[QUOTE=JohnnyMo1;40494853]Maybe only Joey could answer this, but are Smale's paradox and the Poincare conjecture basically statements about the existence of homotopies, i.e. there is a homotopy between the 2-sphere and the 2-sphere of the opposite orientation and there are homotopies between any closed, simply connected 3-manifold and the 3-sphere?
[editline]1st May 2013[/editline]
Not sure about Poincare, but Smale's paradox definitely seems like it could be phrased as the 2-sphere and the inside-out 2-sphere are homotopic.[/QUOTE]
I've not really studied either in any detail (maybe not surprising given that they're quite tricky!) but I'll say what I think I know... For starter's they're both to do with differential topology, which is a bit harder than algebraic topology (since you have to worry about silly stuff like smoothness and derivatives :v: ) but I don't think the statements require you to know more than very basic differential geometry.
Smale's Paradox is not [i]quite[/i] that simple, although intuitively it just asks if you can deform the sphere inside out without creating cusps/singularities.
More precisely I think it asks if there is is a [i]regular[/i] homotopy between the immersion of the 2-sphere and the inverted 2-sphere in R[sup]3[/sup]...
You need to be careful to distinguish a homotopy (between 2 maps) and a homotopy equivalence (between 2 spaces) - although you might say a space X is homotopic to Y, what you (presumably) really mean is that they're homotopy equivalent. Usually embedding everything in R[sup]n[/sup] and imagining deforming one space into the other is good enough, but that's not technically quite the same. It's trivially obvious that the sphere and 'inverted' sphere are homotopy equivalent, since the orientation doesn't mean anything to a homotopy!
The other thing to note is the word regular... I'm not 100% sure but I think it means that the homotopy must always be an immersion (i.e. injective derivative) - which is why you can't have any singularities or cusps when you're turning it inside-out. (Otherwise you could just squash it down to a point and blow it back up again in reverse)
All this boils down to deforming the sphere to the inverted sphere only through R[sup]3[/sup] (the homotopy part), and only in such a way that there are no cusps or singularities at any point (the regular part).
The Poincaré conjecture is a statement about the existence of a homeomorphism between spaces, which is of course much stronger than just a homotopy. Indeed, a weaker (but apparently equivalent - not that I know of a proof) form of it says:
'If a compact 3-manifold is [i]homotopy equivalent[/i] to the 3-sphere, it is actually [i]homeomorphic[/i] to the 3-sphere'
If you know about compact 2-manifolds, you'll know that they have a very nice classification - the orientable ones are just spheres with n holes (e.g. n=0 is the sphere, n=1 is the torus etc.) - hence if you have a compact 2-manifold with no 'holes' (trivial fundamental group) it [i]must[/i] be homeomorphic to a 2-sphere.
The Poincaré conjecture is trying to say the same thing about compact 3-manifolds - if it doesn't have any 'holes' is it homeomorphic to the 3-sphere, or is there some other exotic space which it could be?
[QUOTE=Joey90;40495941]I've not really studied either in any detail (maybe not surprising given that they're quite tricky!) but I'll say what I think I know... For starter's they're both to do with differential topology, which is a bit harder than algebraic topology (since you have to worry about silly stuff like smoothness and derivatives :v: ) but I don't think the statements require you to know more than very basic differential geometry.
Smale's Paradox is not [i]quite[/i] that simple, although intuitively it just asks if you can deform the sphere inside out without creating cusps/singularities.
More precisely I think it asks if there is is a [i]regular[/i] homotopy between the immersion of the 2-sphere and the inverted 2-sphere in R[sup]3[/sup]...
You need to be careful to distinguish a homotopy (between 2 maps) and a homotopy equivalence (between 2 spaces) - although you might say a space X is homotopic to Y, what you (presumably) really mean is that they're homotopy equivalent. Usually embedding everything in R[sup]n[/sup] and imagining deforming one space into the other is good enough, but that's not technically quite the same. It's trivially obvious that the sphere and 'inverted' sphere are homotopy equivalent, since the orientation doesn't mean anything to a homotopy!
The other thing to note is the word regular... I'm not 100% sure but I think it means that the homotopy must always be an immersion (i.e. injective derivative) - which is why you can't have any singularities or cusps when you're turning it inside-out. (Otherwise you could just squash it down to a point and blow it back up again in reverse)
All this boils down to deforming the sphere to the inverted sphere only through R[sup]3[/sup] (the homotopy part), and only in such a way that there are no cusps or singularities at any point (the regular part).
The Poincaré conjecture is a statement about the existence of a homeomorphism between spaces, which is of course much stronger than just a homotopy. Indeed, a weaker (but apparently equivalent - not that I know of a proof) form of it says:
'If a compact 3-manifold is [i]homotopy equivalent[/i] to the 3-sphere, it is actually [i]homeomorphic[/i] to the 3-sphere'
If you know about compact 2-manifolds, you'll know that they have a very nice classification - the orientable ones are just spheres with n holes (e.g. n=0 is the sphere, n=1 is the torus etc.) - hence if you have a compact 2-manifold with no 'holes' (trivial fundamental group) it [i]must[/i] be homeomorphic to a 2-sphere.
The Poincaré conjecture is trying to say the same thing about compact 3-manifolds - if it doesn't have any 'holes' is it homeomorphic to the 3-sphere, or is there some other exotic space which it could be?[/QUOTE]
That's interesting. I don't know why it didn't occur to me that orientation is irrelevant to a homotopy, but now it seems obvious. (probably because it's been >1 year since I did anything with differential geometry) I was going to use the term "homotopy equivalent" in one of my statements in my original post, I don't remember which one, but I switched to homotopic when I realized that homotopy equivalence is different from being homotopic and I don't actually know what homotopy equivalence is. Munkres doesn't actually talk about homotopy equivalence, so we haven't covered it.
But thanks for a rather complete answer!
[QUOTE=JohnnyMo1;40496621]That's interesting. I don't know why it didn't occur to me that orientation is irrelevant to a homotopy, but now it seems obvious. (probably because it's been >1 year since I did anything with differential geometry) I was going to use the term "homotopy equivalent" in one of my statements in my original post, I don't remember which one, but I switched to homotopic when I realized that homotopy equivalence is different from being homotopic and I don't actually know what homotopy equivalence is. Munkres doesn't actually talk about homotopy equivalence, so we haven't covered it.
But thanks for a rather complete answer![/QUOTE]
Strange that you do homotopies but not homotopy equivalences.
One way of looking at it is like a weakening of the idea of a homeomorphism.
For a homeomorphism you need f:X->Y and g:Y->X such that fg = 1 and gf=1
For a homotopy equivalence you need f:X->Y and g:Y->X such that fg ~ 1 and gf ~ 1 (where ~ means 'homotopic to')
(Kinda like how if you have two categories they are 'isomorphic' if there are functors F:C->D and G:D->C such that both composites are equal to the identity, but 'equivalent' if the composites are only naturally isomorphic to the identity)
As an intuitive thing, two spaces are homotopy equivalent if you can continuously deform one into the other. It's also not too difficult to see that two homotopy equivalent spaces have the same homotopy and homology groups.
We haven't done homologies either, unfortunately, and that was one of the topics I've been looking forward to in algebraic topology. Munkres baaarely touches on the first homology group, it's kind of disappointing. I have Hatcher's [I]Algebraic Topology[/I] (since it's available legally for free online!) but I've yet to go through it. Maybe I'll delve in over the summer.
[editline]2nd May 2013[/editline]
Wait, I lied. Munkres does do homotopy equivalence, we just haven't got there. Perhaps we will in the two classes before the end of the semester. :v:
[QUOTE=JohnnyMo1;40503121]We haven't done homologies either, unfortunately, and that was one of the topics I've been looking forward to in algebraic topology. Munkres baaarely touches on the first homology group, it's kind of disappointing. I have Hatcher's [I]Algebraic Topology[/I] (since it's available legally for free online!) but I've yet to go through it. Maybe I'll delve in over the summer.
[editline]2nd May 2013[/editline]
Wait, I lied. Munkres does do homotopy equivalence, we just haven't got there. Perhaps we will in the two classes before the end of the semester. :v:[/QUOTE]
That's a shame :/
Although homology groups require a lot more set up to show they're well defined etc. once you actually start calculating stuff with them they're much nicer than homotopy groups. (Provided you like exact sequences!)
Yeah, I spoke to my professor today and he said it usually takes a semester to build up homology theory. There's a grad-level algebraic topology course starting next semester and I'm gonna try to audit it if I can. It's a year long and I'm graduating the fall so I'll only catch the first half though. Also, you can't take a class for credit after you audit it here and I'll need it if I stay for grad school so I'm gonna speak to the professor first and see if there's some way around that. (there's 5 people in the class and 40 seats at the moment so I may just be able to sit in :v:)
You've probably answered this before, but what are you majoring in, Johnny?
Physics and math, with physics being my primary major. I picked up the math major because I'd like to study string theory in grad school.
[QUOTE=JohnnyMo1;40507399]Physics and math, with physics being my primary major. I picked up the math major because I'd like to study string theory in grad school.[/QUOTE]
It's pretty cool that you're into stuff like Algebraic Topology and Category Theory even though you're doing physicsy stuff too... Most of the people here are very polarised - either entirely Theoretical Physics or entirely Pure!
[QUOTE=Joey90;40508080]It's pretty cool that you're into stuff like Algebraic Topology and Category Theory even though you're doing physicsy stuff too... Most of the people here are very polarised - either entirely Theoretical Physics or entirely Pure![/QUOTE]
I was pretty amazed when I was going through a paper from one of the string theorists here (the one I'd probably like to study under if I get into a PhD program here) and saw that category theory was involved. It just makes me more excited to study it.
Hello there, I'm having a small problem here with a matrix exercise. I have to prove that, knowing that the matrix E from Mn(IR) is representative of a projection endomorphism, if its rank r is such that 0<r<n, E is similar to the block matrix:
(Ir 0
0 0)
So far, I have stated that, knowing that E is of rank r, there are T € Mr(IR) and Z € Mn-r,r(IR) such that E is similar to M=
(T Z
0 0)
Then, using that E is representative of a projection, E²=E and as a result, M²=M, which leads to the equations: T²=T and ZT=Z. Of course, T=Ir and Z=0 is a solution to that, but nothing states it is the only solution.
I'm not really sure if what I'm doing here is correct, and I don't know if it is of any use for the proof. Perhaps there's a better way of proving it.
[QUOTE=_Axel;40512932]Hello there, I'm having a small problem here with a matrix exercise. I have to prove that, knowing that the matrix E from Mn(IR) is representative of a projection endomorphism, if its rank r is such that 0<r<n, E is similar to the block matrix:
(Ir 0
0 0)
So far, I have stated that, knowing that E is of rank r, there are T € Mr(IR) and Z € Mn-r,r(IR) such that E is similar to M=
(T Z
0 0)
Then, using that E is representative of a projection, E²=E and as a result, M²=M, which leads to the equations: T²=T and ZT=Z. Of course, T=Ir and Z=0 is a solution to that, but nothing states it is the only solution.
I'm not really sure if what I'm doing here is correct, and I don't know if it is of any use for the proof. Perhaps there's a better way of proving it.[/QUOTE]
A projection is defined to be a matrix E such that E[sup]2[/sup] = E right? Showing that it is similar to the block matrix is the same as finding a basis for which the matrix has that form.
The crucial part is showing that R[sup]n[/sup] is a direct sum of U = the image of E and V = the kernel of E.
The dimensions of U and V are r and n-r respectively which add up to n, so we only need to show that they intersect trivially. Suppose v was a vector in the intersection... since it is in the image of E, we can write v = E(w) for some w, and now E[sup]2[/sup] = E so v = E[sup]2[/sup](w) = E(E(w)) = E(v), but now v is also in the kernel, so v = E(v) = 0!
Now we just pick a basis for U and a basis for V, since R[sup]n[/sup] is the direct sum of these this forms a basis for the whole space, and then with respect to this basis E has the form you want.
[QUOTE=Joey90;40514743]A projection is defined to be a matrix E such that E[sup]2[/sup] = E right? Showing that it is similar to the block matrix is the same as finding a basis for which the matrix has that form.
The crucial part is showing that R[sup]n[/sup] is a direct sum of U = the image of E and V = the kernel of E.
The dimensions of U and V are r and n-r respectively which add up to n, so we only need to show that they intersect trivially. Suppose v was a vector in the intersection... since it is in the image of E, we can write v = E(w) for some w, and now E[sup]2[/sup] = E so v = E[sup]2[/sup](w) = E(E(w)) = E(v), but now v is also in the kernel, so v = E(v) = 0!
Now we just pick a basis for U and a basis for V, since R[sup]n[/sup] is the direct sum of these this forms a basis for the whole space, and then with respect to this basis E has the form you want.[/QUOTE]
I see, thanks a lot ! I had already established the direct sum of ImE and kerE knowing that it was what we usually did when presented with a projection. I was unsure if it was sufficient to prove that ImE inter kerE was {0}, but as you said the rank theorem gives information about dimensions.
Didn't think of picking a basis for each set, though. Didn't see the link between the direct sum and combining both basis to form a basis of Rn. So let's say (e1,...er) is the basis of ImE and (er+1,...en) the basis of KerE, the image of each vector of KerE is null by definition,which gives us the 0 on the right of the E matrix, and knowing that E²=E, E applied to any vector of ImE is identity, which gives us the Ir on the left, right ?
Since I said I would, [url=https://dl.dropboxusercontent.com/u/4081470/FactorisationSystems.pdf]here[/url]'s the essay/paper thing that I've been working on (I've submitted it now)... I don't really expect anyone to read it, and it won't mean much unless you know some elementary category theory!
Also, if there are any mistakes, please don't tell me - it's too late to fix them :v:
I'll give it a read most likely, probably not for a week or so. End of the semester, exams and all that.
Tina is playing with a tub of building blocks. The tub contains 3 red blocks, 5 blue blocks, 2 yellow blocks, and 4 green blocks. How many different ways can Tina stack the blocks in a single tower in each situation below?
A) There are no conditions
I am trying to figure this out, I though with zero conditions at all, I would just add the number of blocks in total, 14, and solve the problem with 14P14. That isn't right.
What the heck am I forgetting? I've done questions like these before, in success, my brain sucks for remembering how to do math!
Okay, I have been adding instead of multiplying when trying that approach. I will try that one out and edit with whether or not it's correct. I'm sure it will be right.
3P14*5P11*2P6*4P4 is incorrect, gives a math error. I assumed it was because we were using the smaller numbers P larger numbers.
I tried reversing the numbers, ie, 14P3 ... This did not give me the solution I am looking for, which is 2 522 520
You should be using C not P, and yes, you conventionally put the larger number first.
The reason is that it only matters (for example) which 3 spaces the red blocks take, but not which individual red block goes where (the implication is that all blocks of the same colour are identical)
I got 40/40 on my last topology homework for the semester...
...but my professor took off epsilon points on a problem ;_;
I wish my Calculus teacher was cool like that.
Every time I submit my assignments (online) all she ever does is glance at it to make sure I *showed my work*, gives me a 100%, and says Good job, without fail.
There've been a few times where I was confused and unsure as to whether or not I was doing any of the problems right, and made that clear in my submission, only to have it ignored.
I think of all the core courses, math teachers should be the most involved.
Also, turns out we're doing some more basic work with differential equations just before my final exam. This is going to be an interesting few weeks...
Hey, speaking of differential equations, could someone explain this problem to me?
(I'm supposed to separate and solve this separable differential equation for y)
dy/dx = 4ye^(5x)
Now, I was under the assumption that I'd separate it as 1/4y dy = e^5x dx and then integrate for an implicit solution of y^2/8 = e^5x/5 but that gives a really strange answer in terms of y, that doesn't seem to be right. When I try to solve it with Wolfram, it takes the problem as 4e^5x y(x) regardless of how I input it, and the step by step solution doesn't make any sense to me as it separates just y, and takes the integral of y as ln(y) instead of 1/2y^2 like it should, so I'm thinking their answer is off also.
Any help would be greatly appreciated, I have a love hate relationship with these equations.
I think the solution is e^(4/5*e^5x). Unless I've made a mistake verifying it, it should work. I've done it just like you did, except I integrated 1/4y into lny/4. I think you may have thought the y was in the numerator instead of the denominator, I do that kind of mistake all the time.
Whoops, forgot the constant.
Shouldn't it simplify to (e^c1)*(e^(e^5x))^4/5 instead ? Unless I'm reading it wrong.
To be honest my version was fucked up too apparently. Oh well v:v:v
[editline]7th May 2013[/editline]
Wait, no, it's just the constant being a different one.
Hey guys, just want to ask you if you think this proof looks alright:
Prove that the following series is convergent:
[img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty}%20\frac{\sqrt{1-e^{-k}}}{3^k}[/img]
I started off using the following theorem:
Given a convergent series [img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty%20}t_{n}[/img] and given that Sn<=C*Tn for some constant C then [img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty%20}s_{n}[/img] is also a convergent series.
Then I had:
[img]http://latex.codecogs.com/gif.latex?\frac{\sqrt{1-e^{-k}}}{3^k}%20\leq%20\sqrt{1-e^{-k}}%20\leq%201-e^{-k}[/img]
[img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty%20}1-e^{-k}%20=%201-\sum_{k=0}^{\infty}(\frac{1}{e})^k%20=%201-\frac{1}{1-\frac{1}{e}}[/img]
Therefore that series is convergent.
Hence since [img]http://latex.codecogs.com/gif.latex?1-e^{-k}%20\geq%20\frac{\sqrt{1-e^{-k}}}{3^k}[/img] the original series must be convergent.
Is this okay as a proof? I don't think its wrong since I've covered this specific theorem in lectures. My professor used partial sums in the solution, is that a better method?
the sum(1 - e[sup]-k[/sup]) = 1 - sum([1/e][sup]k[/sup]) step is invalid
Also, when you drop the sum sign to show those few inequalities on the functions, restrict the domain of the function. As it stands you have a potentially negative function under that square root.
What's a good site to use for college liberal arts mathematics?
And yes you can rate me dumb, I failed it because I couldn't do mortgage stuff.
I could do algebra, but I'm scared and don't want to fail anymore so I'm retaking liberal arts.
And I want to make sure I pass, so any suggestions for sites?
[QUOTE=agentalexandre;40621813]Hey guys, just want to ask you if you think this proof looks alright:
Prove that the following series is convergent:
[img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty}%20\frac{\sqrt{1-e^{-k}}}{3^k}[/img]
I started off using the following theorem:
Given a convergent series [img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty%20}t_{n}[/img] and given that Sn<=C*Tn for some constant C then [img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty%20}s_{n}[/img] is also a convergent series.
Then I had:
[img]http://latex.codecogs.com/gif.latex?\frac{\sqrt{1-e^{-k}}}{3^k}%20\leq%20\sqrt{1-e^{-k}}%20\leq%201-e^{-k}[/img]
[img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty%20}1-e^{-k}%20=%201-\sum_{k=0}^{\infty}(\frac{1}{e})^k%20=%201-\frac{1}{1-\frac{1}{e}}[/img]
Therefore that series is convergent.
Hence since [img]http://latex.codecogs.com/gif.latex?1-e^{-k}%20\geq%20\frac{\sqrt{1-e^{-k}}}{3^k}[/img] the original series must be convergent.
Is this okay as a proof? I don't think its wrong since I've covered this specific theorem in lectures. My professor used partial sums in the solution, is that a better method?[/QUOTE]
Unfortunately there are a few issues...
[img]http://latex.codecogs.com/gif.latex?\sqrt{1-e^{-k}}%20\leq%201-e^{-k}[/img]
Is false if k > 0 - since the thing under the square root is actually between 0 and 1, so square rooting increases it's value (e.g. the square root of 1/4 is 1/2)
And as Johnny says
[img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty%20}1-e^{-k}%20=%201-\sum_{k=0}^{\infty}(\frac{1}{e})^k%20[/img]
Is also completely false, firstly you can only break apart sums like that if you know each part converges, and far more importantly you go from an infinite sum of 1's to just a single 1!
I think you have basically the right method, but the wrong inequality. From what I said, you have
[img]http://latex.codecogs.com/gif.latex?0%20\leq%20\sqrt{1-e^{-k}}%20\leq%201\quad%20\forall%20k\geq%200[/img]
Hence
[img]http://latex.codecogs.com/gif.latex?\sum_{k=0}^{\infty}%20\frac{\sqrt{1-e^{-k}}}{3^k}\leq\sum_{k=0}^{\infty}%20\frac{1}{3^k}[/img]
And you should be able to show the right hand side converges, and since all terms are positive this means the original sum must converge [i]absolutely[/i].
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