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visits | member for | 5 years |
seen | Sep 21 '12 at 5:09 | |
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Jan 27 |
awarded | Yearling |
Jun 25 |
awarded | Revival |
Jun 25 |
awarded | Excavator |
Jun 25 |
awarded | Pundit |
Jan 27 |
awarded | Yearling |
Sep 20 |
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How many more join-irreducibles can there be in a sub join-semilattice of a finite lattice?
This has a smell of Frankl's union closed sets conjecture about it. Gerhard "Ask Me About System Design" Paseman, 2012.09.20 |
Sep 20 |
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Cylinders dividing $\mathbb{R}^{3}$
@VCF: I think the situation is more complex than that. However, I will cast my vote and then move on. I have learned that Joseph is rather unaffected by the vote, and I appreciate the opportunity to contribute to the question and to disagree with each of you and Joseph. I look forward to more questions as well as more progress on this question. Gerhard "Agreeing To Disagree Agreefully; Agreed?" Paseman, 2012.09.20 |
Sep 20 |
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Origin of square-and-multiply algorithm
To be clear, square for even exponents, square and multiply one more for an odd exponent. Gerhard "Or Some Other Variation Thereupon" Paseman, 2012.09.20 |
Sep 20 |
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Discrete orderings on polynomial rings that violate the universal theory of the integers
If all the axioms of an ordered ring are to be satisfied, then I think there are few choices, as (if I understand correctly) the basic orders that are discrete are determined by the order of x,y and the integers. So I suggest no such order exists that will not satisfy the universal theory. Gerhard "Ask Me About System Design" Paseman, 2012.09.20 |
Sep 20 |
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Discrete orderings on polynomial rings that violate the universal theory of the integers
For clarification, can someone explain why some version of lexicographic order might (or might not) work? I am thinking such an order might be where any polynomial that has a monomial containing y is greater than any polynomial that has no y whatsoever. (If on the other hand, all such orders have to respect the order on Z, then I think it unlikely such an order will be found.) Gerhard "Ask Me About System Design" Paseman, 2012.09.20 |
Sep 20 |
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Cylinders dividing $\mathbb{R}^{3}$
Given the trouble Joseph has taken to help (and to further the spirit of contention), I disagree with your downvote, VCF. I will return to this later today with my voting account and give it my seventh (sixth?) vote. Gerhard "Miss Manners Might Also Disapprove" Paseman, 2012.09.20 |
Sep 20 |
answered | What do we call a set that has one or fewer elements? |
Sep 20 |
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Subwords of cube-free binary words
The number of cubefree words of length 30 is less than twice the 29th Fibonacci number, which is less than 10*7^6, so it is believable. Gerhard "Ask Me About Crude Estimates" Paseman, 2012.09.20 |
Sep 20 |
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Subwords of cube-free binary words
added 1331 characters in body |
Sep 19 |
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Subwords of cube-free binary words
Here is a pseudo-elegant suggestion. Suppose you look at cube free words that avoid 010. Then all blocks after the initial block will be 0, 00, or 11. Cube-free words will then have a suffix having mostly 11's. But there are only finitely many ways to avoid cubes with this restriction. Argue similarly for the other five subwords. Gerhard "I'll Take Working Over Elegant" Paseman, 2012.09.19 |
Sep 19 |
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Cylinders dividing $\mathbb{R}^{3}$
And if you are gracious enough to humor me, how many regions do you get from a slight longitudinal rotation to a 3x1x1 cylinder? Gerhard "I'm Thinking You Get Ten" Paseman, 2012.09.19 |
Sep 19 |
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Cylinders dividing $\mathbb{R}^{3}$
I hope you don't mind this method of asking, Joseph. VCF (in a comment to my answer) posed the possiblity of up to 8 regions with two congruent ellipsoids. For the drama of contention (and through lack of imagination) I say six. Do you know what the answer is? Gerhard "Thank You For Your Attention" Paseman, 2012.09.19 |
Sep 19 |
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Cylinders dividing $\mathbb{R}^{3}$
VCF: I think you should ask Joseph's help with this one. The cylinder one was easy because I could start with two unit cylinders to get 10 regions, and then lengthen each. For the ellipsoids, I can't do that. I think 6 is the most, but Joseph has programs that can likely resolve this for you. Gerhard "My Thinking Isn't That Twisty" Paseman, 2012.09.19 |
Sep 19 |
answered | Multivariable Calculus Lecture Ideas |
Sep 19 |
answered | Subwords of cube-free binary words |