## A look at a few Tripos questions IV

This post belongs to a series that began here. Next up is a question about integration.

11B. Let $f:[a,b]\to\mathbb{R}$ be continuous. Define the integral $\int_a^bf(x)dx$. (You are not asked to prove existence.)

Suppose that $m, M$ are real numbers such that $m\leq f(x)\leq M$ for all $x\in [a,b]$. Stating clearly any properties of the integral that you require, show that

$\displaystyle m(b-a)\leq\int_a^bf(x)dx\leq M(b-a)$.

The function $g:[a,b]\to\mathbb{R}$ is continuous and non-negative. Show that

$\displaystyle m\int_a^bg(x)dx\leq\int_a^bf(x)g(x)dx\leq M\int_a^bg(x)dx$.

Now let $f$ be continuous on $[0,1]$. By suitable choice of $g$ show that

$\displaystyle \lim_{n\to\infty}\int_0^{1/\sqrt{n}}nf(x)e^{-nx}dx=f(0)$,

and by making an appropriate change of variable, or otherwise, show that

$\displaystyle \lim_{n\to\infty}\int_0^1nf(x)e^{-nx}dx=f(0)$.

The first part of this question is something I recommend practising until you can write it out very fast, since it often comes up. (I’ve just checked and it came up in 2010, though in the slightly more complicated case of a function defined on $[a,\infty)$, and not in 2011, so the probability of its coming up this year is reasonably high.)

Are there any decisions to make about what to say and what not to say? In particular, does the fact that the function is assumed to be continuous, as opposed to merely integrable, make a difference? Not really — it just means that we can assume that the function is integrable and concentrate on saying what the integral is. But the two are so closely related that that isn’t much of a saving, if any. So we’d better dive in and write out a chunk of bookwork.

I was just about to write something when I realized that there is in fact a decision to make. For example, would it be enough to write this?

The integral $\int_a^bf(x)dx$ is the supremum of $m_D(f)$ over all dissections $D$ of the interval $[a,b]$, where $m_D(f)$ is the lower sum associated with $f$ and the dissection $D$.

Or should we say what a dissection is and what a lower sum is? (Note that we can get away with not mentioning upper sums, since for continuous functions this really does define the Riemann integral. If we were being asked to define the Riemann integral more generally, we would need to talk about this sup equalling the inf of all upper sums.) This is a bit of a grey area. On the one hand, dissections and lower sums come slightly before the definition of the integral. On the other hand, they aren’t used for much except the definition of the integral, so one can’t really think of them as an earlier part of the course. My guess is that if you were to write a super-brief answer like that, the examiner would be irritated enough to remove a mark or two, and enough irritations like that can end up costing you an alpha.

I would therefore go for an answer more like this.

A dissection of an interval $[a,b]$ is a sequence $a=x_0. If $D$ is a dissection of $[a,b]$ and $f:[a,b]\to\mathbb{R}$, then the lower sum $m_D(f)$ is the quantity

$\displaystyle \sum_{i=1}^n(x_i-x_{i-1})\inf_{t\in[x_{i-1},x_i]}f(t)$.

If $f$ is continuous, then the integral $\int_a^bf(x)dx$ is the supremum of $m_D(f)$ over all dissections $D$ of $[a,b]$.

That doesn’t take too long to write, especially if you’ve practised it. It may be that the notation and terminology you were given in lectures was slightly different from what I’ve just written down. In that case, use what you were given.

For the next part of the question, the annoying process of reading the mind of the examiner is quite hard. What properties of the integral are we allowed to assume? It’s confusing, because if we argue directly from the definition, then we don’t seem to need any additional properties, whereas if we assume a property such as that if $u(x)\leq v(x)$ for every $x$, then $\int u(x)dx\leq\int v(x)dx$, then the question becomes rather easy. Or does it? We also need to know that the integral of a constant function is the constant times $b-a$.

The wording of the question is clearly suggesting that we don’t have to argue directly from the definition, so it is probably inviting us to use properties of the integral that are typically proved immediately after the definition. Perhaps that year the lecturer proved that the integral of a non-negative function is non-negative, and that the integral of $f+g$ is the integral of $f$ plus the integral of $g$. In that case, the following answer would have been appropriate.

I shall use the fact that the integral of a non-negative function is non-negative, and that the integral of the sum of two continuous functions is the sum of their integrals. I shall also use the fact that the integral of the constant function $C$ on the interval $[a,b]$ is $C(b-a)$.

From these facts, we find that $\int_a^b(f(x)-m)dx\geq 0$, and therefore that

$\int_a^bf(x)dx\geq\int_a^bmdx=m(b-a)$.

Similarly, $\int_a^b(M-f(x))dx\geq 0$, and therefore

$0\leq\int_a^b(M-f(x))dx=M(b-a)-\int_a^bf(x)dx$.

This proves that $\int_a^bf(x)dx\leq M(b-a)$.

I am not completely confident that that is the answer that the examiner was hoping for, but I think it would be difficult to justify removing marks for making those assumptions, given how the question was worded.

For the next part it is similarly not quite clear what one is allowed to assume. But since we can get away with the same assumptions that we’ve already made, that seems a sensible choice. So I’d write this.

Since $g$ is non-negative and $m\leq f(x)\leq M$ for every $x$, we have the inequalities $mg(x)\leq f(x)g(x)\leq Mg(x)$ for every $x$. The result follows by integrating from $a$ to $b$.

Actually, I used there the result that if one function is at most another then the integral of the first is at most the integral of the second. But I feel OK about that, since I basically proved it when doing the previous part (or at least demonstrated that I knew how to prove it).

In the next part, we have to make “a suitable choice of $g$“. Sometimes such questions are frightening, but not here. How are we going to choose a function $g$ that will turn the expression in front of us into something to which we can potentially apply the result we have just established? Well, we’re looking at the integral of a product of $f$ with something, so there’s not much choice but to choose $g$ to be that thing. Accordingly, we write

Let $g(x)=ne^{-nx}$ for each $x\in[0,1/\sqrt{n}]$.

Now we plug what we’ve got into the previous result, writing this.

By what we have just proved,

$\displaystyle \int_0^{1/\sqrt{n}}nf(x)e^{-nx}dx$

lies between

$\displaystyle m\int_0^{1/\sqrt{n}}ne^{-nx}dx$

and

$\displaystyle M\int_0^{1/\sqrt{n}}ne^{-nx}dx$

where $m$ and $M$ are the infimum and supremum of $f$ on the interval $[0,1/\sqrt{n}]$.

There I used the time-honoured technique of saying, “I want the minimum and maximum values of $f$ in that interval, but this is analysis so to be on the safe side I’ll call them the infimum and supremum.” As it happens, in this case it would have been fine to say minimum and maximum since we have a continuous function defined on a closed bounded interval. The slightly wordy “lies between” above is there merely because I wasn’t confident of displaying the full three-way inequality on one line — it’s not what I would have written in an exam.

What should we do next? Probably a good move would be to calculate the integral, since it’s clearly not a difficult one and the answer we are aiming for doesn’t involve an integral. So we write

$\displaystyle \int_0^{1/\sqrt{n}}ne^{-nx}dx=[-e^{-nx}]_0^{1/\sqrt{n}}=1-e^{-\sqrt{n}}$

It’s now fairly easy to see why the result is true: since $f$ is continuous, both $M$ and $m$ will be close to $f(0)$ once $n$ is large (and hence $1/\sqrt{n}$ is small. At the same time, $1-e^{-\sqrt{n}}$ is tending to 1. I think we can be pretty brief here.

Since $f$ is continuous, both $m$ and $M$ tend to $f(0)$ as $n\to\infty$. Since $1-e^{-\sqrt{n}}\to 1$, it follows that the upper and lower bounds obtained above both converge to $f(0)$, which proves the result.

We now come to the last part of the question. What is the change of variable that the examiner has in mind? It’s quite strange because the function we’re integrating is precisely the same. Still, we could try making a substitution that would change the range of integration from $[0,1]$ to $[0,1/\sqrt{n}]$. The obvious one is to set $y=x/\sqrt{n}$. I’ll try that.

Let $y=x/\sqrt{n}$. Then

$\displaystyle \int_0^1nf(x)e^{-nx}dx=\int_0^{1/\sqrt{n}}n^{3/2}f(y\sqrt{n})e^{-n^{3/2}y}dy$

I don’t see how that is supposed to help. I’ll come back to that question, but since I don’t really like these look-for-the-magic-key style answers, I think what I’d actually do in an exam is consider what the “or otherwise” possibilities are. For some reason, there are many questions where the method suggested is so overwhelmingly superior to anything else that the words “or otherwise” are almost a joke. It’s as though they’re saying, “If you don’t see the way you’re supposed to do this, then you’re going to be wasting the next half an hour and a lot of paper.” But this question is not like that. Given what we’ve already shown, it’s enough if we can show that the bit we’ve added on to the integral tends to zero. That is, we want to prove that

$\displaystyle \lim_{n\to\infty}\int_{1/\sqrt{n}}^1nf(x)e^{-nx}dx\to 0$

How does one show that an integral is small? One way is to calculate it exactly, obtain a formula, and prove that the resulting expression is small. That doesn’t work here, because $f$ is an arbitrary continuous function.

Another method is the very crude one of finding bounds for the integrand and multiplying them by the length of the interval. It’s surprising how often that works. (A slightly more complicated method that also works well is to divide the range of integration up into two parts, one quite wide where the function is very small, and one very narrow where the function may be rather bigger.)

What bounds do we know here? Well, we don’t know any explicit bound on $f$, but we do know that it is a continuous function defined on a closed bounded interval, so at least it is bounded. What about $ne^{-nx}$. That is a positive decreasing function, so on the interval in question its maximum value is $ne^{-\sqrt{n}}$. Ah, that looks extremely small when $n$ is large. And now we have the ingredients for a very short and completely rigorous proof, which goes like this.

Since $f$ is continuous on the closed bounded interval $[0,1]$, there exists $C$ such that $|f(x)|\leq C$ for every $x\in[0,1]$. Also, the function $ne^{-nx}$ is decreasing and positive, so on the interval $[1/\sqrt{n},1]$ it takes a maximum value of $ne^{-\sqrt{n}}$. Hence,

$\displaystyle |\int_{1/\sqrt{n}}^1nf(x)e^{-nx}dx|\leq Cne^{-\sqrt{n}}$

for every $n$. Since the right-hand side tends to 0, so does the left-hand side. The result follows by adding to the previous integral.

Further remarks.

That argument was easy enough that I’m tempted not to look for the magic substitution that solves the problem. But I’m curious to know what was intended, so I’ll press on.

Looking at where things went wrong before, the difficulty was that the range of integration didn’t relate in the right way to the constant inside the exponential function. So let’s try another important trick in mathematics: make your guess less precise until you know more about what it needs to do.

In this case, we don’t have to substitute $y=x/\sqrt{n}$. We could simply substitute $y=\lambda x$ and decide later what $\lambda$ suits us best. If we do that, then we find that

$\displaystyle \int_0^1nf(x)e^{-nx}dx=\int_0^\lambda (n/\lambda)f(y/\lambda)e^{-ny/\lambda}dy.$

To relate this to the first integral, we’d like to be able to write $n/\lambda$ as $m$ and $\lambda$ as $1/\sqrt{m}$. That tells us that $n/\lambda$ should equal $1/\lambda^2$, and therefore that $\lambda$ should be $1/n$. In other words, we substitute $y=x/n$. Doing that, we get the integral

$\displaystyle \int_0^{1/n}n^2f(ny)e^{-n^2y}dy$.

Now everything is wonderful apart from one thing: the fact that we’ve got $f(ny)$ instead of just $f(y)$. Can’t we just define $g(y)=f(ny)$? No, because then $g$ depends on $n$, which isn’t allowed in the first result.

I’m forced to admit defeat here: I cannot see what the examiner had in mind.

By the way, if you happen to be the person who set that question, let me make clear that although I have been somewhat critical of the way it was worded, I am also very much aware of how difficult it is to word questions in a way that makes it clear what is intended. I have often written what seemed to me like crystal clear questions that were given some interpretation I didn’t expect by a significant percentage of people.

Update. I have just been told by someone who has looked at the examiner’s model answer for this question what the examiner’s intended solution was. It was precisely the solution I attempted just now where I substituted $y=\lambda x$ and ended up choosing $\lambda=1/n$. There’s an important moral here: Tripos questions can sometimes contain mistakes. It’s very unfortunate when they do, but it’s also very hard to avoid. It’s a difficult judgment to make, but if you really really really are confident that what you have written is correct, despite the fact that it disagrees with what the question claims, then you should probably move quickly on to a new question, perhaps putting a little note saying “question wrong?”. If it is early on in the exam, you should also inform the invigilator, which will result in the examiner being contacted urgently. In this case it’s an even more difficult judgment to make, since there is the “or otherwise” option, so the question isn’t so much wrong as accidentally misleading.

To finish off, here’s an example of a fairly common style of misinterpretation, but this time the fault is with the candidate rather than the examiner. Consider the following question.

Let $f$ be a continuous function defined on the interval $[0,1]$. Show that $f$ is bounded.

It is not unknown for people to give answers of the following kind to that kind of question.

Let $f$ be the function $f(x)=x^2$. Then $f$ is continuous. Since $0\leq x^2\leq 1$ for every $x\in[0,1]$, it is bounded, as required.

If you think you might feel that sort of temptation, the following rule of thumb should help: if the question wants you to look at just one example of your choice, it will say so very explicitly; if it doesn’t say so explicitly, then it is asking you to look at a general object.

The words “any” and “arbitrary” make this confusion more likely. Suppose, for example, that the question says, “Let $f$ be an arbitrary continuous function defined on the interval $[0,1]$.” Is that an invitation to choose your own function? Well, it could just about be interpreted that way, but if there is the slightest room for doubt, then don’t interpret it that way. Similar remarks apply to “Let $f$ be any continuous function defined on the interval $[0,1]$.” In each case you are being asked to prove a result for all continuous functions and not just one.

### 10 Responses to “A look at a few Tripos questions IV”

1. Vania Mascioni Says:

Interesting to hear that even in “Cambridge English” there is a potential for confusion when using “any” or “arbitrary” in questions, and your final advice is surely sound! Way too often in my exams I have had students just pick their own example rather than carrying out the general argument I was asking for, and it’s very annoying to have to make the questions longer just to eliminate that possibility.

2. alja Says:

I always liked doing the first part of this question by coming up with an alternative definition of the integral of a continuous function in terms of antiderivatives (which is equivalent to the standard definition since all continuous functions have an antiderivative).

The next two parts are then immediate from the Mean-Value Theorem and the Cauchy Mean-Value Theorem.

I don’t think I got any marks, but it’s the most use I’ve gotten from the Cauchy Mean-Value Theorem so far.

3. Wednesday Highlights | Pseudo-Polymath Says:

[…] Tripos (whatever that is) entertainment continues. […]

4. woutervandoorn Says:

You seem to spend a lot of time thinking about what the examiner thinks about what is and what is not allowed. E.g. can I assume fact X, or use theorem Y, or do I have to be more explicit somewhere? Do you have any ideas on how to make an exam so that students don’t have to spend the majority of their time dealing with these meta-questions?

• gowers Says:

I’ve certainly written a lot about that, but that’s because it is something that has often worried people I’ve supervised. It certainly doesn’t reflect the amount of time I would recommend thinking about it in an exam: by and large I would hope that a bit of common sense and some sensibly worded questions would be enough.

So how does one word questions sensibly? I think there are certain conventions (that I have mentioned in the posts — things like that it’s OK to assume something from earlier in the course, or that there is a presumption that if you can use an earlier part to solve a later part then that is what the examiner wants), and that if there is any reasonable doubt, even taking into account those conventions, then the setter of the question should be explicit about what can and can’t be assumed — as Tripos examiners often are.

5. Richard Baron Says:

In response to the second part (Suppose that m, M are real numbers …), my first thought, as a non-mathematician, was to sketch some axes and vertical lines at arbitrarily chosen x = a, b, draw any old wiggly line for f(x) between those lines, then draw horizontal lines at y = m, M such that they do not cross the wiggly line, and point out that as you squeeze the horizontal lines inwards without crossing the wiggly line, you must preserve the desired inequalities. But I don’t suppose I would get any marks for that, even if mathematicians have in the past expressed approval of proof by picture. (There is an example in Littlewood’s Miscellany: a secondary source tells me it is on page 55).

So let me say something a bit more formal that is related to the pictorial approach, and that argues directly from the definition that is given of an integral of a continuous function.

1. Given any dissection, its lower sum will not be less than m(b – a) and will not exceed M(b – a). (I was going to rely on pictorial imagination to justify this, but I don’t need to because I am about to place no reliance on the claim.)

2. That is not enough, because the supremum of the lower sums won’t usually be any particular lower sum. You usually have to carry on dissecting more and more finely, without any end to that process.

3. But what we can say is that for any dissection of f(x) over [a, b], if we had matching dissections of g(x) = M and of h(x) = m, over [a, b], where “matching” means that the xi are the same, then the lower sums would comply with the inequalities we wish to prove. (I think I can have this on the basis of elementary arithmetic, without relying on pictorial imagination.)

4. So for any sequence of finer and finer dissections, we can have corresponding sequences of matching dissections of g(x) and of h(x), and at any point in the sequence, the inequalities would hold.

5. Am I now allowed to say that the inequalities would still hold when we moved on from particular dissections to the supremum? Intuition says yes, but rigour would doubtless demand a better justification, presumably based on the formal properties of the reals.

On the terminological point at the end, I do not see that there is any scope for doubt. Either of “Let f be an arbitrary function” and “Let f be any function” must require a proof of the result for all functions, not for a particular function that the candidate happens to like. (I conclude that on the basis of my grasp of English, not on the basis of any supposition as to the cruelty of examiners.)

• gowers Says:

In response to the last point, there is scope for doubt, but not scope for reasonable doubt. Some people sometimes think as follows: the question says, “Let $f$ be an arbitrary function”; the function $f(x)=x^2$ is pretty arbitrary; so I’ll go for that.

6. Steve Collyer Says:

This is a very nice series of articles. I wish you’d written them thirty years ago, though.

Re: the question of ambiguity: why not simply choose a wording like: “Prove that for all continuous functions defined on blah, X is true” ?

• gowers Says:

I agree: wording that is open to misinterpretation, even if it is an unreasonable misinterpretation, is not as good as wording that is completely unambiguous. But it can be hard to predict when an unreasonable misinterpretation will occur, so it’s just an unfortunate fact that wording of Tripos questions is not always optimized.

7. Steve Collyer Says: