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#23 |
"Bob Silverman"
Nov 2003
North of Boston
22·1,877 Posts |
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Both of you are just bandying words.
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#24 | |
"Bob Silverman"
Nov 2003
North of Boston
22×1,877 Posts |
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the density function is not ergodic. When we say the gaps are Poisson distributed, we mean that they are LOCALLY Poisson (i.e. in the neighborhood of any finite index). However, as p-->oo the gaps become longer and longer. The number of expected M_p in [p, 2p] is exp(gamma). Note that the interval [p, 2p] grows in length as p grows, yet the number of expected M_p in the interval is constant. |
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#25 | |
"ร
ke Tilander"
Apr 2011
Sandviken, Sweden
23616 Posts |
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Last fiddled with by aketilander on 2012-02-23 at 16:24 |
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#26 |
"William"
May 2003
Near Grandkid
1001010001102 Posts |
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How about the statistical question of whether the existing data is consistent with a Poisson process? The graph gives a pretty convincing "Chi-by-Eye," but it would be fun to have some "statistical confidence" statements. I've not seen anybody attempt this.
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#27 | |
"Phil"
Sep 2002
Tracktown, U.S.A.
25·5·7 Posts |
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#28 |
Aug 2003
Snicker, AL
26×3×5 Posts |
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The human brain is a marvelous thing. It seeks patterns, even when patterns do not exist. For example, if you watch the waves on the seashore long enough, you would think you could discern the pattern of large vs small waves. The waves represent a pseudo-random physical process which is not the same thing as a mathematically random process (which as RDS notes is not truly defined either).
So to add to the cnofusion :), I'll stipulate that in an effort to perceive a pattern, your mind throws up possibilities which then have to be waded through to determine that there is no discernible pattern. Here is a different way to see this and it will illustrate the lack of pattern. Instead of expressing the known Mersenne Primes in base 10, try converting them to binary and then look at the distribution. If you are really good, you might see something amazing. DarJones p.s. RDS, would you care to have a discussion about whether or not there are any truly random mathematical processes? Note that I can prove that there are random physical processes, a la Heisenberg Uncertainty Principle. |
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#29 | |
"Bob Silverman"
Nov 2003
North of Boston
1D5416 Posts |
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Take a time sample from a signal. Assume the measurments have some small error. Now, take the Fourier Transform of the signal. The errors transform as well. One can prove that the product of the errors in the original signal times the errors in the transform are bounded from below. This is what happens when one tries to measure position and velocity vectors at the same time. One can be viewed as the Fourier Transform of the other and the product of the errors is bounded from below. |
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#30 |
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
11100001101012 Posts |
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More generally/abstractly, all waves and wave-like things have some form of uncertainty principle attached to them. For quantum particles, it just happens to be position and momentum. The only reason this became such a big deal is because of the classical counterintuitiveness of the result, which is exactly the same as saying that wave-particle duality is classically counterintuitive.
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#31 | |
Aug 2003
Snicker, AL
26·3·5 Posts |
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Lets postulate that it does exist. We will call it the MUP (Mathematical Uncertainty Principle). Now an inherent property of MUP would be that some Mathematical Process would be truly random, therefore if you can identify a truly random mathematical process, you have inherently proven MUP. The reason I am even opening up for this is: if the distribution of Mersenne Primes is governed by MUP, then they are truly random and if not, then they inherently can't be random and therefore there is a pattern to them. And yes, I'm just having fun on a late Thursday evening. DarJones |
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#32 |
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
11100001101012 Posts |
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Define "mathematical process". I have no problem saying that the propogation of a wave is a mathematical process determined by the wave equation. Also, just because you have a random process does not mean that there is a "MUP". There might be something other than a "MUP" that implies randomness of some process.
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#33 | |
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
22·5·11·53 Posts |
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A physically observable quantity must be represented by a Hermitian operator, because only such operators have real eigenvalues. Two observables may have simultaneously precisely defined values if and only if the corresponding operators commute. Proof of necessity is simple. Proof of sufficiency is a bit trickier because of the possibility of degenerate eigenfunctions but not particularly difficult. Two operators which do not commute must necessarily correspond to observables which can not be simultaneously measured to arbitrary precision. Examples of such pairs of operators include Incidentally, those two show that space and time are already on an equal footing in this picture, something that was noted very early on in the development of QM. Last fiddled with by xilman on 2012-02-24 at 09:57 Reason: Fix \TeX |
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