A look at a few Tripos questions I

When I was a mathematics undergraduate, I became aware of a huge cultural difference between mathematicians and engineers. That sounds like the beginning of a joke you’ve heard twenty times already, but it isn’t. The difference was that when mathematicians were set questions, they were expected to work out how to solve them, and if they couldn’t do so then it was too bad — the best they could do about it was ask their supervisors. But engineers had model answers for everything, available with the latest technology, which in those days was microfiche. In case you have no idea what I’m talking about, the answers were reduced in size by a factor of about five in each direction and printed on to some kind of transparent plastic that you could look at through a magnifying machine. There were a couple of the machines in our college library, and they were nearly always in use.

Model answers have always seemed to me to be a bad idea in mathematics, because it is hard to learn how to think for yourself when you are given the answers to all the problems you tackle. So it might seem a bit odd that in this post I’m going to attempt to help people preparing for Part IA of the Cambridge Mathematical Tripos by providing some model answers.

However, my aim is not just to give the answers. Rather, I want to explain in as much detail as I can (without getting tedious) how I come up with the answers. I haven’t yet started thinking about the questions, so I’m not sure how that is going to work out, but what I’m hoping is that if I describe my thoughts as I tackle the questions, then I’ll transmit some general messages about exam technique and dealing with the rather characteristic kind of question that gets set in Cambridge. You will of course benefit far more from this if you do the questions first (or try hard to do them) before reading what I have to say about them.

To minimize the chances of interfering with revision supervisions, I am taking my questions from quite some time ago — 2003, to be precise. Also, I’m concentrating on only two subjects: Analysis I and Numbers and Sets. I hope it helps at least some people with their exam preparation.

In this post I’ll look at some Section-I questions, but I’ll get on to the harder questions in due course. The thing to remember with Section I is that the questions are not supposed to involve serious thought. So if you find yourself thinking seriously, then there are three possibilities, of which I hope the third is the most likely:

(i) you haven’t learnt something that you should have learnt;

(ii) the examiners misjudged the difficulty of the question;

(iii) you are going about things the wrong way.

The real point I’m making here is that the fact that the question is supposed to be straightforward can often be a big clue, since it narrows down the kind of answer you should be looking for.

I’ll put the bits of the questions in italics and not all at the beginning.

3B. Define what it means for a function of a real variable to be differentable at $x\in\mathbb{R}$ .

Absolutely no excuse for not being able to do this. The answer I would give is this.

A function $f:\mathbb{R}\to\mathbb{R}$ is differentiable at $x$ if $\frac{f(x+h)-f(x)}h$ tends to a limit as $h\to 0$ .

A couple of remarks about this. Note first that the question didn’t give a name to the function, so the first thing I did was remedy that by calling it $f$ . That was obviously the right thing to do here (nobody is going to write phrases like “the number you get by dividing the difference between the value the function takes at $x+h$ and the value it takes at $x$ by $h$ “) but in other questions it is easier to forget.

Secondly, I had various choices to make here. Should I write what I wrote, or should I have written something more like “there exists $l$ such that $\frac{f(x+h)-f(x)}h\to l$ as $h\to 0$ “? And should I have explained what it means to converge to a limit?

In general, the answer to questions of the last type is that if the tripos question is about one part of the course, then you are free to use facts from earlier on in the course but not facts from later. Another principle to keep in mind is that if the fact or definition you assume leaves you with almost no work to do, then you may have assumed too much. (I say “may have” here because if you’re doing an early part of the question, it may be clear that it’s just a quick preliminary to be got out of the way.)

A subtlety here is that it matters that $h$ is not allowed to be zero. That is, the statement that $\frac{f(x+h)-f(x)}h\to l$ as $h\to 0$ means that for every $\epsilon>0$ there exists $\delta>0$ such that for every $h$ such that $|h|<\delta$ and $h$ is not zero we have $|\frac{f(x+h)-f(x)}h-l|<\epsilon$ .

Should you bother to explain this to the examiner? I would say no in this instance, since the definition of limits for functions does specify that we don’t look at what happens at the actual value. And limits of functions are an earlier part of the course than derivatives. So even if other people do explain about the precise meaning of the limit, the examiner cannot reasonably penalize you for not doing so. And if that’s the case, then you’ve got better things to do than gaining an extra zero marks.

Prove that if a function is differentiable at $x\in\mathbb{R}$ , then it is continuous there.

This should be filed away in your mind as “easy result”. Ideally, you’ll have practised writing out the proof quickly, but if you haven’t, you just need to keep your head and remember that as long as you are sensible, then nothing will go badly wrong.

It’s worth stopping to think whether there is some simple high-level proof that doesn’t require you to mess about with the definitions of differentiability and continuity. But there doesn’t seem to be. Another thing that’s worth thinking about (always always always) is which definitions you want to use. The obvious definition of continuity at $x$ is

$\forall\epsilon>0\ \exists\delta>0\ \forall y\ |x-y|<\delta\implies|f(x)-f(y)|<\epsilon.$

So you should write that down … shouldn’t you? No — just allow yourself a few seconds to think, because your expected gain of time is positive. Is there any other way of defining continuity? Yes, there are several ways. Is there one that’s likely to be more suitable than the epsilon-delta definition? Yes there is. The definition of differentiability was expressed in terms of limits of functions, so if we use a limits-of-functions definition of continuity, it is much more likely to dovetail neatly with the definition we’ve already written down. Accordingly, we note to ourselves that our aim is to prove that $f(y)\to f(x)$ as $y\to x$ . (I’m assuming you’ve been given this as an equivalent way of saying that $f$ is continuous. If you don’t know this alternative definition, then obviously you can’t use it, but the general principle of thinking before you write out definitions is still very important.) If it helps to write this down, then we can, but it may be better just to hold it in our heads.

Now let’s see whether we can manage without writing down any epsilons and deltas. The reason for this is not that there’s anything too bad about epsilon-delta proofs. It’s just that they take a little time, and time is what you are trying to save.

What do we know? We know that there is some $l$ such that

$\displaystyle \frac{f(x+h)-f(x)}h\to l$

as $h\to 0$ . How can we relate this to what we are trying to prove? Well clearly we can think of $x+h$ as our $y$ . So what we want to prove can be reformulated as the statement that $f(x+h)\to f(x)$ as $h\to 0$ .

How can we get that? Well, we want to have $f(x+h)-f(x)$ on the left-hand side, so it makes sense to multiply by $h$ . But is that allowed? Here’s where another very important principle comes in: use those little facts you’ve been taught! In this case, we know that

$\displaystyle \frac{f(x+h)-f(x)}h\to l$

and also that $h\to 0$ . But … little fact coming up … the limit of a product is the product of the limits (quotable because it belongs to an earlier part of the course), from which it follows that

$f(x+h)-f(x)\to 0.l=0$ .

And from that it follows that $f(x+h)\to f(x)$ (by adding $f(x)$ to both sides).

I spent a long time explaining all that, but what I actually write is very short indeed. It goes like this.

Since

$\displaystyle \frac{f(x+h)-f(x)}h\to l$

as $h\to 0$ , it follows, from the fact that a product of limits is the limit of the product, that

$f(x+h)-f(x)\to 0$

as $h\to 0$ , and therefore that $f(y)\to f(x)$ as $y\to x$ . Therefore, $f$ is continuous at $x$ .

You can probably even get away with not saying “from the fact that a product of limits is the limit of the product”, since it will be clear to the examiner that this is the justification you have in mind. But you have to be a little careful: if in doubt, put in just enough to make it absolutely clear that you know how to justify what you’ve written. This is particularly important if you occasionally get things wrong, at which point the examiner will lose faith in you and start removing marks for insufficient explanation.

Show directly from the definition that the function $f$ with $f(x) = x^2\sin(1/x)$ when $x\ne 0$ and $f(0)=0$ is differentiable at 0 with derivative 0.

Here we can be very concise indeed. Plugging $x=0$ into the definition, we see that the question is asking us to show that $f(h)/h\to 0$ as $h\to 0$ . So we write this.

If $h\ne 0$ , then $f(h)/h=h\sin(1/h)$ . Since $|\sin x|\leq 1$ for every $x$ , it follows that $|f(h)/h|\leq|h|$ , and therefore that $f(h)/h\to 0$ as $h\to 0$ .

Show that the derivative $f'(x)$ is not continuous at $0$ .

Here we have a slight dilemma. In most analysis courses, a formal treatment of trigonometric functions comes after the definition of differentiability. So are we allowed to assume things like that the derivative of $\sin(x)$ is $\cos(x)$ ? Can we assume the chain rule?

The only suggestion I can give here is to bear in mind facts like these.

(i) This is a Section I question.

(ii) Deriving elementary properties of trigonometric functions would take quite a bit of time and effort, as would proving the chain rule.

(iii) The question is not about trigonometric functions or the chain rule, and the examiner has not given any sign of expecting you to prove those facts.

Basically, it’s safe to differentiate $x^2\sin(1/x)$ as though you were still at school. So here goes. First, let’s compute the derivative. So we write this.

When $x\ne 0$ , the derivative of $x^2\sin(1/x)$ is $2x\sin(1/x)-\cos(1/x)$ .

Now how are we going to show that the function that equals that when $x\ne 0$ and $0$ when $x=0$ is not continuous? Again, choose the right definition! It will save you time. The rough reason this function doesn’t converge is that it oscillates faster and faster between approximately -1 and approximately 1 as $x\to 0$ . The main definitions we could use are the epsilon-delta definition, the limit definition, and the sequences definition. The nice thing about the sequences definition is that it lets us concentrate on the tops and bottoms of these oscillations (if we sort of ignore the first term) in a nice way. Actually, if we’re feeling very clever, we might observe that the function $2x\sin(1/x)\to 0$ as $x\to 0$ , so it is sufficient to prove that the function $-\cos(1/x)$ is not continuous at 0. Why? Because if $2x\sin(1/x)$ is continuous and $2x\sin(1/x)-\cos(1/x)$ is also continuous, then so is $-\cos(1/x)$ . (In general, a sum of a continuous function and a discontinuous function is discontinuous, just as the sum of a rational number and an irrational number is irrational. The proofs are the same.)

How do we prove that $\cos(1/x)$ doesn’t tend to 0? It’s enough to find a sequence $a_n\to 0$ such that $\cos(1/a_n)$ does not tend to 0. But that’s easy: just choose $a_n$ such that $1/a_n$ runs through multiples of $2\pi$ , where $\cos$ takes the value 1. So the eventual answer goes like this.

Since $x\sin(1/x)\to 0$ as $x\to 0$ it is enough to prove that $-\cos(1/x)\not\to 0$ . For each $n\in\mathbb{N}$ let $a_n=1/2\pi n$ . Then $a_n\to 0$ , but $-\cos(1/a_n)=-1$ for every $n$ . This proves the result.

If we hadn’t thought of the clever trick for removing the first term, we might have written this.

For each $n\in\mathbb{N}$ let $a_n=1/2\pi n$ . Then $a_n\to 0$ , but $-\cos(1/a_n)=-1$ for every $n$ . Also, $|a_n\sin(1/a_n)|\leq 1/2\pi\leq 1/2$ , so $f'(a_n)\leq -1/2$ for every $n$ . Therefore, $f'(a_n)\not\to 0$ as $n\to\infty$ , so $f'$ is not continuous at 0.

4C Explain what is meant by the radius of convergence of a power series.

OK, the decision here is how much you should explain. Here are a couple of answers one might give.

(i) For any power series there exists $R\in[0,\infty]$ such that the series converges at every $z$ with $|z|<R$ and diverges at every $z$ with $|z|>R$ . This $R$ is called the radius of convergence of the power series.

(ii) The radius of convergence of a power series $\sum_na_nz^n$ is the supremum of $r$ such that the series converges for some $z$ with $|z|=r$ .

The second of these is correct as a formal definition, but it doesn’t include the information that if the power series converges for some $z$ with $|z|=r$ then it converges for all $w$ with $|w|<r$ . So it doesn’t really explain where circles and radii come in. On the other hand, it is quite concise. If you go for the first definition, you might wonder whether you are expected to justify the fact that some $R$ with that property exists. But the wording of the question clearly indicates that you are not: you are asked to explain what it means, rather than proving anything, so simply stating as a known fact that there is such an $R$ is OK.

Find the radius of convergence $R$ of each of the following power series:

(i) $\displaystyle \sum_{n=1}^\infty n^{-2}z^n$ ,

(ii) $\displaystyle \sum_{n=1}^\infty(n+1/2^n)z^n$ .

In each case, determine whether the series converges on the circle $|z|=R$ .

OK, let’s have an informal think about part (i). Since $n^{-2}\to 0$ , and, even better, $\sum_{n=1}^\infty n^{-2}$ converges, we can be pretty confident that the series will converge when $|z|\leq 1$ . (How exactly to justify this we’ll come back to in a moment.) What about if $|z| > 1$ ? Well now $|z|^n\to\infty$ exponentially fast, which ought to be enough to cancel out the shrinking of $n^{-2}$ . So it looks very much as though the radius of convergence is 1.

Now let’s think how to prove it as quickly and concisely as we can. We want to avoid arguing from first principles (because we’re in a hurry), so let’s try to use one of the various tests we know. For power series, the ratio test is often good, since the ratios of successive powers of $z$ are constant, and therefore nice. OK, I’ll dive in.

The ratio $(n+1)^2z^{n+1}/n^2z^n$ equals $(1+1/n)^2z$ . This converges to $z$ as $n\to\infty$ . Therefore, the series converges when $|z|<1$ and diverges when $|z| > 1$ , by the ratio test, which shows that $R=1$ .

That still leaves what happens when $|z| = 1$ . It looks as though the series converges, but how do we prove that? Again, use facts that you’ve been told. The rough reason it converges is that the terms go to zero like $n^{-2}$ and $\sum_{n=1}^\infty n^{-2}$ converges. What are we doing here? We’re considering another series. That suggests the comparison test. But that’s about positive series and here we’ve got complex numbers that needn’t be positive. But we could look at their moduli. But the modulus of $|n^{-2}z^n|$ is just $n^{-2}$ when $|z|=1$ . But that’s great — it shows that the sum of the moduli converges. And that property has a name: the series converges absolutely. And we’ve been taught that a series that converges absolutely must converge.

OK, enough thinking: let’s write.

When $|z| = 1$ , $\sum_{n=1}^\infty |n^{-2}z^n|=\sum_{n=1}^\infty n^{-2}$ , which converges. Thus, the series converges absolutely, so it converges.

Note how little I had to write. Note also that I was happy to assume that $\sum_{n=1}^\infty n^{-2}$ converges and that absolute convergence implies convergence. That is partly for “proved earlier in the course” reasons, but partly for a reason that I haven’t yet got round to mentioning, namely this.

It’s OK, and even positively good, to assume as much theory as you like when you are trying to prove something about an example.

Let’s do the second part now. What can we say about the series $\sum_{n=1}^\infty(n+1/2^n)z^n$ ? Well, the coefficients have a big part, $n$ , and a tiny part, $1/2^n$ . So we’d expect the big part to be the one that affects convergence. How about going for the ratio test again? Here’s the first sentence I might write.

The ratio of successive terms is $(n+1+1/2^{n+1})z/(n+1/2^n)$ .

There’s a trick here, and if you haven’t already internalized it, then I recommend doing so now. We want to determine the limit of a fraction. That’s difficult if the top and the bottom both tend to infinity. It’s also difficult if the top and the bottom both tend to zero. So divide through so that at least one of them tends to a finite non-zero limit. The obvious nice thing to divide by here is $n$ . So I’ll extend that first sentence as follows.

The ratio of successive terms is $(n+1+1/2^{n+1})z/(n+1/2^n)$ , which equals $(1+1/n+1/n2^{n+1})z/(1+1/n2^n)$ .

Once we’ve done that, we’ve got a fairly unpleasant looking fraction, but we can rapidly make it nicer by using the second component of this general approach, which is to use limit theorems. Here, there are a number of quantities that clearly tend to 0 (and that “clearly” means that the proofs that they tend to 0 are very easy). So we can say this.

The numerator tends to $z$ and the denominator to $1$ . Therefore, the fraction tends to $z$ .

I’ve used the limit theorems for sums, products and quotients without explictly saying that that was what I was doing, since it’s obvious what I was using and I want to save time.

Next sentence.

Therefore, the series converges when $|z|<1$ and diverges when $|z| > 1$ , so $R=1$ .

What about when $R=1$ ? Well, the terms seem to be getting bigger and bigger. In fact, they are getting bigger and bigger. A minimal requirement for a series to converge (but not a sufficient one) is that the individual terms should tend to zero. So we can write this.

When $|z| = 1$ , $|(n+1/2^n)z^n|=n+1/2^n\to\infty$ , so the series does not converge.

Note how little you have to write if you use the right bits of theory. You can use this to your double advantage: in general, trying to think of the right bit of theory that will make the question come out quickly is easier (if you’ve had a bit of practice) than trying to work out the details of a clumsier proof from first principles, and the more theoretical proofs are much quicker to write out.

Since radii of convergence come up year after year, and since theoretical questions about them tend to be answered rather badly (as I know from experience as a IA examiner a few years ago), let me end by discussing how to prove that the radius of convergence exists. Suppose, then, that you had the following question as part of a tripos question. (Who knows? It might even come up this year. I’m free to say this, since I have nothing to do with the exams this year. I note that last year the short question just asked you to calculate the radii of convergence of two series, and the year before asked for the definition but not a proof that it exists.)

Suppose then that you are explicitly asked to prove that there is some $R$ such that the power series converges when $|z|<R$ and diverges when $|z|>R$ . You probably remember a number of the ingredients: a funny geometric series that comes up, the appearance of the words “converges absolutely”, and so on. But a lot of people seem to end up mixing the ingredients together in a fairly haphazard way.

A great way to start is to focus on the following statement. It contains the technical part with the geometric progression, and once you’ve done it, a bit of straightforward theory will finish things off. The statement is this: if $\sum_{n=1}^\infty a_nz^n$ converges, and $|w|<|z|$ , then $\sum_{n=1}^\infty a_nw^n$ converges.

If you’ve remembered that somewhere we get a geometric progression with common ratio $|w/z|$ , then it’s easy to work out what to do here. We somehow want to compare the two series. So we write this:

$\displaystyle \sum_{n=1}^\infty a_nw^n=\sum_{n=1}^\infty (w/z)^na_nz^n$

That’s supposed to make it obvious that the series converges. But why is that? The rough reason is that the series $\sum_{n=1}^\infty (w/z)^n$ is a geometric progression with ratio of modulus less than 1, so converges, and the other part of the term, $a_nz^n$ , doesn’t mess things up. But it’s difficult to make sure of all that unless we take the modulus of everything in sight. So instead of that first line, we write,

$\displaystyle \sum_{n=1}^\infty |a_nw^n|=\sum_{n=1}^\infty |w/z|^n|a_n||z^n|$

Now what can we say about $|a_n||z^n|$ that will make it clear that it doesn’t cause trouble? Well, we know that $\sum_{n=1}^\infty a_nz^n$ converges, from which we can deduce that at the very least the terms $a_nz^n$ are bounded in modulus. So there is some $C$ such that $|a_n||z^n|$ is always at most $C$ . So a well-organized proof will now go like this.

Since $\sum_{n=1}^\infty a_nz^n$ converges, there is some $C$ such that $|a_n||z^n|\leq C$ for every $n$ . But then the series

$\displaystyle \sum_{n=1}^\infty |a_nw^n|=\sum_{n=1}^\infty |w/z|^n|a_n||z^n|$

converges by the comparison test, since the series $\sum_{n=1}^\infty C|w/z|^n$ converges (because it is a geometric progression with positive ratio less than 1).

We’re sort of done now, but not quite, because we actually wanted to show that $\sum_{n=1}^\infty a_nw^n$ converges. But the statement we have is precisely the statement that this series converges absolutely. (A series converges absolutely if the series you get by replacing each term by its modulus converges.) So all we have to do is tag the following line on to the end of what we’ve written.

This shows that the series $\sum_{n=1}^\infty a_nw^n$ converges absolutely, and therefore that it converges.

Finally, what about the radius of convergence? It’s at this point that we just write down the definition.

Let $R$ be the sup of all $|z|$ such that $\sum_{n=1}^\infty a_nz^n$ converges. Then if $|w|<R$ , we can find $z$ with $|w|<|z|$ such that $\sum_na_nz^n$ converges (or else there would be a smaller upper bound). By our earlier result, it follows that $\sum_{n=1}^\infty a_nw^n$ converges. And if $|w|>R$ , then $\sum_{n=1}^\infty a_nw^n$ must diverge, since otherwise $R$ wouldn’t be an upper bound at all.

This entry was posted on April 28, 2012 at 12:28 am and is filed under Cambridge teaching, IA Analysis. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

20 Responses to “A look at a few Tripos questions I”

typo service Says:
April 28, 2012 at 9:07 am | Reply
There are two “|z| = 1″s where I believe a “|z| < R" is intended: one in the paragraph (i) beginning the discussion of the radius of convergence, and significantly later in the paragraph beginning "Suppose, then, that you are explicitly asked to prove…"
typo service Says:
April 28, 2012 at 9:09 am | Reply
Bizarrely, the “alt text” for the images of the rendered TeX read correctly, but the images themselves do not. So perhaps this is an issue with my own browser’s cache.
- gowers Says:
  April 28, 2012 at 9:19 am
  Some weird things were going on with me too. I thought I had carelessly typed $|z| = R$ when I meant $|z| < 1$ and it turned out that I had typed the right thing and it was coming out wrong — but not nonsensically wrong. It seems to be a very strange WordPress bug. I’ve found a fix for it, which is to insert a few spaces where I would ordinarily not need to.
A look at a few Tripos questions II « Gowers's Weblog Says:
April 28, 2012 at 11:11 am | Reply
[…] Gowers's Weblog Mathematics related discussions « A look at a few Tripos questions I […]
Piero Giacomelli (@pierogiacomelli) Says:
April 28, 2012 at 10:24 pm | Reply
Prove that if a function is differentiable at , then it is continuous there. This one is something that I have always struggle with. Considering you previous definition the following function is differentiable in zero but it is obviously not continuos in it

f(x) = 1 for all x >= 0
f(x) = 0 otherwise
- gowers Says:
  April 29, 2012 at 10:53 am
  You say, “Considering you previous definition the following function is differentiable in zero.” However, that is not true. (Hint: $h$ can be negative.)
espressonator Says:
April 29, 2012 at 7:35 pm | Reply
Posted in my blog are my comments about Professor Gowers’ Model answers to the question (paraphrased here and in my post) “Prove that differentiability implies continuity.” To access my comments go to the following new blog: http://espressonator.wordpress.com/
Piero Giacomelli (@pierogiacomelli) Says:
April 29, 2012 at 10:33 pm | Reply
You say, “Considering you previous definition the following function is differentiable in zero.” However, that is not true. (Hint: can be negative.).

In this case you mean that if h \in \mathbb{R}, then f is diffrentialible in a point x if the limit for h->0^+ or h->0^- is finite, but again in the function above both limits are still finite and again the function is not continuos .

Sorry I meant it was always one of my concern about limit’s definition and for me the exception still remains.
Do not ment to be irritant or boring
- gowers Says:
  April 29, 2012 at 11:39 pm
  I don’t quite understand what you are saying, but here is a proof that the function $f(x)=1$ when $x\geq 0$ and 0 otherwise is not differentiable at 0. Consider the quantity $(f(h)-f(0))/h$ . When $h\geq 0$ it is 0, but when $h<0$ it is $-1/h$ , which tends to infinity. So your statement that both limits are finite is simply false.
  
  A small additional remark is that the limits from above and below have to be equal.
Monday Highlights | Pseudo-Polymath Says:
April 30, 2012 at 4:11 pm | Reply
[…] blog series is a lot of fun. Here’s part 1, part2, and part […]
Stones Cry Out - If they keep silent… » Things Heard: e219v1 Says:
April 30, 2012 at 4:12 pm | Reply
[…] blog series is a lot of fun. Here’s part 1, part2, and part […]
How to tackle an exam question… | Degree of Freedom Says:
May 1, 2012 at 10:15 am | Reply
[…] my attention to a sequence of posts on Tim Gowers’s excellent blog, in which Prof. Gowers lets us into his thought processes as he tackles Cambridge maths exam questions. Well worth a read, if only as an antidote to the habit of reading every exam question as a request […]
A look at a few Tripos questions IV « Gowers's Weblog Says:
May 2, 2012 at 3:36 pm | Reply
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torridtornado Says:
May 3, 2012 at 7:57 pm | Reply
Dear Mr Gowers,I am sure your discussion reaches out to a much wider audience than your own countrymen.We will be obliged if you, as a professional mathematician can post something like this at intervals.Thank you!
Antony Says:
July 13, 2012 at 4:01 pm | Reply
I would like to add my voice to those who have said this series of questions has been invaluable. I am an OU student, so I don’t directly benefit from discussing Cambridge exam preparation, but there is an awful lot to learn and try to follow in the Mathematics and the arguments you lay out. I am sure your time is in short supply, but if you ever do a series like this again I will happily read along and rest assured will be very, very grateful!
PL Says:
December 5, 2012 at 4:41 am | Reply
Firstly, I’d like to apologize for not knowing LaTeX.

In the example of finding the radius of convergence, in part i, because the exponent of n is -2, wouldn’t the ratio test of the (n+1)st term over the nth term result in z(n^2)/[(n+1)^2] rather than x[(n+1)^2]/(n^2)? I’m not sure if I’m missing something or if this was just a typo. Thanks for taking the time to read this:)
Analysis I: Lecture 14 « Theorem of the week Says:
February 18, 2013 at 12:31 pm | Reply
[…] Gowers wrote some blog posts about past Tripos questions. At the end of one of those posts, he wrote about what is for us Lemma […]
Cambridge Math Tripos Sample Questions | Math Online Tom Circle Says:
September 2, 2013 at 5:54 am | Reply
[…] https://gowers.wordpress.com/2012/04/28/a-look-at-a-few-tripos-questions-i/ […]
Tim Gowers (Fields Medalist) explains How to Answer Cambridge Tripos questions | Delightful & Distinctive COLRS Says:
July 2, 2016 at 7:13 pm | Reply
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