3. Homework feedback#
Here is some general feedback comments about each piece of homework for MAS2004/9. Individual feedback can be viewed on Crowdmark — let me know if you have any difficulty finding it.
Responding to individual feedback is the fastest way to improve as a mathematician. Hopefully your feedback will feel useful and make sense to you, but if it ever doesn’t, please get in touch and I am more than happy to look at it with you.
Feedback for Homework 5#
Problems: 41, 47, 4 from ESQs — see Blackboard (MAS2004, MAS2009) ; Solutions: 41, 47, 4 from ESQs — see Blackboard.
Problem 41#
This question led you through a proof that the function \(f:\mathbb{R}\to \mathbb{R}\) given by \(f(x) = \begin{cases} x\sin\left(\frac{1}{x}\right) & \text{if }x \neq 0,\\ 0 & \text{if }x = 0.\end{cases}\) is differentiable on the whole of \(\mathbb{R}\).
Part (i).
A common error with this part was to argue that because you can calculate a formula for \(f'(x)\) at every \(x\neq 0\) using the chain and product rules, \(f\) must be differentiable on \(\mathbb{R}\setminus\{0\}\). In fact, the logic should flow as follows:
Note that \(1/x\), \(\sin(x)\) and \(x\) are all differentiable at every \(x\neq 0\) by standard derivative facts.
Therefore, by the chain & product rules (Theorems 4.3(ii) and 4.4 in the notes) \(f\) is differentiable at every \(x\neq 0\).
If you want, you can also write down a formula for \(f'(x)\) when \(x\neq 0\). But this doesn’t directly contribute to the proof that \(f\) is differentiable on \(\mathbb{R}\setminus\{0\}\).
This is an important distinction as sometimes you will get a formula for \(f\) that looks like it has a standard derivative. For example, even in this question, we have \(f(x)=0\) when \(x=0\), from which you could (incorrectly) conclude that f’(x)=0 when x=0. But \(f\) is not differentiable at \(x=0\), for reasons explored in part (ii).
Another place this kind of error often creeps in is piecewise-defined functions. For example the modulus function has domain \(\mathbb{R}\) and rule
Applying standard derivatives to each formula might lead you to think this function has derivative \(1\) for \(x\geq 0\) and \(-1\) for\( x<0\). But of course, we know that the modulus function is NOT differentiable at \(0\) (too pointy).
Part (ii).
Most people had the underlying idea for how to solve this problem, but there were a couple of recurring issues, as follows:
Try to avoid writing \(f'(0)\) or \(\displaystyle\lim_{h\rightarrow 0}(...)\) unless you know these limits exist. In 41(ii), you actually know they don’t exist, so you definitely shouldn’t write them out this way. This is similar to the principle that you should avoid writing \(\frac{1}{0}\) or \(\ln(0)\) — these expressions aren’t mathematically meaningful.
Instead: manipulate \(\frac{f(h)-f(0)}{h}\) on its own, without the \(\lim\)’s, and then show that the limit doesn’t exist as \(h\rightarrow 0\).
For a rigorous proof that \(\sin(1/h)\) does not converge as \(h\rightarrow 0\), you should quote (or write out again) the solution to Problem 9.
Problem 47#
Part (i).
This was generally answered well. :-)
Part (ii).
A few people misread the “and” in the statement of the question, which has quite a big impact on the answer. In fact, the statement is false, and the quickest way to prove it is by counter-example (e.g. \(f:[-1,1]\to\mathbb{R}\); \(f(x)=x^2\) does not attain its maximum value inside the open interval \((-1,1)\), but is differentiable, with \(f(-1)=f(1)\).).
While the extreme value theorem (Theorem 3.5) guaruntees maximum and mininimum values of \(f\) are attained in \([a,b]\), they may lie at the end points and not the interior \((a,b)\). But, since \(f(a)=f(b)\), either \(f\) is a constant function, or max and min cannot both be at the end points. In either case, \(f\) has a maximum or minimum value attained inside \((a,b)\).
Some people tried to use Rolle’s theorem for 47(ii), which requires some care. Remember, finding a point \(c\in(a,b)\) such that \(f'(c)=0\) does not guaruntee \((c,f(c))\) is a turning point — it may be a point of inflection. Even if it is a turning point, this only tells us about a local behaviour of \(f\), and the question asked about global extreme values. If you want to use Rolle’s theorem at all, it’s its proof that is the most use here — have another read here to see what I mean.
Q4 from ESQs#
In 4(i), most people were able to show that the left and right limits exist at \(x=1\) and are equal, but quite a few people forgot to also check they equal \(f(1)\). When checking continuity using limits, remember it isn’t enough to show the limit exists — you have to show it exists and is equal to the value of the function at that point.
In 4(ii), a concerning number of people wrote something like “\(f'(x)=\begin{cases} 2 & x\leq 1 \\ 2x & x>1 \end{cases}\). This doesn’t work. Differentiability requires a 2-sided limit to exist, and the formula for \(f(x)\) is different depending on whether \(x\geq 1\) or \(x<1\). So to decide if \(f\) is differentiable at \(x=1\), you need consider convergence from both sides of \(x=1\) — using the \(x\leq 1\) formula on its own is not enough.
Some people made the opposite mistake: they considered behaviour at \(x=1\) but forgot consider differentiability/continuity when \(x\neq 1\). This can often be brief — e.g. “Note the restrictions \(f|_{(-\infty,1)}\) and \(f|_{(1,\infty)}\) are both polynomial functions, so are continuous/differentiable.” is plenty.
Feedback for Homework 3#
Problems: 18, 25 ; Solutions: 18, 25
This homework was generally answered well. I’m particularly pleased to see proofs and mathematical presentation improving, so well done for your hard work there.
Remember to take the time to carefully review the feedback your work has received on Crowdmark, and seek help interpreting or applying feedback, if anything is unclear.
General presentation remarks#
As I say, presentation is definitely getting better, which is great! Where there were issues, it’s largely the same comments as previous times:
It is essential to use words/short sentences to express a mathematical argument, not just equations.
From another marker:
Arguments which are just mathematical symbols are difficult to read, often unclear and ambiguous, and often betray a failure to properly understand what you are writing. (This will also possibly lead to lower marks in an exam.)
Always introduce variables/functions before referring to them in a proof.
If you introduce a function, remember to say what the domain is, as well as the rule. Different domains mean different functions, even when the rule is the same. See below for more on this.
If you use a theorem, check all the conditions explicitly (e.g. continuity of \(g\) in Q25).
If you are still getting comments on your work about presentation, it is probably worth your while to come to office hour and we can look over things together. I won’t judge — these things are just easier to explain in person.
Exam-related reminder: You don’t really get marks for equations in analysis, only mathematical arguments. So it will benefit you materially to seek help with this if you need it.
Problem 18#
This question was generally answered well. The most frequent comments are below.
Functions, domains and continuity.
The following comment comes from another marker, but I noticed it too:
The students have a tendency to discuss “the value of \(f\) at \(x\)” when \(x\) is not part of the domain but is a limit point. They also tend to call this point a “discontinuity of \(f\)” to rewrite a function with the same name but now \(x\) is part of the domain and they prove that it is continuous. Basically there seems to be confusion about the fact that the continuous extension of \(f\) is a different function than \(f\) with a bigger domain. Even when they seem to understand what is happening and their reasoning is right.
I saw this most often in 18(ii), where quite a few people fell into the trap of claiming \(f\) can be defined on \(A=\mathbb{R}\setminus\{-2\}\), because of the simplification \(f(x)=\frac{x^3-8}{x^2-4}=\frac{x^2+2x+4}{x+2}\). However, this simplification is only valid on \(\mathbb{R}\setminus\{\pm 2\}\), and this needs to be the set \(A\). We have to use the formula for \(f\) that we are given, and so \(f(2)\) is not meaningful.
Unfortunately, if you made this mistake, the rest of 18 generally didn’t work out. The progression of ideas is: \(f\) has domain \(A:=\mathbb{R}\setminus\{\pm 2\}\), has a continuous extension to \(\mathbb{R}\setminus\{-2\}\), but cannot be extended to any continuous function with domain \(\mathbb{R}\). In particular, \(f\) does not have a discontinuity at \(-2\), because \(-2\) does not belong to the domain of \(f\).
See Section 2.1 in the notes for more revision of functions, or click here for more on continuous extensions.
Please give justification for limits.
This can be brief, e.g. “by continuity” or “by algebra of limits” (as appropriate).
In particular, it isn’t necessary to use \((\varepsilon-\delta)\) arguments every time now, as we have proven so many functions are continuous, and have results like Theorem 3.2 to combine them.
On the other hand, quite a few wrote something like “\(\lim_{x\rightarrow-2}(...) = \frac{4}{0}\)”. This is not mathematically meaningful, especially in an analysis course. You need to show that the limit does not exist.
L’Hôpital’s rule
I’m afraid we can’t use it at this stage, as we haven’t proved it. But you shouldn’t need it, either — for example, in Q18(i), you can simplify \(f(x)=\frac{(1+x)^2-1}{x}=x+2\) and then send \(x\) to zero (using continuity of \(x+2\)). No need for L’Hôpital.
More generally: unless explicitly stated otherwise, you are welcome to use any result that has been proven rigorously in this module or previously, provided you indicate clearly that this is what you are doing. Mathematical rigour is like building a house. We can only put bricks (proofs) on top of bricks already laid underneath.
Problem 25#
Most people worked out that they needed to put \(g:[a,b]\to\mathbb{R}\), \(g(x)=f(x)-x\), but quite a few forgot to say what the domain of \(g\) is. This is important, since Proposition 3.1 only applies to functions with certain kinds of domain. Once you have what \(g\) needs to be, the solution is a straightforward application of Proposition 3.1 — just make sure you check all the conditions hold.
This question was quite all or nothing, in that there was a trick to spot before you could really get anywhere. The good news is, there are only so many of these analysis “tricks” you need to know. And this is one of them! Similar things come up from time to time — see for example question 24, and questions 3 and 5 in the extra analysis exam-style questions (this last one requires the mean value theorem, which at time of writing (2/5/25) we have not yet covered in lectures).
If you didn’t get the trick for this question, have another go now you know what function \(g\) to pick. The solutions are also available for when you are ready to see the full argument and check your answers.
Feedback for Homework 2#
Problems: 8, 9, 19, 22 ; Solutions: 8, 9, 19, 22
Thank you for your hard work completing this rather long set of exercises.
Remember to take the time to carefully review the feedback your work has received on Crowdmark, and attend office hours if you would like help interpreting or following the advice you are given.
General presentation remarks#
It’s fine to write out definitions, but be clear that’s what you are doing by clearly delineating it from your proofs. If it is an \(\varepsilon-\delta\) definition that you are trying to prove is true, try to convert it into a goal (e.g. “Given \(\varepsilon>0\), we need to find \(\delta>0\) such that…”) rather than just rewriting the definition — you haven’t proven it applies yet!
Make sure you introduce variables before using them (\(\varepsilon\), \(\delta\), \(x\), \(n\), etc.). This was a common issue in 8(b) — see below.
If you skipped questions because you were stuck or short of time, I’m happy to still look at them with you in office hours.
There were still a few instances of people defining \(\delta>0\) dependent on \(x\). See above for why this doesn’t work! And/or ask me about it.
Problem 8#
In this question, you had to compute left and right limits of the given functions and comment on when they disagreed. You then had to prove you were right using Definition 2.4, or the equivalent definitions involving sequences.
Nearly everyone was able to compute the limits in parts (a) and (c). For part (b), a lot of people looked at \(\lim_{x\rightarrow n^+}g(x)\) and \(\lim_{x\rightarrow n^-}g(x)\) without saying what \(n\) is. If you are assuming \(n\) is an integer, make sure you say so first. In fact, you get different answers depending on whether \(n\) is a positive integer or a negative integer.
In the examples given, all functions had jump discontinuities (either a finite or countably infinite number of them). Proving that a function has a jump discontinuity requires more work than proving it is discontinuous at a point. For the former, you need to prove that two marginal limits exist, but disagree. If you are using sequences to do this, they need to be general — picking a couple of specific sequences isn’t enough. On the other hand, if all you aim to show is that a function is discontinuous somewhere, then you don’t need a general sequence — you just need one (or occasionally, two) examples that show the definition of continuity fails.
In some cases, people had computed limits without any attempt at proving them. Be careful here! We already know you can compute limits: this was core syllabus at Level 1. The “new” content in this module is all about formal proofs — make sure seek out help with this as much as you need.
Finally, many people made this question unecessarily hard for themselves by not drawing pictures. I especially recommend doing so for parts (b) and (c)!
Problem 9#
This question was answered well on the whole, so well done! Presentation was also generally better for the question than the \(\varepsilon-\delta\) proofs. I think the main cases where people didn’t get the question out are because they didn’t follow the hint. Please come to office hours for help with this if you struggled with this bit.
Problem 19#
Here, nearly everyone I saw who attempted the question knew they needed to put something like \(\varepsilon=\frac{g(a)}{2}\) into the definition of continuity of \(g\) at \(a\), which shows good intuition! However, many struggled to turn this into a mathematical argument.
Remember, an inequality calculation on its own does not constitute a proof, you need to explicitly connect it in some way to the assumptions of the question (in this case, that \(g\) is continuous and positive at \(a\)). In fact, most people did have the bare bones of what they needed to complete the question correctly. Remember, you can always ask for help with this. Once you get the trick for turning this kind of working out into a proof, it is the same nearly every time, but it does take most people some time for it to “click” (ask anyone).
If you are getting comments like “there is no argument here” or similar, that is a sign that you may need to come to an office hour for more support with your proof writing.
Problem 22#
Those who attempted this generally got it out, so well done! In future, if a “weird” question like this one is assigned as homework like this one, I’d much prefer you come and get some help with it than leave it blank. But I also appreciate that this time around, there was a lot to do.
Feedback for Homework 1#
Problems: P2, P3, 2, 3; Solutions: P2, P3, 2, 3.
Well done for completing the first analysis homework! Here are some general comments.
General comments on mathematical writing#
The absolute best thing you can do is look at the individual comments on your work on Crowdmark, and get help interpreting them as needed. Analysis proofs are notoriously picky, and all mathematicians go through this at some stage or other, so please don’t feel embarassed about needing help with your proof writing.
When using a definition to solve a problem, it’s fine to write out the general definition, but make sure you also explicitly link it to the question at hand. So, for example, when solving Problem 3, you could (though don’t have to) begin by copying out Definition 2.2 from the notes. If you do that, make sure you then say something like “Therefore, given \(\varepsilon>0\), we seek \(\delta>0\) such that …” and link it to the specific functions and limits you are working with.
Remember to only use notation/variables after you have introduced them. This sometimes relates to the above point — Definition 2.2 has a general limit \(L\) in it, but there’s no need to carry it around in your solution if you know the actual numerical limit you’re aiming for (e.g. \(5/2\)).
Problems P2 and P3#
These were generally answered well, and most people were able to find the intended theorems from their MAS107/MAS117 notes in order to re-prove the Bolzano-Weierstrass theorem. We’ll be using this theorem more than once this semester, so this revision will come in handy.
In P2, there were some issues with mathematical writing, so make sure you go and look at the individual comments on your work and see what comments you can use to improve this.
Problem 2#
For this question, it’s fine to just write down what the limit points are, but of course we like proofs in this module. If you did include a proof, well done! And take a look for comments about its structure and presentation. The solutions do now include proofs, if you would like to see how they go.
Problem 3#
The three parts of this question were not of equal difficulty, and this was intentional. I would say part (iii) is definitely the hardest, so well done if you got that one out. If you didn’t, take a look at the individual feedback and see if you can make some progress, and/or, bring it to an office hour. You may find it easier to tackle now you have had more time to practise.
One of the more common issues people had in Problem 3(iii) was defining a \(\delta\) that depended on \(x\) as well as \(\varepsilon\). This is an understandable error: once you get to the stage
it can be hard to know how to deal with the \(\left|1-\frac{1}{2x}\right|\) factor. However, letting \(\delta=\frac{\varepsilon}{|1-\frac{1}{2x}|}\) does not work here — we need a \(\delta\) that is independent of \(x\) (though it is allowed to depend on \(\varepsilon\)). One way to see why is to look at the order that different variables appear in the definition of the limit:
Here, \(\varepsilon\) appears first, and so cannot depend on anything, and \(\delta\) appears second, so can only depend on \(\varepsilon\). In particular, \(\varepsilon\) and \(\delta\) appear before \(x\), so neither are allowed to depend on \(x\).
To progress past the point of (3.1) with a \(\delta\) independent of \(x\), the thing to do take a \(\delta\) that is the minimum of two values, with on value controlling each factor. For example, if \(|x-2|<1\), then \(0<1-\frac{1}{2x}<1\), so put \(\delta=\min\{1,\text{?}\}\). Now find “\(\text{?}\)” (or have a look at the solutions).