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I presume that this is a well-known result, but I couldn't find it. What is the greatest number of points which can be marked on an n by n square grid so that no three are collinear? Drawing successive cases suggests that for n=0, 1, 2, 3 and 4, the answer is 0, 1, 4, 5 and 6 - is this correct so far, and what's the general result?— 86.132.235.208 ( talk) 18:21, 13 July 2009 (UTC)
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Is it possible to know what the answer to that is - in which case what is it and how did you work it out - without laboriously adding them up? -- JackofOz ( talk) 20:58, 13 July 2009 (UTC)
"Trovatore" didn't really state his answer except for its bottom line number. Here's the rest: The average of the ten digits 0 through 9 is
That's Trovatore's answer.
That's approximately correct if the ten digits occur equally frequently in the long run. But here's a hard question: Do the ten digits really occur equally frequently?
For that we have only statistical evidence, not a mathematical proof.
In one sense, the answer is clearly "no": if the sum is 4500057062 instead of 4500000000 (i.e. 4.5 billion) then it deviates slightly from exact equality. But it is conjectured that you can get as close as you want to equality of those ten frequencies by making the number of digits big enough.
If you want to get into statistical evidence, then we'd also talk about pairs of consecutive digits, and triples of consecutive digits, and so on. The number cited above, 4500057062, does not deviate from 4.5 billion by more than would be predicted by the full-fledged conjecture dealing with pairs, triples, etc. occurring equally frequently. One could get into details of how that conclusion was reached as well.
But the way Trovatore came up with 4.5 is just that it's the average of the ten digits 0 through 9. Michael Hardy ( talk) 23:25, 13 July 2009 (UTC)
It is conjectured but unproven that pi is a normal number (Trovatore's answer being wrong would have been interesting evidence against the conjecture). However, if you just want the billionth digit without having to compute all the preceding ones, there is a beautiful spigot algorithm for obtaining it, at least if you don't mind a hexadecimal rather than decimal expansion. 70.90.174.101 ( talk) 01:08, 14 July 2009 (UTC)
During a physical-world discussion of a subject that has caused much distress here, the person I was discussing it with suggested that it would be much easier to look at from the point of view of picking a door from an infinite set of doors. This has caused me great distress independent of the problem. Let's look at the odds of randomly selecting one element from N with equal probability, that is P(1) = P(k), for any k in N. By the definition of probability, P(1) is either 0 or an element of (0..1]. The sum of all P(x) (=P(1)), x in N, equals 1, and by the Archimedian principle, if P(1) > 0, there exists some n such that P(1) * n < 1, so P(1) = 0. Thus P(1..n) = n * P(1) = 0, and the limit as n goes to infinity of 0 is 0. The only conclusion I can come to is that it's not meaningful to speak of selecting an element from N with equal probability?!? I'm not sure what I'm missing here, but I think I'm missing something.-- Prosfilaes ( talk) 23:17, 13 July 2009 (UTC)
Mathematics desk | ||
---|---|---|
< July 12 | << Jun | July | Aug >> | July 14 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
I presume that this is a well-known result, but I couldn't find it. What is the greatest number of points which can be marked on an n by n square grid so that no three are collinear? Drawing successive cases suggests that for n=0, 1, 2, 3 and 4, the answer is 0, 1, 4, 5 and 6 - is this correct so far, and what's the general result?— 86.132.235.208 ( talk) 18:21, 13 July 2009 (UTC)
**--- *--*- ---** -**-- --*-*
**---- ---*-* -*--*- *-*--- ----** --**--
Is it possible to know what the answer to that is - in which case what is it and how did you work it out - without laboriously adding them up? -- JackofOz ( talk) 20:58, 13 July 2009 (UTC)
"Trovatore" didn't really state his answer except for its bottom line number. Here's the rest: The average of the ten digits 0 through 9 is
That's Trovatore's answer.
That's approximately correct if the ten digits occur equally frequently in the long run. But here's a hard question: Do the ten digits really occur equally frequently?
For that we have only statistical evidence, not a mathematical proof.
In one sense, the answer is clearly "no": if the sum is 4500057062 instead of 4500000000 (i.e. 4.5 billion) then it deviates slightly from exact equality. But it is conjectured that you can get as close as you want to equality of those ten frequencies by making the number of digits big enough.
If you want to get into statistical evidence, then we'd also talk about pairs of consecutive digits, and triples of consecutive digits, and so on. The number cited above, 4500057062, does not deviate from 4.5 billion by more than would be predicted by the full-fledged conjecture dealing with pairs, triples, etc. occurring equally frequently. One could get into details of how that conclusion was reached as well.
But the way Trovatore came up with 4.5 is just that it's the average of the ten digits 0 through 9. Michael Hardy ( talk) 23:25, 13 July 2009 (UTC)
It is conjectured but unproven that pi is a normal number (Trovatore's answer being wrong would have been interesting evidence against the conjecture). However, if you just want the billionth digit without having to compute all the preceding ones, there is a beautiful spigot algorithm for obtaining it, at least if you don't mind a hexadecimal rather than decimal expansion. 70.90.174.101 ( talk) 01:08, 14 July 2009 (UTC)
During a physical-world discussion of a subject that has caused much distress here, the person I was discussing it with suggested that it would be much easier to look at from the point of view of picking a door from an infinite set of doors. This has caused me great distress independent of the problem. Let's look at the odds of randomly selecting one element from N with equal probability, that is P(1) = P(k), for any k in N. By the definition of probability, P(1) is either 0 or an element of (0..1]. The sum of all P(x) (=P(1)), x in N, equals 1, and by the Archimedian principle, if P(1) > 0, there exists some n such that P(1) * n < 1, so P(1) = 0. Thus P(1..n) = n * P(1) = 0, and the limit as n goes to infinity of 0 is 0. The only conclusion I can come to is that it's not meaningful to speak of selecting an element from N with equal probability?!? I'm not sure what I'm missing here, but I think I'm missing something.-- Prosfilaes ( talk) 23:17, 13 July 2009 (UTC)