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May
22
comment A series involving harmonic numbers
Some of these symbolic algorithms have "certification methods" attached to them -- e.g., the Wilf-Zeilberger method. There are some add-on packages for Mathematica that show such certificates. This is probably not what you mean exactly by "derivation method". But the fact is that it does become a fully automated proof. Some computer algorithms allow such certifications, others do not. Once should always check numerically that the symbolic result returned agrees with the numeric approximation, of course.
May
19
comment free groups of rotations
So that assertion is too strong. What about this: If A and B are noncommuting rotations of infinite order, then G, the group they generate, has a free subgroup of rank 2.
May
19
comment free groups of rotations
Many thanks. It might be nice to have a specific examples of the 3x3 matrices that do not commute but satisfy a word such as you suggested, but I don't really need that. It is enough to know that the theorem is false, and so this adds to the value of the many theorems that say that certain pairs of rotations are independent.
May
18
comment free groups of rotations
30 years on, I am working (with G. Tomkowicz) on revising the BTP book. Big job. Almost finished. That is where this question arose.
Dec
21
comment If $X$ is $G$-paradoxical then $G$ is $G$-paradoxical. Is my proof correct?
Sorry if I did not read it in detail, but I guess all is well. And I guess my old hint is acceptable. But, as noted, we are now working on a second edition so if you see anything that is obscure, or could benefit from an elucidating phrase, (or is wrong!) please tell me. You can find my e-address atop by web page, stanwagon.com
Dec
16
comment free group generated by polynomials
Thanks to all for the quick and complete answers!
May
28
comment Fractional Edge Coloring of Snarks
Excellent. Many thanks.
Feb
9
comment Why can a Venn diagram for 4+ sets not be constructed using circles?
Given that it is an MAA journal (nonprofit organization) I don't mind and I doubt anyone would. Still they do own the copyright, so this it not a trivial question. But I won't worry about it.
Dec
9
comment Fractional chromatic number of fullerenes
Checking over my database I found that 3 of the 100-200 fullerenes I examined did not have 5/2... So this is intriguing. If your proof, Angela, holds water, then there will be something funny about these 3 cases. I double-checked with others that my fractional chi is correct on these 3. Example:
Dec
5
comment Fractional chromatic number of fullerenes
The ones I checked are up to 50 vertices. I am told that perhaps things don't get interesting until they are much larger, so maybe there is nothing to this "conjecture" about 5/2.