Should I Become a Mathematician?

[homeomorphic]

Another problem is that you will most likely have to move several times for your career. You might have to move for grad school. Once you get your PhD, you'll most likely have to move a couple of times for post docs. And after that you'll probably have to move again for a tenure-track job, and then maybe again after that.

All this instability will make having a long-term relationship more complicated.

It can also make getting the relationship more complicated.

My most common excuse to chicken out on making the moves is the girls that I meet are often grad students who are not in synch with me as far as graduating at the same, as well as later difficulties after that, so I just rationalize it by telling myself it will never work out anyway. When it comes to chickening out, I'll take any excuse I can get.

 

[mathwonk]

none of these difficulties is insuperable. i know mathematicians who work not only in different cities from their spouses but in different countries and even different continents.

 

[epenguin]

How do mathematicians find time to have a girlfriend?

Unless maths doesn't consume most of their time, I guess.

Hardy said he could work on math for only 3 or 4 h a day. Which would have left time for girlfiends - if he had been interested in having one. But Hilbert was, and was a counterexample of the conjecture being made here.

I am not a mathematician but find this impossibility hard to believe. Is it not true mathematicians more than other scientists have time for politics?

Painlevé had time for Madame Curie. What is harder to imagine is her having time for him. Which is the point. Surely experimental scientists have more demanding, longer, less flexible time commitments, demanded by both the experiments themselves and often by the teamwork which is less common for mathematicians. If mathematicians don't have time who does? (And they can even be secretly working while they are out with their girlfriends which experimentalists can't.)

 

[mathwonk]

i assure you that if you think about math while with your girlfriend that she will notice it.

 

[MathematicalPhysicist]

I wrote after my question, "unless maths doesn't consume most of their time".

I didn't say that I'm consumed by maths, I have other interests, but it does seem difficult to find a girl who is nice looking, interested in a similar subject and is unattached already.

I guess I need to compromise...

 

[Mépris]

It can also make getting the relationship more complicated.

My most common excuse to chicken out on making the moves is the girls that I meet are often grad students who are not in synch with me as far as graduating at the same, as well as later difficulties after that, so I just rationalize it by telling myself it will never work out anyway. When it comes to chickening out, I'll take any excuse I can get.

I can relate to/imagine that (in that, this will probably be me (too) in a few years).

You should also consider the possibility of female grad students who share your view while still being open to the occasional "fling". While we're at it, I strongly suspect female undergraduate students could be a possibility as well.

As was once rightly said by some misbegotten fool, "A morning of awkwardness is far better than a night of loneliness" and I'll add to that my opinion that company for a limited amount of time is better than no company at all!

---

Math question this time.

Can any budding mathematician/physicist get away with not being formally acquainted with Euler's "Elements of Algebra" or a similar higher algebra book? I'm already familiar enough with algebra and I can use it well, in my opinion but I may be wrong.

The thing is, I'd rather get started on Spivak or Apostol as soon as I can. On my to-do list, is reviewing some geometry, trig, probability and combinatorics, all of which I should be done with by the end of the next week. Starting Spivak or Apostol at around that time would be lovely as it would help me greatly with my main exams which will be in May/June. (A-Levels - around the same level as freshman maths)

I've actually completed high school but I want better grades at A-Levels, which is why I aim at writing them again in the coming months.

The following is to get an idea of my command of algebra. I can't think of anything else to explain what I know or don't know.

Let a and b be positive integers. Show that 2^1/2 lies between a/b and (a+2b)/(a+b).

I was able to prove that a/b is less than 2^1/2 fairly quickly but I had to refer the the worked example to be able to "see the trick" for the other half of the "puzzle". I'm currently working through problems like that and my review of the other chapters will be from the same book. So, would I be able to "get away" with learning as I do or would I be further complicating things by skipping the reading of this book?

I can also do reasonably well on http://www.cie.org.uk/docs/dynamic/41859.pdf . (note: get some trouble with complex numbers and vectors)

Thoughts on this, gentlemen? (and ladies...if any :-))

 

[mathwonk]

your answer makes no sense to me. a/b can be either greater or smaller than sqrt(2). one can prove that a/b < sqrt(2), if and only if (a+2b)/(a+b) > sqrt(2), however. And there really is no "trick", just obvious rearrangements of fractions (and squaring).

indeed if you cannot prove this yourself without any help, then your algebra skill seems rather weaker than an average high school algebra student's should be, and much weaker than euler's book would teach.

I was about to say go on to spivak, but after this example, I think you need more practice in algebra. Indeed success in standard college calculus is more closely related to skill with algebra than any other thing. Spivak requires also logical ability and creativity, but algebraic manipulation is still crucial.

Oh yes and in many college classes no calculators are allowed. That test looked like the sort of depressing standardized tests they give for AP scores in US, no concepts, no definitions, or proofs, just tedious calculation.

 

[Mépris]

your answer makes no sense to me. a/b can be either greater or smaller than sqrt(2). one can prove that a/b < sqrt(2), if and only if (a+2b)/(a+b) > sqrt(2), however. And there really is no "trick", just obvious rearrangements of fractions (and squaring)

Either I didn't correctly express myself or I really did something wrong.

Here's how I worked it out:

If $$\frac ab < \sqrt2 < \frac{a+2b}{a+b}$$

Then,

$$\frac ab < \frac {a+2b}{a+b}$$

So,

$$a(a+b) < b(a+2b)$$

Thus,

$$a^2 < 2b^2$$

Therefore,

$$\frac ab < \sqrt2$$

---

That was how I figured this out a couple of hours ago. It felt like a great deal to me and I was smiling to myself while half walking, half jumping around the room*. My over excitement was quite short-lived as I couldn't figure out how to proove the second part, i.e, $$\sqrt2 < \frac {a+2b} {a +b}$$.

After looking at the book, I understood that I might have approached the question the wrong way. (this is actually an "example question" w/answer from the book) At the top of page 14 of this book is the solution to the question, which (obviously) is a good way to approach this. Anyway, I hope that I've explained myself clearly enough this time.

I'll ask the question again, if I have no formal algebra knowledge and mainly learned via doing, could I still pull this off? :-)

*Consider that before having done "Example 1 & 2" on that book, the hardest mathematics I had ever been in contact with was that "monkey math exam" I linked to in the previous page. And yes, you're correct, in that this test is essentially an AP equivalent although it covers more material than AP Calculus and Stats. Even then, I'm not sure what good this does, if any at all...

 

[jbunniii]

I was able to prove that a/b is less than 2^1/2 fairly quickly

If you proved that, then your proof can't be true in general. It's not true, for example, if a = 5 and b = 1.

The key thing to note is that

"Show that 2^1/2 lies between a/b and (a+2b)/(a+b)."

means that EITHER

$$\frac{a}{b} < \sqrt{2} < \frac{a+2b}{a+b}$$

OR

$$\frac{a+2b}{a+b} < \sqrt{2} < \frac{a}{b}$$

i.e. there are two cases to consider:

$$\frac{a}{b} < \frac{a+2b}{a+b}$$

or

$$\frac{a+2b}{a+b} < \frac{a}{b}$$

 

[Mépris]

^
Yes and thank you. :-)

I saw this after I looked at the worked out solution. It was my second time working through this type of problem. Now I understand that I should be considering both possibilities when such a problem presents itself. What is the name of this kind of question, assuming it has one? It would make it easier for me to gain a higher level of understanding if I were to work out some more.

 

[mathwonk]

you are asked to show that a/b < sqrt(2) iff (a+2b)/(a+b) > sqrt(2).

The only thing to do here is square both sides, so we are asked to prove that

a^2/b^2 < 2 iff (a+2b)^2/(a+b)^2 > 2.

Then the only possible thing to do is multiply out the denominators,

so we are asked to show that

a^2 < 2b^2 iff

(a^2 + 4ab + 4b^2) > 2(a^2 + 2ab + b^2) = 2a^2 +4ab + 2b^2.

subtracting, we want to show that 4b^2 > a^2 +2b^2, iff a^2<2b^2,

i.e. that 2b^2 > a^2, iff a^2 < 2b^2.

that is a tautology. Since this looks hard to you, you need to practice your algebra.

this type of problem has no special name, it is just basic algebraic manipulation.

this is the sort of skill everyone had 50 years ago and has been lost perhaps due to overuse of calculators today.

you also apparently do not understand the logic of the problem since you begin with the wrong assumptions.

I strongly recommend you take our advice and actually read and work through one
of the algebra books we have recommended.

 

[Mépris]

you also do not understand the logic of the problem since you begin with the wrong assumptions.

Yes, I understand why I was wrong.

As far as algebra is concerned, should Euler's book be enough to prepare me for Spivak or Apostol then?

EDIT:

The other thing is I don't think my problem is "not knowing any algebra" but is more a question of me never getting my hands dirty with it. With this in mind, what books could I get that's filled with challenging algebra exercises? Euler (Elements of Algebra)? Allendoerfer and Oakley (Principles of Math)? Lang (Basic Mathematics)? Hall and Knight? (Higher Algebra)

These are the books I've heard of (mostly on here) and I have a copy of Hall and Knight around.

 

[mathwonk]

you seem to have plenty of recommendations. as you said it is time to stop asking advice and do some homework. Obviously a book written by a genius like euler is superior say to one written by me, or some other contemporary average author.

 

[Mépris]

http://scratchpad.wikia.com/wiki/Euler's_Elements_of_Algebra

Would these exercises be adequate to supplement my reading of the book? Or would I be better served if I tried to make up my own as I go along? :-)

I'm only on page 8 and it's a joy to read. I wonder how the original version "reads like".

 

[wisvuze]

Yes, I understand why I was wrong.

As far as algebra is concerned, should Euler's book be enough to prepare me for Spivak or Apostol then?

I think it's absurd to suggest reading a book by Euler just to practise algebraic manipulations. Allendoefer's book is great in general to learn from though. Also, the first couple of chapters in Spivak's book will have enough exercises for you to practise from. It's just a matter of practising, that's it.

 

[Mépris]

Why would it be absurd? What's wrong with Elements of Algebra? Is it because it's dated? (I've seen people speak thus of it) I tried a more "recent" book, Hall and Knight and I didn't like the presentation at all. In fact, it was more or less just presentation and little explanation/understanding involved.

Now, don't get me wrong. While I find Euler's book very interesting (I actually find it fun/challenging, unlike other "mainstream" math books I've read), if I could delay its reading for a few months, it would be in my advantage. As linked in the page before, my A-Level exam requires a superficial understanding of math concepts and when I have chemistry, physics and literature to take care during the coming months, this alternative looks more appealing. After May/June, when I'm done with these pesky exams, I could devote more time to Euler's work. But anyway, that's all under the assumption that you're correct and that I can pull this off. If my recent form is any indicator, then I shouldn't be able to - I feel very stupid and apathetic these days.

But as, I quote, a great man once said: "Form is temporary, class is permanent."

blergh, I'm rambling now...

 

[mathwonk]

wisvuze, i respect your right to your opinion. But I still stand by my advice, based on my experience. You might yourself learn something from euler. i know i did.

 

[mathwonk]

here are some notes from day 1-3 of honors calc, one variable first semester.

 

[mathwonk]

here is another set of early calculus notes.

 

[mathwonk]

here are some proofs of big theorems in differential calculus.

 

[battousai]

thank you ! very much appreciated !

 

[mathwonk]

here is another set on neighborhoods, and a summary note of the 4 main principles of differential calculus.

 

[mathwonk]

here is a little discussion of how functions are constructed from more elementary ones that is seldom seen in beginning courses, plus a brief sketch of why the easy proof of the chain rule is completely rigorous.

I.e. the easy proof of the chain rule uses the product rule for limits. It is often objected that this argument fails in certain special cases because it requires dividing by zero. See e.g., Thomas/Finney, 9th edition, p. 156, where they say a different proof is "required". That is not true however since at the points where one must divide by zero to give the product rule argument, the limit is already known for a more obvious reason.

I.e. we want to prove that the limit of [f(g(x))-f(g(a))]/[x-a] equals the product f'(g(a).g'(a). the easy proof is to note that [f(g(x))-f(g(a))]/[x-a] = [f(g(x))-f(g(a))]/[g(x)-g(a)] . [g(x)-g(a)]/ [x-a], as long as the denominators are not zero, and use the product rule.People who object say that there are cases where the denominator [g(x)-g(a)] is zero on every neighborhood of a. But that does not matter for the following reason. If that happens then g'(a) = 0, and we are then trying to prove that the limit of [f(g(x))-f(g(a))]/[x-a] is zero as x-->a. But at all points where g(x) - g(a) = 0, we also have [f(g(x))-f(g(a))] = 0, so the fraction we are looking at, namely [f(g(x))-f(g(a))]/[x-a], is also zero.

You thus do not have to factor the fraction out to prove it converges to zero at those points, since it is already zero. I.e. you only have to argue the fraction is approaching zero at those points where it is not already dead zero. So you can ignore the points where g(x) - g(a) = 0.

This was well known over 100 years ago, in old books, but forgotten more recently, with probably G.H Hardy to blame.

 

[mathwonk]

to say this again briefly, if we want to show that [f(g(x))-f(g(a))]/[x-a] is near zero when x is near a, there are two cases.

1) If the fraction equals zero, as it does whenever g(x)-g(a) = 0, then it is certainly near zero.

So we only need consider x such that g(x)-g(a) ≠ 0. In that case the usual easy product rule argument works fine.

Thus the claim in thousands of books that this argument is inadequate are wrong. This is another example of book authors copying the content of other books for decades without thinking about the material.

The easy proof was correctly given for years before 1900 until G.H Hardy decided it was useful to give another "linear approximation" argument in about 1908. Ever since then, most modern books have copied his proof and many have even claimed incorrectly that the old argument was invalid.

At the bottom of pages 154-155, vol.1, Courant however notes that the usual proof is valid but requires a special argument when g'(a) = 0, hence the one in the text is "preferred".


I also attach notes on implicit functions, an application of the chain rule.

 

[thrill3rnit3]

mathwonk, what do you know about algebraic k-theory?

 

[mathwonk]

here is another set of notes, with some repetition.

 

[mathwonk]

ok, k theory, let's see, as i recall grothendieck introduced a group built on vector bundles, on a manifold. one has an operation on bundles by tensor product. this let's us regard them as basically a monoid like the natural numbers. then we can imitate the procedure of constructing all integers from the natural numbers, to construct a group whose "positive" elements are the vector bundles. this is called the K group on that manifold.
uh... I'm running out of things to say, so obviously i know squat about this subject.

well i hate to admit that so let's go algebraic, in which case a vector bundle is replaced by a projective, i.e. locally free, module over a fixed ring R (or even locally free sheaves on a topological space).

So I guess we could consider all projective modules over R, or all locally free shaevs over X, up to isomorphism, and then try to construct a group whose "positive" elements are the isomorphism classes of projective modules.
well I'm really stuck now. sorry.

but i have lots of friends who know this subject inside out. i will eventually remember their names. thank you for helping me fight alzheimers by your questions.
do you have a specific question about K theory that i can help find an answer to?

I do have a copy of atiyah's book on k theory. i guess i could open it.
this is clearly one of the many subjects i know almost nothing about.

ah yes! it always helps to recall why a subject was introduced. this topic was defined to refine the riemann roch theorem.

remember the rrt is an equation between the dimension of the space of sections of a vector bundle and some more computable, i.e., coarser, group of invariants of that bundle. the traditional theorems used as that coarser group the group of integers or the cohomology group iof the space, i believe the grothendieck version uses the K group.

im sorry, i really don't know much about this.

 

[morphism]

Algebraic K-theory is very daunting...

Notice that what mathwonk calls the K-group of a manifold is really just a K^0 in a family of "naturally-defined" (i.e. defined via reasonable geometric considerations, and in such a way that they're all related) K^n groups.

So his passage to modules over functions on the manifold really only defines an algebraic K_0. The geometric process that produced the other K^n's doesn't have a clear algebraic analogue. The search was on to try to find appropriate definitions of higher algebraic K-theory, and this was achieved most successfully (?) by Dan Quillen, who got a Fields medal for his work. I've never seriously tried to study this stuff, but what I've seen of it is fairly intense.

Anyway, there are some curious relations between algebraic K-theory for number fields and special values of zeta functions... This is something that has fascinated me for a long time, but which I've always had trouble getting into it because of the really high barrier to entry.

 

[homeomorphic]

ok, k theory, let's see, as i recall grothendieck introduced a group built on vector bundles, on a manifold. one has an operation on bundles by tensor product. this let's us regard them as basically a monoid like the natural numbers. then we can imitate the procedure of constructing all integers from the natural numbers, to construct a group whose "positive" elements are the vector bundles. this is called the K group on that manifold.

Correction: the operation is direct sum, not tensor product.

I'm no expert on K-theory, either, but I know a thing or two about it. This might get technical in places, but maybe Mathwonk and morphism will be interested if nothing else, and it's good for me to try to summarize what I know to clarify the ideas for my own sake. You can always look at Wikipedia for definitions. Also, I may get sort of speculative as I go on because my knowledge will peter out pretty quickly.

If you want it to be a group, somehow you have to find a way to get an identity element and inverses. That doesn't look like it's going to happen if you just take vector bundles with direct sum because when you direct sum them, they just get bigger and bigger. So, what you have to do is consider stable equivalence classes of vector bundles--i.e. you declare the vector bundles to be equivalent if they become isomorphic after direct summing with a big enough trivial bundle. Any vector bundle over paracompact base is trivial after direct summing with some other vector bundle (by a trick kind of like Whitney embedding theorem or for bundles over manifolds, you could use the Whitney embedding theorem itself).

So, that's topological K-theory, and it was Atiyah and Hirzebruch who started the subject. Actually, Grothendieck started algebraic K-theory before the topogical K-theory, and Atiyah and Hirzebruch were inspired by that. The algebraic version is K^0 of a ring.

For a geometrically-minded guy like me, the next place to look would be simple homotopy theory, which is concerned with K^1 of group rings. Simple-homotopy theory is sort of a cell-by-cell geometric approach to homotopy theory. If you have a CW complex (actually, Whitehead invented CW complexes in this context because of their technical advantages over simplicial complexes), you might wonder if you can do the homotopies one cell at a time. It turns out you can't do it in general, but the obstruction to being able to do it is determined by groups called Whitehead groups. There's a geometric definition of the Whitehead group of a CW complex, but also an algebraic one as K^1 of the group ring of the fundamental group (actually, maybe it's the reduced K^1, which is a quotient of K_1 by something). The place to read about this is Cohen's book, Simple Homotopy Theory. Very nice and well-motivated book. Before Whitehead torsion, there were other torsions, like Reidemeister torsion, which was introduced in order to classify some 3-manifolds called Lens spaces. So, the theory of K^1 goes back to the 30s, I guess, although it wasn't called K_1 until while later.

So, apparently there is some kind of relation between K^0 and K^1 of a ring. Some kind of similarity. I'm not quite sure what it is. It may be cheating to say K^0 and K^1 at this point because I doubt they were called that originally. But, anyway, someone noticed some similarity there, and with that in mind, Milnor defined another group, K^2. And then the question arose as to whether K_0 and K_1 were part of a sequence, K^n. My guess is that this was conjectured with classifying spaces and cohomology theories in mind.

Given a ring, you can form the group of n by n matrices over the ring, GLn(R). And there's an inclusion map from GLn(R) into GLn+1(R). If you keep going and take the union of all those (direct limit), you get a big group called GL(R). Given a group, you can form a classifying space, BG, and the bundle EG over it. The significance of BG is that you can get any principal G-bundle by mapping the base into BG and pulling back EG. Homotopy classes of maps into BG correspond to isomorphism classes of principal G-bundles.

So, for example, homotopy classes of maps into BGL(R) give principal GL(R) bundles. So, it looks like that ought to be related to K groups of a space, somehow, since maybe GL(R) could act as symmetry groups for stable equivalence classes of vector bundles if R is a field, for example. So, building on those kinds of ideas, Quillen introduced his Quillen-plus construction. The plus construction was method of Kervaire for modifying the fundamental group without changing homology and cohomology, but Quillen applied it to BGL(R). I guess you just attach some cells to it. I'm not sure what the accomplishes, but evidently, it's pretty important, since he won a Fields medal for it. So, to define K^n, you take the nth homotopy group of the Quillen plus contruction.

If it were just BGL(R) without the plus construction, it's giving you principal GL(R) bundles over spheres, except that that would correspond to free homotopy classes of maps of spheres into BGL(R), whereas the homotopy groups are maps with basepoints. The point of the plus construction seems to be to kill off the commutator subgroup of the fundamental group, so you have free homotopy classes, eliminating the basepoint dependence.

Somehow, this is supposed to be related to some algebraic geometry and number theory stuff that I know almost nothing about.

Topologically, again, I'm guessing maybe if you want to deal with stable bundles and want to study them in a cell by cell way, perhaps, you would care about maps from spheres into the Quillen plus construction. If you want to look at one cell, any bundle is trivial over the cell, but you do care about the how the bundle over the cell is glued to the rest of the bundle. So, I could see it being relevant there. That may be what K theory as a cohomology theory is telling you about (there are K groups associated to rings and K groups associated to spaces which form a cohomology theory).

Stable bundles are the kind of thing maybe a high-dimensional topologist might care about, so I could see it coming up in surgery theory or something (took a class from a surgery theory guy who is interested in algebraic K-theory). And also, of course, algebraic topologists would be interested, too. High dimensional topology is very homotopy-theoretic now due to surgery theory tools that boil a lot of it down to homotopy theory.

It always seemed like kind of an obscure subject to me. I went to a talk about it last year and it was pretty much over my head. But after writing this and putting some strands together, it actually seems like there's something pretty cool going on there.

 

[mathwonk]

"Correction: the operation is direct sum, not tensor product."

Thank you for this correction, homeomorphic!here are my calculus notes summarizing the results of differentiability.

 

[mathwonk]

here are notes on rates of change and derivatives, and on exponentials and logs.

 

[Inertigratus]

Hello,
I'm currently studying engineering in electronics (undergrad). A programme that has a lot of courses in physics and math. Soon I will have to choose a path (master) and recently I've been thinking more and more about choosing a master in mathematics. I like mathematic theory, fascinating stuff. However, practically, I'm not that good at solving problems. I'm more into theory then actual problem solving so to say. My grades are not too good either.
Is it a bad idea then to choose a master in math?
I'm thinking that perhaps that is what will make me a better mathematician.

 

[DivisionByZro]

Hello,
I'm currently studying engineering in electronics (undergrad). A programme that has a lot of courses in physics and math. Soon I will have to choose a path (master) and recently I've been thinking more and more about choosing a master in mathematics. I like mathematic theory, fascinating stuff. However, practically, I'm not that good at solving problems. I'm more into theory then actual problem solving so to say. My grades are not too good either.
Is it a bad idea then to choose a master in math?
I'm thinking that perhaps that is what will make me a better mathematician.

What courses have you taken in mathematics? Graduate school in math is no joke, you can't just BS your way through. You'll need at least one or two courses in analysis, one or two in abstract algebra, topology, you'll need to be very familiar with set theory, you'll need lots of linear algebra, and a lot more too (to be competitive). Also, good grades go without saying.

 

[Inertigratus]

What courses have you taken in mathematics? Graduate school in math is no joke, you can't just BS your way through. You'll need at least one or two courses in analysis, one or two in abstract algebra, topology, you'll need to be very familiar with set theory, you'll need lots of linear algebra, and a lot more too (to be competitive). Also, good grades go without saying.

Analysis in one variable, analysis in multiple variables, linear algebra, vector analysis (shared), complex functions/analysis (shared), some system theory, ODE (shared) & PDE and have had courses involving transforms (Laplace, Fourier) and Fourier series.
By shared I mean that they squeezed two things in one course, for example the complex functions/analysis and vector analysis was in one and the same course.

The master in mathematics has specializations within various fields of math and topology was in there, so I'm not sure if that's a pre-requisite.

 

[DivisionByZro]

You seem to have some background in analysis, but nowhere did you mention abstract algebra. That's a problem as abstract algebra pops up in a lot of places. I'm not too far into it yet, but I can assure you you'll need to be very comfortable with it. For one, it's an integral part of the GRE.

http://www.ets.org/gre/subject/about/content/mathematics