Mathematician Chat V.floor(π) - General

[QUOTE=JohnnyMo1;42324391]Hrm. How does the free product with amalgamation change things, and is that what is being used in Seifert Van-Kampen? Because Seifert-Van Kampen says that the fundamental group of the figure eight is Z*Z, and I'm just not positive what distinguishes that from Z.[/QUOTE] It doesn't really change things, but with amalgamation you will (almost) definitely be doing some reductions, so it's even easier to get confused. It's just a really picky way of examining the definition - for the Z*Z example, if the both copies of Z contain ...,-1,0,1,2,... then the definition states you have 'words' of elements in each group, e.g. 1|2|4|1|-1|3 (the pipes here are just to divide the elements), but then you are supposed to 'reduce' according to the group laws in each group, the problem is since all the elements are in *both* groups, you should just reduce the whole lot down (in this case, to 10) What you [i]really[/i] mean when you write Z*Z is two isomorphic but distinct copies of Z. Say the first has elements ...,-1,0,1,2,... and the second ...,-1',0',1',2',... Then 1|2'|4'|1'|-1|3 would reduce to 1|7'|2 - which is less trivial. The most clear way of doing it is to pick generators for each - x and y - so you just get powers of x and y which clearly aren't compatible. This means that the best way to think of Z*Z (otherwise known as the free group on 2 generators!) is as all words containing x, y, x^-1 and y^-1 where they combine in the obvious way. So the last example would be written x y^7 x^2 Amalgamation is what you are doing in Seifert Van Kampen, it's phrased weirdly because it's quite general and you need normal subgroups, but in some sense you're identifying different bits of each group and you can use that relation to further simplify the words in the free product with amalgamation. In terms of fundamental groups and SVK, if the two parts meet at a single point, you (iirc) just have the free product, but if the two parts meet nontrivially you need to somehow identify the fact that you might be able to translate a loop in one half to a loop in the other, and that's how you identify bits of the fundamental groups of the two parts. Some of the confusion is the way we say tend to say the fundamental (or homology, galois etc.) group of something [i]is[/i] Z, when what we mean is it is [i]isomorphic[/i] to Z. We know what we mean, and it would be silly not to do it like that, but it does occasionally make a difference. For the figure 8 you split it into 2 circles, each with fundamental group [i]isomorphic[/i] to Z (but crucially, they do not have the same elements!) The fundamental group of the whole 8 is the set of sequences of loops in one side or the other (with order being important) - which is exactly the free product Z*Z as above. (And is certainly not just Z) So for example, x y^7 x^2 would correspond to looping round the top half, then the bottom half 7 times, then the top half twice.

[QUOTE=Joey90]What you [i]really[/i] mean when you write Z*Z is two isomorphic but distinct copies of Z.[/QUOTE] That's what I was missing. Thanks. Hatcher does not make that clear.

What is the subject called (if it has a name) where you look at complex analysis type stuff except with quaternions instead of just complex numbers?

[QUOTE=ThisIsTheOne;42376007]What is the subject called (if it has a name) where you look at complex analysis type stuff except with quaternions instead of just complex numbers?[/QUOTE] [url]http://en.wikipedia.org/wiki/Quaternionic_analysis[/url] I have no experience with this, but I guess this is what you mean? I'd imagine it doesn't quite work out as nicely as complex analysis does though (not much does!)

Well you don't have commutativity, but at least you do have anticommutativity with i,j,k, so i could see that if it's nice like complex analysis is (big if) then it shouldn't get all [I]that[/I] messy. That page is a start, I was hoping to see more on contour integrals over the quaternionic plane, the article mentions an equivalent of the cauchy integral formula, that'd be interesting.

I really ought to take complex analysis at some point. Probably I will when I go to grad school.

I've taken a second year course in it, and should be taking a 4th year course in it next year. It was definitely one of the best courses I've taken. Complex analysis is such a beautiful subject, you probably wouldn't think it would work out as nicely as it does. For example the contour integral along a closed path which encloses a singularity can be found by computing a derivative of a similar function through the Cauchy integral formula, and its value can also be found by considering certain coefficients of its Laurent series (which is similar to Taylor series but allows you to find a series outside of the standard radius of convergence) So really, differentiation, integration, and series expansions are all even more closely linked in complex analysis.

Man this is a good algebraic topology homework. First problem is finding the fundamental group of the plane with n points deleted. Second problem is finding the fundamental group of the 2-sphere with n point deleted. (Same as plane with n-1 points deleted. Woohoo, the stereographic projection is a homeomorphism!) Third problem is finding the fundamental group of R3 with n lines through the origin deleted. (Same as sphere with 2n points deleted. Woohoo, deformation retracts!) It's like, three problems for almost the price of one. [editline]9th October 2013[/editline] Then only one problem left after those are done.

Well I had a real sexy proof of the first one by induction + seifert-van kampen all formulated in my head and was about to write it down when suddenly I realize... it's not the plane with n points deleted. It's R3 with n points deleted. Derp.

[QUOTE=JohnnyMo1;42473821]Well I had a real sexy proof of the first one by induction + seifert-van kampen all formulated in my head and was about to write it down when suddenly I realize... it's not the plane with n points deleted. It's R3 with n points deleted. Derp.[/QUOTE] Shame that R3 with points removed isn't very exciting from a fundamental group level... For the others, depending on how clear you can make your homotopy equivalences you can go even further - a plane with one point removed is homotopy equivalent to S1, remove 2 points and it's equivalent to S1vS1 etc. This makes it pretty much trivial to work out the fundamental group (provided you know the fundamental group of S1vS1v...vS1) That's kinda what I found the most fun about finding homotopy/homology groups - seeing how much I could deform the space into ones I already knew! Obviously it's not always possible, and to be honest, it's often more straightforward not to try and be too clever, just apply SVK to what you have! Nothing like spending an hour sketching silly pictures to finally realise you actually have a very simple space though :v:

[QUOTE=Joey90;42475266]Shame that R3 with points removed isn't very exciting from a fundamental group level...[/QUOTE] Yeah. At least I still got to use induction + SVK. It was a decently fun proof. Now we're done with fundamental group and it's on to homology. I'm excited.

Because no question is stupid, I have a stupid question for you all. Simplify this, with some explanation please. 4/7ln(1/6)+3/7ln(6) [editline]17th October 2013[/editline] Figured it out, glad I embarrassed myself.

I think I finally understand how manifolds are defined. I've only been doing GR for two years now, no big deal. Physicists are awful at explaining mathematical concepts.

[QUOTE=JohnnyMo1;42560500]I think I finally understand how manifolds are defined. I've only been doing GR for two years now, no big deal. Physicists are awful at explaining mathematical concepts.[/QUOTE] my thermodynamics teacher told us matlab can do linear interpolation as a fcn of matlab...... [sp]it doesn't[/sp] i finally had to track down the actual fcn and put it into my calculator...now i understand how to do linear interpolation...... [editline]18th October 2013[/editline] just to clarify, he never actually used matlab in any demonstrations of how to use matlab....he has a mac and for some reason has never heard of bootcamp yet he expects us to use matlab on the tests for linear interpolation

I have never used matlab once in my life. I didn't become a math major to work with numbers, dammit.

The only time I've enjoyed using computers for maths is when looking at dynamical systems, using it to plot the strange attractor of the Henon map was pretty cool.

So should physics posts go in here, or the science thread?

I'd say here.

I guess it depends on how theoretical it is. [editline]18th October 2013[/editline] Sounds about right [IMG]http://i40.tinypic.com/2r3jbqc.png[/IMG]

[URL]http://thatsmathematics.com/mathgen/[/URL] Reminds me of that.

Is there an infinite set of primes of the form {p, (2^p) -1, 2^(2^p - 1) - 1, ...}?

[QUOTE=Krinkels;42570055]Is there an infinite set of primes of the form {p, (2^p) -1, 2^(2^p - 1) - 1, ...}?[/QUOTE] I don't know, but at least probabilistically it seems unlikely - almost all Mersenne numbers are not prime. Generally not a lot is known about Mersenne primes - For starters it's not even known if there are an infinite number of them! So certainly we can't prove that there [i]is[/i] an infinite set like that. In some sense the best candidate would be the Catalan-Mersenne numbers (i.e. p=2) - the first 5 terms are prime, but after that they become so huge that it's infeasible (with current methods) to test if they're prime!

[QUOTE=Joey90;42612461]Generally not a lot is known about Mersenne primes - For starters it's not even known if there are an infinite number of them![/QUOTE] Huh. Here I was thinking we knew that! Crazy.

As a part of my IB diploma, I will soon, as a part of my course work, ensue a "mathematical exploration"-project. I am doing Mathematics at Standard Level. When I finish my IB diploma, I am thinking of studying Computer Science, therefore I thought this would be an excellent chance to do my project on math on some topic which could relate to the field of Computer Science So my question is, if anyone has an idea of what kind of topic I could explore? Thank you!

IB is something you take in secondary school, yes? There's plenty of neat stuff you could do, though I'm not exactly sure what would be relevant to CS. Maybe you could take a look through Concrete Mathematics by Graham, Knuth, and Patashnik. Some of it is a little bit beyond me but there's some interesting stuff in there, and it has computer science in the title. I especially liked the part on Stern-Brocot trees. The work of Alan Turing is also worth mentioning. He did the Enigma machine, the Church-Turing Thesis, and LU-decomposition, and many things I've never even heard of. There are more abstract things you could do too, like set theory, geometry in more than three dimensions, hyperbolic geometry, fractional calculus, or Euler's formula. I have no idea what you'd be interested in unless I had a better idea of why you find computer science interesting, but I'd say a lot of these are worth a look.

I had an abstract algebra exam this morning. The final question was interesting -- let G be an Abelian group. Let H = {g in G | g^2=e}. Prove that H is a subgroup of G. I started working my way through the properties -- associativity, identity, inverses, and closed under the binary operation. Associativity was easy. It'll be inherited. The identity was likewise easy -- by definition, the identity is ae=ea=a for all a in G. Since ee=e^2=e, we know that e is an element of the set H. Inverses followed a similar argument. By definition, the inverse an element a is defined as a^(-1) and has the property that aa^(-1)=a^(-1)a=e. Well, if we let a = g, we know that g^2=e. Therefore, every element is its own inverse. The problem was proving the subgroup was closed. That is to say, for any elements a and b, ab=c and ba=d, where c and d are elements of H. After a few minutes of thought, I figured it out though. We know the group is Abelian, so the subgroup will inherit that. Therefore, ab=c=ba=d, simplify to ab=ba=c. All we have to do is prove that c is in the group. I hit on the idea of squaring the equation. (ab)^2=(ba)^2=c^2. As it is an Abelian group, we can distribute the exponents and rearrange terms: a^2b^2=b^2a^2=c^2. We know that for any element a and b in H, a^2=e, b^2=e. So a^2b^2=b^2a^2=ee=e=c^2, therefore c^2=e. By definition, that would have to be an element of the group H. Thus, H is a subgroup of G. Q.E.D. Thoughts?

Looks good. [editline]24th October 2013[/editline] Why are category theorists so bad at English? They think cocoa is the first letter of the alphabet.

Can anyone who's doing discrete math help me out with this? Let f: R => R be the function defined by f(x) = x^2 - 4X + 6 How would I find the the range and check if the function is onto and/or one to one?

[QUOTE=B!N4RY;42632968]Can anyone who's doing discrete math help me out with this? Let f: R => R be the function defined by f(x) = x^2 - 4X + 6 How would I find the the range and check if the function is onto and/or one to one?[/QUOTE] What level of math is this? I assume proofs aren't necessary? The range is every value the function takes on. You can use the time-honored method of looking at the graph or if you know some calc you can find the minimum and take the limit as x goes to infinity and all. A function is onto if every point in the codomain has something mapped to it by the function. Does every point of R show up in the range? A function is one-to-one if every point on the y-axis has only one x-value at most that gets mapped to it, i.e. if you draw any horizontal line on the graph, it will touch the graph of the function at most once.

This is a university level math course and it's all about proving, so yes I need to proof why they're onto and 1-1 Getting the answer visually by graph isn't an option because of so.