Activity for ziggurism
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #290305 |
Yes, the 11- and 57- cells apparently do have those very high dimensional realizations. And apparently with the right definitions every abstract polytope has a realization? Ok, but I would expect the general realization of a general abstract polytope to be pretty divorced from their geometry.
Wha... (more) |
— | 11 months ago |
| Comment | Post #290318 |
the set of normal numbers in $[0,1]$ is measure 1. Which means the non-normal numbers are a null set. Of course, contradictorily, the set of numbers which we can prove to be normal is probably countable. (more) |
— | 11 months ago |
| Edit | Post #290305 |
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— | 11 months ago |
| Edit | Post #290305 |
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— | 11 months ago |
| Comment | Post #290305 |
Ok I have read some of the sources linked on wikipedia, like [these two](https://people.eecs.berkeley.edu/~sequin/PAPERS/2007_SIGGRAPH_57Cell.pdf) and [pdfs](https://people.eecs.berkeley.edu/~sequin/PAPERS/2007_ISAMA_11Cell.pdf) by Sequin and Hamlin, and a [related powerpoint] and [this](https://en.w... (more) |
— | 11 months ago |
| Comment | Post #290305 |
ok if that's what the atlas link shows, then my answer is completely wrong. I notice that the atlas link lists neither the 11-cell (schäfli symbol {3,5,3}) nor the 57-cell (schäfli symbol {5,3,5}) so I don't know what to make of that list (more) |
— | 11 months ago |
| Comment | Post #290305 |
right, i think i may have overlooked the class of projective polytopes. thanks for the correction. But the point remains that unshackling yourself from any specific geometry unlocks two new regular polytopes. (more) |
— | 11 months ago |
| Edit | Post #290305 |
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— | 11 months ago |
| Edit | Post #290305 | Initial revision | — | 11 months ago |
| Answer | — |
A: What is special about the 11-cell and 57-cell? The abstract polytopes which admit realizations are special, and the very concept of an abstract polytope exists to generalize and categorize the polytopes first understood through their realizations. The classical proof by Euclid that there are only five regular convex polyhedra is considered so ... (more) |
— | 11 months ago |
| Comment | Post #290291 |
i tried to answer the question "why is homological orientation equivalent to the smooth definition of orientation". But upon re-reading the question, I'm wonder whether what you're really asking about is Betti numbers? (more) |
— | 11 months ago |
| Comment | Post #290291 |
OP asked for a first principles explanation, but i have left a lot of details vague in my answer. It is not complete at this stage and I would like to come back later and flesh it out. (more) |
— | 11 months ago |
| Edit | Post #290291 | Initial revision | — | 11 months ago |
| Answer | — |
A: Is the nth Betti number determined by orientability? If you are an algebraic topologist then you might take $\betan(M)=1$ as the definition of orientable. So to answer your question, we first have to decide on a different definition of "orientable" than $\betan(M)=1.$ Usually the alternative notions of orientability are calculus based, about either ... (more) |
— | 11 months ago |
| Comment | Post #287820 |
the Lagrange interpolation formula is an explicit formula for the polynomial of degree $n$ matching $n$ given values. https://en.wikipedia.org/wiki/Lagrange_polynomial (more) |
— | almost 2 years ago |
| Comment | Post #287674 |
You may think that 60 is too unwieldy a number to count to, to learn the positional number system, but if you're willing to use a sub-base, like apparently Kaktovik does, so that needn't be a problem.
According to wikipedia, the motivating reason for using Kaktovik is compatibility with Inupiaq la... (more) |
— | almost 2 years ago |
| Comment | Post #287674 |
Apparently the the Kaktovik numerals are a base 20 system with a sub-base of 5. Personally changing my own numeral system would be a huge task that I wouldn't undertake unless there were huge benefits. And of course the threshold to change the standards used for education and communication with other... (more) |
— | almost 2 years ago |
| Edit | Post #287658 |
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— | almost 2 years ago |
| Edit | Post #287658 | Initial revision | — | almost 2 years ago |
| Answer | — |
A: What did James Stewart mean by "the line integral reduces to an ordinary single integral in this case" ? A line integral is integrated with respect to arc parameter. If the path you're integrating along is the $x$-axis, then the arc parameter can be taken to be just $x$, and so the integral is identical to an ordinary integral with respect to $x$. You are right that it's a bit improper to write $f(x,... (more) |
— | almost 2 years ago |
| Comment | Post #286907 |
Although @#53398 is correct, that you cannot use mathematical induction to prove an identity for all real numbers, what you can do is prove Bernoulli's inequality for the natural numbers, using mathematical induction, which is straightforward. Then you can extend it to negative exponents, and rationa... (more) |
— | about 2 years ago |