most recent 30 from http://mathoverflow.net 2013-05-22T17:34:42Z http://mathoverflow.net/feeds/user/7043 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32011/direct-proof-of-irrationality/49003#49003 Michael Burge 2010-12-11T01:19:19Z 2010-12-11T01:19:19Z

<p>Most common axiom systems I've seen are a list of $\forall$ and $\exists$ axioms. If you look at a minimal underlying logic, most of the common rules for transforming these axioms shouldn't change the $\forall$ or $\exists$ into a $\neg \exists$. So you could fix a logic system, and argue that the only method that results in a $\neg \exists$ statement is the equivalent of proof by contradiction.</p> <p>You haven't fixed a logic system in your original question, but the "proof by direct substitution" method won't be sufficient.</p>

http://mathoverflow.net/questions/46311/transfinite-sums-related-to-a-sequence Michael Burge 2010-11-17T01:06:15Z 2010-11-17T23:03:01Z

<p>Hello,</p> <p>Given a sequence $S$ indexed by the finite ordinals, a limit ordinal $\alpha$, and $k \in \mathbb{N}$, define $S_{\alpha+k}$(the extension of $S$ to $\alpha+k$) to be the sum over the products of all $k+1$ tuples of elements taken from {$S_\beta| \beta &lt; \alpha$} if the sum exists. Also, define $(S+k)$ for $k \in \mathbb{N}$ to be the sequence <code>$(S+k)_n = S_{k+n}$</code>, essentially "dropping" the terms up to index $k$.</p> <p>For example, $S_\omega$ is the standard sum of a series from Calculus, $S_{\omega+1} = \sum_{n=1}(S_n \cdot \sum_{k=n+1}^\infty S_k)$, and in general $S_{\alpha+1} = \sum_{n=1}S_n (S+n)_\alpha$.</p> <p>I've managed to convince myself it has a few self-consistent properties:</p> <ul> <li>If $S_\alpha$ exists, then $(S+k)_\alpha$ exists.</li> <li>If $S_\alpha$ exists and $\beta &lt; \alpha$, then $S_\beta$ exists.</li> </ul> <p>My questions are:</p> <ol> <li><p>Has this concept been studied before, and if so can you provide references? I'm more concerned about the formal properties of these sequences, such as the algebra it generates. I suspect there's some paper or book on Generating Functions that would cover it. </p></li> <li><p>Given a set S of Complex numbers and an ordinal $\alpha$, you can create a new set $S_\alpha$ containing $s_\beta$ for every $\beta &lt;= \alpha$ and every sequence $s$ of numbers taken from $S$. Is there a paper that gives more detail on this operation? Interesting choices for $S$ might be the set of all roots of unity, or the set of zeros from some analytic function.</p></li> <li><p>Except for the constant 0 sequence, is it true that if you repeatedly transfinitely extend a sequence you will find a large enough ordinal where the sum does not exist? If so, what is the largest ordinal(or the supremum of all such ordinals) possible to extend a sequence to?</p></li> <li><p>My definition should work for countable ordinals, but I'm not entirely sure it's well-defined for anything larger. Is there a more natural/general definition I should be using that captures the concept better?</p></li> </ol> <p>Thank you,</p> <p>-- Michael Burge</p>

http://mathoverflow.net/questions/29982/has-this-notion-of-product-of-graphs-been-studied/30012#30012 Michael Burge 2010-06-30T06:11:08Z 2010-06-30T06:16:40Z

<p>Section 6.3 in "Algebraic Graph Theory" by Chris Godsil and Gordon Royle covers products of graphs; 6.6 covers colorings of these products. I also vaguely remember a reference to it in Douglas West's "Introduction to Graph Theory" when he was constructing a counterexample to something involving colorings, but it's been too long since I took my Graph Theory course out of that book and I couldn't find it.</p>

http://mathoverflow.net/questions/29490/how-many-surjections-are-there-from-a-set-of-size-n/29502#29502 Michael Burge 2010-06-25T11:40:32Z 2010-06-25T12:45:48Z

<p>If f is an arbitrary surjection from N onto M, then we can think of f as partitioning N into m different groups, each group of inputs representing the same output point in M. The Stirling Numbers of the second kind count how many ways to partition an N element set into m groups. But this undercounts it, because any permutation of those m groups defines a different surjection but gets counted the same. There are m! such permutations, so our total number of surjections is</p> <p>m! S(n,m)</p> <p>To look at the maximum values, define a sequence S_n = n - M_n where M_n is the m that attains maximum value for a given n - in other words, S_n is the "distance from the right edge" for the maximum value. Computer-generated tables suggest that this function is constant for 3-4 values of n before increasing by 1. If this is true, then the m coordinate that maximizes m! S(n,m) is bounded by n - ceil(n/3) - 1 and n - floor(n/4) + 1.</p> <p>I have no proof of the above, but it gives you a conjecture to work with in the meantime.</p>

http://mathoverflow.net/questions/46785/a-learning-roadmap-for-additive-combinatorics Michael Burge 2010-11-21T02:35:45Z 2010-11-21T02:35:45Z

Just apply e^x to whatever you're studying and then you'll be doing multiplicative combinatorics, the theory of which is already well-known.

http://mathoverflow.net/questions/46311/transfinite-sums-related-to-a-sequence Michael Burge 2010-11-17T02:56:25Z 2010-11-17T02:56:25Z

You're right; changed to Complex Numbers, but the underlying set doesn't matter too much.

http://mathoverflow.net/questions/45951/sexy-vacuity Michael Burge 2010-11-15T03:07:07Z 2010-11-15T03:07:07Z

&quot;...it will lead you into all sorts of mistakes like 2 + 2 = 5, etc...&quot; But I know all sorts of silly people who believe things like &quot;the Axiom of Choice is false&quot;, yet they've managed to convince themselves they won't run into any mistakes in ZF&#172;C.

http://mathoverflow.net/questions/40011/pythagoras-theorem-euclidean-geometry Michael Burge 2010-09-26T10:57:03Z 2010-09-26T10:57:03Z

A proof using pure fundamental physics would be a monumental breakthrough in the field of Euclidean Geometry. I'm surprised you would even mention it on MO, because people here will surely try to steal your proof and the credit for it. You should try contacting Craig Feinstein; I think he'd be a reliable person to discuss your proof with.

http://mathoverflow.net/questions/8846/proofs-without-words/38664#38664 Michael Burge 2010-09-14T08:47:14Z 2010-09-14T08:47:14Z

How can you be sure that you're eventually covering all the points with irrational or transcendental coordinates? And giving a sequence of curves which fill more and more of the plane isn't the same as giving a single curve that does it all at once - it's not clear that such a limiting curve exists just looking at the pictures.

http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/31914#31914 Michael Burge 2010-08-15T02:19:10Z 2010-08-15T02:19:10Z

There is a particular balance scale that can be used at most 3 times, but the puzzle doesn't say we can't use a different scale!

http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/29568#29568 Michael Burge 2010-07-26T12:30:37Z 2010-07-26T12:30:37Z

&quot;anything outside of a padlocked box is guaranteed to be stolen.&quot; - So if I send a locked box, it'll be stolen? It is, after all, not inside a locked box.

http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/29576#29576 Michael Burge 2010-07-25T12:29:56Z 2010-07-25T12:29:56Z

Wow - love the solution to that first ant problem in the Wordpress link, Gil Kalai. Very elegant.

http://mathoverflow.net/questions/32397/vector-spaces-without-natural-bases/32398#32398 Michael Burge 2010-07-19T03:33:58Z 2010-07-19T03:33:58Z

I think you need a little bit more than this: Defining said basis and proving that your definition actually works are different things. Hypothetically, you could define a very natural basis for some vector space V, and then only use AC at the very end to show V was actually all of R.

http://mathoverflow.net/questions/30659/establishing-zeta3-as-a-definite-integral-and-its-computation/30698#30698 Michael Burge 2010-07-06T02:00:02Z 2010-07-06T02:00:02Z

Very interesting - is there a nice expository paper for these motives? Why are the $\zeta(2n)$ values so easy compared to the $\zeta(2n+1)$ values, in these geometric terms?

http://mathoverflow.net/questions/30143/is-there-a-name-for-the-projection-of-a-function-under-argmax Michael Burge 2010-07-01T06:31:58Z 2010-07-01T06:31:58Z

I think it's the &quot;argument that maximizes&quot;; that is, fix a value of x and ask which y leads to the largest value of f(x,y).

http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/29329#29329 Michael Burge 2010-06-25T14:42:31Z 2010-06-25T14:42:31Z

This is probably my favorite puzzle. Anybody who hasn't seen this before should really work it out; it's a great one.