Which Course is Right for Me: Math 25 or 55 at Harvard?

[TheLastMagician]

i'll be attending harvard next year (deferred entry) and am planning to use this year to study calculus and linear algebra in preparation for either math 25 or 55. (See here for the various courses: http://www.math.harvard.edu/pamphlets/beyond.html ) I know math 23 will be too easy because I tried the problem sets and read some of Ross's Elementary Theory of Calculus and it seemed a tad too easy for me in that he explains too much and none of the problems are challenging.

However while going over the descriptions for 25 and 55 I noticed something contradictory which I hope someone can elaborate, especially if they have first-hand experience with either of the courses. The full description is:

"Math 25 and 55 are both full-year advanced courses designed for students with a very strong interest in theoretical mathematics. Each covers multivariable calculus, linear algebra, and some additional topics from a rigorous and advanced point of view. The students in these courses are frequently committed to concentrating in mathematics and are asked to put in extensive work outside the classroom. Many have had more than one year of college mathematics while in high school or have participated in various summer math programs. However, it is not necessary to have had multivariable calculus before taking 25 or 55. Although the syllabus of Math 25 is similar to that of Math 23, students will usually have had more preparation in math.

Math 55 is a faster paced course and covers topics more deeply. It is designed for students who arrive at Harvard with an extensive background in college level math. Math 25 and 55 differ from Math 23 in the level of outside work required: homework assignments in Math 25 and 55 are typically very time consuming. Math 23, 25 or 55 all provide an excellent foundation for further study of mathematics."

Notice that it says:

"However, it is not necessary to have had multivariable calculus before taking 25 or 55."

Then is goes on to say:

"It [Math 55] is designed for students who arrive at Harvard with an extensive background in college level math."

(my emphasis)

Now this is confuding; on the one hand, it says you can take 25 or 55 without multivariable calculus, which is really equivalent to "without college level math" (since most high schools cover everything below multivariable calc). On the other hand, it is saying that math 55 is designed for people with extensive background in college level math.

I don't think this was very smart of them because it leaves people very confused when trying to figure out which course suits them the most. Therefore, I am hoping someone can elaborate slightly on this. Also, my background in math is up to single variable computational (or "mechanical" as some people call it) calculus, but I have extensive (at least 2 years) first-hand experience with proofs from dabbling in competition math, and I have solved some very tough olympiad problems during my time (including several of modern IMO caliber), so I know the basics of some discrete math topics (mainly number theory and combinatorics - functional equations and inequalities as portrayed by competition math never appealed to me, having a sort of cheap plastic feel to them) as well as various proof strategies and general problem solving tricks.

This is probably why I found the course 23 not appealing: the course isn't challenging for me at all, but it has tons to offer with respect to knowledge, so it can probably teach me the most. On the other hand, this may back-fire because I'll find it boring due to the lack of challenge.

Any advice is highly appreciated!

 

[wotanub]

55 in my opinion is really for people who already know what they are doing and want an advanced treatment of the subject. Yes 55 does include multivariable calculus, but I wouldn't try that one unless you already knew multivariable calculus. Think of it as polishing the knowledge rather than teaching the subject.

Do not worry too much about which you will take because at Harvard you can enroll in more than and you do not have to decide which one you are officially sticking with until you have had time to do homework and a test too, I think. Your academic advisors will help you and make sure you find the right fit. This is because people are often conflicted about which they should enroll in.

 

[mathwonk]

I agree with wotanub. The statement that you do not need multivariable calculus before 55 probably means that logically math 55 will cover all the topics and not assume them. However the fact that most students will have already had extensive college math means that the material will be covered so fast and so abstractly that, even though it technically is complete and self contained, the pace will be so fast that you will have a very hard time absorbing it if this is your first encounter with it. So if you go for this course as your first multivariable course, you will be very hard pressed to keep up, and you will also propbably not see a lot of the elementary material that would illustrate what is going on. The only student I personally know to have taken math 55 recently had taken numerous advanced college and graduate level math courses while in high school in Georgia, and finished highly on the collegiate level Putnam exam while doing so, all before college. To be sure, this student was likely not typical and may even have been one of the best in the course. (you may learn something of interest by reading the wikipedia article on this class.
https://en.wikipedia.org/wiki/Math_55 )

I myself took math 55 in 1963-64 without having seen multivariable and without even knowing single variable calc all that well. I did know more linear algebra than usual. I benefited, in a survival sense, from taking it from a very smooth lecturer, Lynn Loomis, in whose course one could pass exams by doing a lot of memorizing of definitions and proofs. I also did well on his homework sets. About the only thing I really got out of the course was that the Frechet differential of a map is a linear map that approximates the original map locally to within "little oh" approximation, and that this makes sense in any Banach space. That is good valuable information, but leaves out a lot. I did not realize for instance that, although the definition makes sense in any Banach space, one has little ability to calculate a differential in infinite dimensions, whereas in finite dimensions one uses partial derivatives to do so quite explicitly. He also presented the theory of content in integration, which was quite beautiful but left me still needing to learn Lebesgue integration. We covered a very abstract version of "Stoke's theorem" but saw nothing of the low dimensional versions of that basic result, like Green's theorem or Gauss's theorem. It took me many years of teaching and study to fill in the many gaps left by math 55, especially useful for this being reading through Mike Spivak's little book, Calculus on Manifolds, supplemented by Courant's vol. 2, and Wendell Fleming's book Calc of several variables, not to mention teaching several times from the classic by George B. Thomas, plus Edwards and Penney, and Stewart.Another specific topic covered was the inverse function theorem and implicit function theorem. I must say that after learning the abstract Banach space version, involving factoring a map with split-surjective differential through a diffeomorphism and a projection, I was completely puzzled when someone said the implicit function theorem just says you can solve for some of the variables in an equation in terms of the others.

If you take math 55 for the rush, I would suggest leaving a lot of time for that course, supplementing it by reading other more traditional books, with more examples, and establishing a working/studying relationship with some other student or students to help each other. I also learned a lot of one variable background and theory by auditing math 112, which was an undergrad version of math 212, i.e. an intro to real analysis, for instance covering metric spaces thoroughly, given by George Mackey. I may have done this the previous semester, or simultaneously with math 55a or b.edit: I have a copy of the notes for math 55 from 2006 and just looked back at them, reminding myself how ridiculously ambitious they are, covering linear algebra, field theory, Galois theory, metric spaces, some functional analysis, differential and integral calculus of several variables, multilinear algebra (tensors), manifolds, etc etc...

as to the claim you do not need to know multivariable, here is a quote from one of the problems:

"This problem involves the notion of gradient, divergence, curl, cross product, and triple
product. If you do not know them, please look them up in a book about vector calculus."

but that may be an isolated example. As pointed out above, Harvard apparently makes it easy to get out of math 55 early if things are not going well. Here is another direct quote from their webpage though (maybe last year's version of the one you linked):
"Meanwhile Math 55 should be taken only by students with extensive college level math backgrounds."You could probably also take math 23 for credit and learn a lot, and audit math 55 for stimulation, but in practice that usually results in not keeping up in the audited course and eventually dropping out of it without benefiting much.

As a final remark, it makes a difference who is teaching the course as well. When I was choosing, the regular advanced calc course was taught by a famous and brilliant elderly mathematician, who had poor handwriting and a weak voice, so he was hard to hear and take notes from beyond the front row. Loomis was the opposite, although prematurely white haired, he was energetic with a clear voice and board style, and perfectly organized presentation. (I learned much later that, ironically, one sometimes learns more from a brilliant but less organized professor, in this case Sternberg, who nonetheless shares more insight, but getting that benefit can take more effort.)

Notice the three courses math 23, 25 and 55 this year have very different teachers. Paul Bamberg in math 23 is a well known senior faculty member of exceptional ability who has been teaching at Harvard since the 1960's, but as far as I know is not a research mathematician. (It seems he has a PhD in theoretical physics however. https://www.extension.harvard.edu/faculty-directory/paul-g-bamberg )

The instructor for math 25 is a Benjamin Peirce fellow, which is a junior level temporary position filled by very promising rising young researchers. Math 55 is taught by Professor Siu, one of the most well known senior analysis researchers in the world, and also an exceptionally polished lecturer, or at least he was several decades ago when I heard him speak, and that is unlikely to have changed for the worse. In general it is probably a pattern that math 55 will always be taught by a well known researcher, and hopefully an excellent lecturer, but you should look at who it will be and perhaps take that into account. The other courses are also taught by excellent people, as is usual at Harvard, but they are perhaps chosen differently. It is of course not clear that these differences will be especially visible to students taking an introductory class.

aha. there seem to be videos available of bamberg teaching this class:

https://www.extension.harvard.edu/academics/courses/linear-algebra-real-analysis-i/13313

 

[mathwonk]

After all that blather, let me make a hopefully more concretely useful suggestion. If you want to have a fighting chance at passing math 55 next fall, try to work through as much as possible of the first two volumes of Apostol's Calculus books, including problem sets.

http://www.abebooks.com/servlet/SearchResults?an=apostol&sts=t&tn=calculus

 

[QuantumCurt]

This is where the distinction between formal prerequisites and real prerequisites is important. One could conceivably take any of these classes without having any prior knowledge of multivariable calculus. However, one would have to be a very gifted student. Math 55 especially is an infamously challenging and rigorous math course. Although it does build things from first principles, it's at a level of depth and rigor that the average student really is not prepared for.

 

[TheLastMagician]

After all that blather, let me make a hopefully more concretely useful suggestion. If you want to have a fighting chance at passing math 55 next fall, try to work through as much as possible of the first two volumes of Apostol's Calculus books, including problem sets.

http://www.abebooks.com/servlet/SearchResults?an=apostol&sts=t&tn=calculus

Thanks! The long post was very helpful.

Apostol looks extremely good and comprehensive too. Maybe I'm getting ahead of myself but suppose I finish it long before the year is up, what would you suggest working on then? An algebra book perhaps?

 

[mathwonk]

It seems to me highly unlikely that you will finish both volumes of Apostol in only one year, if you actually work problems and learn proofs and theorems. This is not a mechanical course where the goal is just read examples, do the trivial homework, and skip ahead to the next section. The goal is know all the definitions by heart, and to know and understand the proofs, and to work hard problems. It also seems to contain some linear algebra, but yes you might work on an algebra book.

The very fact that you assume you may complete Apostol in much less than a year however has me concerned about how to communicate to you how high the level of mastery is that is expected in a theoretical course like this. I.e. it reminds me that you say your background is in computational type courses where mastery is not even attempted. It should be your goal to be able to teach the course afterwards.

Although it seems very unlikely to me, if somehow you do finish Apostol, you might look at Spivak's Calculus on Manifolds, or maybe the book Foundations of Modern Analysis, by Dieudonne', which has been in the past a sort of model for math 55. But I believe the 2 volumes of Apostol are together something over 1200 pages. And they are substantive, so they deserve thorough attention, and will repay it. Good luck.

PS: as a check on your grsp of one variable calculus theory, can you take pemncil and aper and write out a complete proof of the fundamental theorem of calculus? E.g. can you define a continuious function with epsilon and delta and prove that on a closed bounded interval such a function must have a finite maximum? If so that is a good sign.

 

[verty]

I think you have enough information now to be able to tell us which class you will be taking. If you tell us it will be Math 55, you will get suggestions on how to prepare. But now you must decide.

 

[TheLastMagician]

I think you have enough information now to be able to tell us which class you will be taking. If you tell us it will be Math 55, you will get suggestions on how to prepare. But now you must decide.

Math 25 since I can get started on it immediately which will be more challenging than mastering something like Apostol in preparation for 55, and if I make enough progress on 25 this year I may even be able to take 55 come next year. But if I had to pick now definitely 25.

 

[verty]

Math 25 since I can get started on it immediately which will be more challenging than mastering something like Apostol in preparation for 55, and if I make enough progress on 25 this year I may even be able to take 55 come next year. But if I had to pick now definitely 25.

Okay, I know I pressed you for a decision, I actually thought you were going to say, "Math 55 all the way!". I wanted you to have said that so that the thread would serve its purpose, as it were. I thought you were hesitant on that but you were selling yourself as mathematical because you wanted to believe that you could do it, you wanted that reassurance.

And making a choice like this because someone reassures you is usually a mistake. One has to make one's own decisions and be responsible for them. For that reason, even though I pushed you for a decision, you can change it at any time. You don't owe me the decision, if that makes sense. I just thought you would say you want to do Math 55 and then you would be helped in that direction.

 

[TheLastMagician]

Okay, I know I pressed you for a decision, I actually thought you were going to say, "Math 55 all the way!". I wanted you to have said that so that the thread would serve its purpose, as it were. I thought you were hesitant on that but you were selling yourself as mathematical because you wanted to believe that you could do it, you wanted that reassurance.

And making a choice like this because someone reassures you is usually a mistake. One has to make one's own decisions and be responsible for them. For that reason, even though I pushed you for a decision, you can change it at any time. You don't owe me the decision, if that makes sense. I just thought you would say you want to do Math 55 and then you would be helped in that direction.

No problem at all. I think I know what to do though: study as much of math 25 as possible in my "year out", then take math 55 next year.

 

[verty]

e.g. read the definition of continuity, and the definition of a least upper bound, and the least upper bound axiom for reals, and then either using the least upper bound property of reals, or just the fact that a real number is given by an infinite decimal, try to prove that a continuous function on the interval [0,1] must achieve a maximum value somewhere.

I have an argument. For each $x \in [0,1]$, let $x_i$ be an interval $(x - \delta, x + \delta)$ such that $\delta$ is maximal such that there is an $n(x)$ in $[0,1] - (x - \delta, x + \delta)$ such that, for $z \in (x - \delta, x + \delta)$, $f(n(x)) > f(z)$. Notice that $\delta$ can be zero. Let $n(x)$ be implicitly defined by this construction.

Then consider the set $[0,1] - \bigcup x_i$. Clearly every point in $I$ is a maximum. We must show that $I$ is nonempty.

Suppose $I$ is empty. Then $n(x)$ is defined on the whole of $[0,1]$. Let $L(x)$ be a shorthand for this sequence: $x, n(x), n(n(x)), ...$, continuing until a term is found to be undefined. Now consider the following algorithm:

1. Start with $L(0)$. Note that it is strictly increasing.
2. If the sequence is infinite, it must converge to a point $y$. By left-continuity, $f(y)$ is the maximum on $[0, y]$. Consider $L(y)$ and goto step 2.
3. If the sequence is finite but not vacously empty, the final term is a point of maximality.
4. If the sequence is vacuously empty, the parameter to $L$ is a point of maximality.

This algorithm terminates, absurd. Therefore $I$ is not empty and the function attains a maximum.

----------------------------------------------------------------------------------------------------------------------

I suppose this shows why I am merely a math hobbyist. To prove something, one is inevitably going to come up with some sort of ad hoc thing like I did here, that isn't easily converted into a proof. And ultimately, how would this even be converted into a proof? And yet I think it is pretty convincing. So the problem is that there are tons of convincing arguments that probably no one would call a proof.

A proof is something different. It is not merely convincing, not merely true. It has other properties like being elegant, concise, using as few concepts as possible, etc. A true argument, by itself, is not a proof, IMHO, unless you think what I wrote above is a proof (supposing I haven't made a mistake).

PS. Okay, I see my "argument" has some holes but I'm pretty sure there's an argument of this sort that works.

 

[mathwonk]

@verty: I think there some really good ideas in there, even if it seems a little complicated. a few questions; how is n(x) defined? by construction it seems it could equal either f(x-delta) or f(x+delta), and in particular it is always defined everywhere, since when delta = 0, it can equal f(x). But these don't seem like big problems, since whenever n(x) = x, i guess x is a maximum.

I apologize for not yet studying it more thoroughly, as it seems to truly deserve it. Note that this little result is the hardest most fundamental result in all of first year calculus, and implies (with some work) all the other big results, like rolle's theorem, mean value theorem, and fundamental theorem of calculus, which are all much easier than this.

oh yes, there is one other big result, that a continuous function f on [0,1] such that f(0) = -1 and f(1) = 1, must equal zero somewhere between 0 and 1. that is actually a little easier than this. you might try that too.notice by the way that you are making some assumptions in your argument, that "largest" numbers of a certain sort exist, which is usually true for real num,bers, and is implied by the "least upper bound" principle.

 

[mathwonk]

verty: if it is correct that one can always consider n(x) to be defined, and one is thus looking for a fixed point of n, i.e. an x where n(x) = x, can you just say that the limit of the sequence 0,n(0), n(n(0)),... is such a fixed point?

 

[verty]

@verty: I think there some really good ideas in there, even if it seems a little complicated. a few questions; how is n(x) defined? by construction it seems it could equal either f(x-delta) or f(x+delta), and in particular it is always defined everywhere, since when delta = 0, it can equal f(x). But these don't seem like big problems, since whenever n(x) = x, i guess x is a maximum.

A possible change is $z \in \{x\} \cup (x - \delta, x + \delta)$. But then it runs into...

verty: if it is correct that one can always consider n(x) to be defined, and one is thus looking for a fixed point of n, i.e. an x where n(x) = x, can you just say that the limit of the sequence 0,n(0), n(n(0)),... is such a fixed point?

... this. There could be too many points. The algorithm needs to climb the mountain until it reaches the top. But the number of steps needs to be countable, at least this was my idea.

I'd like not to say more on this topic.

 

[verty]

Okay, I did fix the argument and Mathwonk does have a point, there is some merit to trying to prove these. I can't deny that there is use to doing that for a mathematician.

1. Let u = 0, v = 1.
2. Let p = (u+v)/2.
3. If there is a z in [u, p] greater than any point in (p, v], let v = p and goto step 2.
4. If there is a z in (p, v] greater than any point in [u, p], let u = p and goto step 2.
5. Otherwise, let z be a point in [u, p] greater than or equal to any point in (p, v], let v = p and goto step 2.

 

[mathwonk]

verty, I like your algorithmic approach. Here is another version of your argument that also seems to work:

for each x in [0,1] consider the supremum of the values of f on the intervals [0,x] and [x,1]. consider all x in [0,1] such that: supf on [0,x] < supf on [x,1]. If there are no such x, then supf on [0,0] is ≥ supf on [0,1] and 0 is a maximum. Notice that as x moves to the right from 0 to 1, the sup of f on [0,x] increases from f(0) to the sup of f on [0,1] which is ≥ f(1), while the sup of f on [x,1] decreases to the value f(1).

Moreover the functions taking x to each of these sups are continuous as well, if f is. Hence the set of all x such that supf on [0,x] < supf on [x,1] is an interval [0,c), and c is the first point where supf on [0,c] = supf on [c,1]. Claim: f(c) = both of these equal sups, since if f(c) were less, then the sups would remain equal for all x near c, contradicting the choice of c.

Hence c is a maximum. (Note it is allowed here that the sups could be infinite, but it follows from the argument that at the last step they are both = f(c) hence are always finite.)

I especially like this argument, which is indeed yours, i.e. simply a rewording of your ideas, because it proves simultaneously that f is bounded on [0,1] and assumes a maximum there. and I had not seen it before in quite this succinct form. so you have taught me something new about an old subject. thank you! This is what I meant about finding your own proof and possibly obtaining a new argument.

 

[verty]

Onto other matters now, this thread has kind of veered off what its purpose was. It was for TLM to get advice and then choose a class. If he thinks he is going to be a mathematician in future, getting into that advanced class may be worth it because he will meet a top professor and colleagues who will push him as well. And he will have a tremendously good base for getting into research early. And I'm thinking of people like Andrew Granville and Richard Taylor who are famous number theorists. Perhaps TLM will end up like them, who knows. The best place for him to do that is in the advanced class.

So that was my reasoning on that. TLM seems to have decided what to do and I've refocused the thread on that topic in case he returns later with questions.

 

[berkeman]

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