bio | website | gowers.wordpress.com |
---|---|---|
location | ||
age | ||
visits | member for | 5 years, 9 months |
seen | May 22 at 10:59 | |
stats | profile views | 32,635 |
Mathematics professor at Cambridge
Jun 1 |
comment |
Why is differentiating mechanics and integration art?
The reason, I think, is that the analogy between the two situations isn't all that good. When you solve numerically you are trying to find a number. That coincides with what you are looking for when you solve a quadratic equation, whereas when you are antidifferentiating you are looking for a function, and numerical methods just give you function values. |
May 31 |
comment |
Why is differentiating mechanics and integration art?
I'm not sure I buy your argument that the linearization "is somehow a simplification". True, linear functions are simple, but the linearization we perform when differentiating varies from point to point in a nonlinear way, and it's the entire derivative we're interested in here rather than just the derivative at a point. (This is particularly clear if we look at functions of more than one variable, when the derivative is a decidedly more complicated object than the original function.) |
May 25 |
comment |
nontrivial theorems with trivial proofs
Some might call that a trivial theorem with a non-trivial proof ... |
May 25 |
comment |
n-partite n-clique
I haven't checked, but it seems likely to me that a random graph will be a counterexample: you need a very high edge probability to get an n-clique and I think probably a lot lower to satisfy your conditions with high probability. |
May 25 |
comment |
n-partite n-clique
FWIW here is a reformulation (I think). Let G be a graph with vertex set the edges of the complete bipartite graph K(n,n). Suppose that each vertex of G is contained in a perfect matching in K(n,n) and is joined to all the other edges in that perfect matching. Prove that G contains an n-clique. |
May 25 |
comment |
Is anything known about this braid group quotient?
In a very elementary way I observed that when thinking about it. Using the third relation (the one that takes a string and passes it round all the other strings) and applying it to each string in turn, you get not a single twist but a double twist. |
May 23 |
comment |
Is anything known about this braid group quotient?
Looking again at the paper by Gillette and van Buskirk, I see that (if I understand correctly) my group is the mapping class group but not the spherical braid group: the mapping class group adds an extra relation that allows you to twist the entire bottom of the braid through a full turn. So it's really the mapping class group that I'm interested in, though obviously the two are closely related. |
May 23 |
comment |
Is anything known about this braid group quotient?
Ian Agol says in a comment below that Mosher showed that mapping class groups are automatic and hence that there is a polynomial-time algorithm for the word problem. Am I missing something here, or does that mean that the problem you mention was solved in 1995 (the date of Mosher's paper)? |
May 18 |
comment |
Efficient computation of the least fraction with square denominator greater than the square root of 2.
Is it really the very best approximation you want to find, or would you be content with a non-trivially good one? (I'm not saying I have an answer even to that, but it feels quite a lot more feasible.) |
May 14 |
comment |
Almost orthogonal vectors
Almost a year later I've just seen your comment. I had in mind that each point you put in rules out a spherical cap of spherical radius $\pi/2-\epsilon$. Since this has exponentially small volume, you can just greedily add exponentially many points. I think we must be talking at cross purposes though ... |
May 14 |
comment |
A Bijection Between the Reals and Infinite Binary Strings
I should add that you need to encode whether the digit is before or after the decimal point. |
May 14 |
comment |
A Bijection Between the Reals and Infinite Binary Strings
How about just encoding the non-terminating decimal expansion as a sequence of 0s and 1s digit by digit? |
May 13 |
comment |
Rolling-ball game
I've just tried to find a proof of finiteness by using the pigeonhole principle to say that there are two long bits of path that take the same steps and start in almost exactly the same place with almost exactly the same orientation. It feels promising but I haven't managed to push it through. |
May 13 |
comment |
Expectation of Gowers norm
Try thinking about the expectation of the $2^k$th power of the norm. |
May 11 |
comment |
Is anything known about this braid group quotient?
I have seen that survey and I like it very much. Perhaps you can answer another question: is the list of methods for solving the word problem given in that paper essentially complete (in the sense that all methods are small variants of one of the methods mentioned there)? |
May 11 |
comment |
Probability Problem Involving e
Any reason you go for that rather than the umbrella problem? |
May 11 |
comment |
Probability Problem Involving e
One thought is generating functions. If you define f(m,n) to be the expected maximum if you start with m whole pills and n half pills, then f(m,n)=(m/n)f(m-1,n+1)+(1-m/n)f(m,n-1), except that when m or n is zero then you have to change the right hand side. Whether that leads to anything useful I don't immediately see. |
Apr 21 |
comment |
Examples of eventual counterexamples
@Vectornaut, while I think your point is valid, it needs to be adjusted slightly, because the sequence is far from random. For example, in a pattern like that you won't get any primes unless the final digits are odd, and that increases the chance that any individual term is prime, which in turn decreases by quite a bit the chance that 137 terms are composite. |
Apr 11 |
comment |
Why do Bernoulli numbers arise everywhere?
@Alex, I take it that by "simply connected" you mean "connected in a simple way". |
Apr 10 |
comment |
Proofs without words
If you look at the picture in detail you can see that you are defining a sequence of continuous functions that converge uniformly. It's also clear from the picture that the image is dense. Therefore the limiting function exists and its image (being dense and compact) is the whole square. Of course, this proof isn't 100% visual but the non-visual part -- the basic facts about uniform convergence and compactness -- can be regarded as background knowledge. So I think it's a nice example. |