otherwise directed by the question) and use the given equations to ﬁnd ${x}_{1}$ in terms of ﬁrst ${x}_{2}$, then ${x}_{3}$, then ${x}_{4}$ and eventually in terms of ${x}_{1}$ itself. That should give me an equation I can solve, and I should be able to ﬁnd out what conditions are needed on the ${x}_{i}$. Try it. You may need to guess a formula for ${x}_{1}$ in terms of ${x}_{i}$ from a few special cases, then prove it by induction.14 You will ﬁnd it useful to deﬁne a sequence of numbers ${F}_{i}$ such that ${F}_{0}={F}_{1}=1$ and ${F}_{n+1}={F}_{n}+{F}_{n-1}$. (These numbers are called Fibonacci numbers.15)

You should ﬁnd that if ${x}_{n}={x}_{1}$ for any $n$ (greater than 1), then ${x}_{n}=\frac{1±\surd 5}{2}\phantom{\rule{0.3em}{0ex}}$.

13 It is just the sort of thing that is used in university texts; but I’m not sure that it would be used in STEP papers nowadays.

14 I went as far as ${x}_{7}$ to be sure of the pattern: I found that ${x}_{7}=\frac{8{x}_{1}+5}{5{x}_{1}+3}$.

15 Fibonacci (short for ﬁlius Bonacci — son of Bonacci) was called the greatest European mathematician of the Middle Ages. He was born in Pisa (Italy) in about 1175 AD. He introduced the series of numbers named after him in his book of 1202 called Liber Abbaci (Book of the Abacus). It was the solution to the problem of the number of pairs of rabbits produced by an initial pair: A pair of rabbits are put in a ﬁeld and, if rabbits take a month to become mature and then produce a new pair every month after that, how many pairs will there be in twelve months time?