And here are the final lines after an overnight run. Note that accumulating billions of doubles in this fashion loses precision; I should have done my summing more carefully. But at least five of the digits are probably good. At 8589934592 pts 4.0000236480264 area 2.66669753208071 perim 6.2832343008038 At 17179869184 pts 4.00001316709677 area 2.66668286448053 perim 6.28322083261956 At 34359738368 pts 4.00000967484084 area 2.66667855572835 perim 6.28320124178968 On Wed, Aug 7, 2019 at 10:42 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Here's my code, and its results.
That perimeter sure looks close to tau.
-tom
At 67108864 pts 3.99990034103394 area 2.6665184797472 perim 6.28303082552975
At 134217728 pts 3.9999551102519 area 2.66652574548337 perim 6.28309803073879
At 268435456 pts 3.99998798593879 area 2.66669666495159 perim 6.28329936437473
On Wed, Aug 7, 2019 at 9:47 PM Brad Klee <bradklee@gmail.com> wrote:
Andy,
In this case the function doesn't have poles, so you are probably right. I still think it's weird that a complete integral can be taken on an incomplete domain, so I'm willing to worry more.
Another "problem" is that, after more thought, it seems plausible that the configuration space R would need an odd volume element to match Jim Propp's preferred standard. I think it's still "okay" to start with Cartesian, but would not be surprised if this led to statistical diversity.
Also, nice reading of Keith's algorithm, I was confused by the part about "carving", but now I can see from your perspective that it actually sounds right. Is Keith's algorithm on Mathworld? Maybe we could get EW to add a few more equations?
Please don't blame Fred for listening to me. We are all trying to have fun and learn more, and it doesn't hurt to double or triple check for quality assurance!
Cheers --Brad
On Wed, Aug 7, 2019 at 11:28 PM Andy Latto <andy.latto@pobox.com> wrote:
If what we're interested in is the integral of a function with respect to a particular measure, there is nothing tricky about ignoring measure zero sets. If a set has measure 0, the integral is unchanged by integrating a function that has different, arbitrary, values on that set.
Andy andy.latto@pobox.com
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