April 11

If you look at the list of the first few nontrivial Riemann zeta zeroes of the form 1/2 + i x for x > 0 (see below), you see quite a few that are close to integers. E.g. values for x that are approximately within 1/100 of an integer should occur on average once in every 50 entries, however in the first 10 entries we already see 2 of them. In the later entries we don't see many of these values so close to integers, so the total number of the values in the table containing 151 entries within 1/100 of an integer isn't that large (there are 3 of them, the probability of getting more than 2 is approximately 0.08). However, the entries close to integers are still so close that you wouldn't have expected to see them at all in the table. So, x = 48.005150881 or an x closer to an integer should occur with probability of approximately 1/100 (probability for this to occur once or more within the first ten entries is about 1/10 ), x = 146.000982487 or a x closer to an integer should occur with probability 1/500 (probability for this to occur once or more in the table is approximately 1/10). So, it looks as if most entries are random, but you have some that are anomalously close to integers.

Table:

    14.134725142
    21.022039639
    25.010857580
    30.424876126
    32.935061588
    37.586178159
    40.918719012
    43.327073281
    48.005150881
    49.773832478
    52.970321478
    56.446247697
    59.347044003
    60.831778525
    65.112544048
    67.079810529
    69.546401711
    72.067157674
    75.704690699
    77.144840069
    79.337375020
    82.910380854
    84.735492981
    87.425274613
    88.809111208
    92.491899271
    94.651344041
    95.870634228
    98.831194218
   101.317851006
   103.725538040
   105.446623052
   107.168611184
   111.029535543
   111.874659177
   114.320220915
   116.226680321
   118.790782866
   121.370125002
   122.946829294
   124.256818554
   127.516683880
   129.578704200
   131.087688531
   133.497737203
   134.756509753
   138.116042055
   139.736208952
   141.123707404
   143.111845808
   146.000982487

Count Iblis (talk) 01:09, 11 April 2011 (UTC)[reply]

It does seem slightly more than would be expected:

                  VALUES       VALUES        VALUES 
                  WITHIN       WITHIN        WITHIN
                  0.10         0.03          0.01
                  ---------    --------      ------
PERCENT FOUND     13/51=25%    6/51=12%      2/51=4%   
PERCENT EXPECTED        20%          6%           2%

However, presumably you stopped where you did to maximize this effect. If we stopped one sooner, the results would be:

                  VALUES       VALUES        VALUES 
                  WITHIN       WITHIN        WITHIN
                  0.10         0.03          0.01
                  ---------    --------      ------
PERCENT FOUND     12/50=24%    5/50=10%      1/50=2%   
PERCENT EXPECTED        20%          6%           2%

This is close enough to just be a coincidence. StuRat (talk) 04:19, 11 April 2011 (UTC)[reply]

Probably you want to do statistical hypothesis testing that the fractional parts of the x-values have (not) a continuous uniform distribution. Bo Jacoby (talk) 12:14, 11 April 2011 (UTC).[reply]

Per the post below, if we assume that the string was preferentially selected in this way:

                                                     VALUES       VALUES        VALUES 
                                                     WITHIN       WITHIN        WITHIN
                                                     0.10         0.03          0.01
                                                     ---------    --------      ------
                                     PERCENT FOUND   13/51=25%    6/51=12%      2/51=4%   
                 PERCENT EXPECTED IN RANDOM STRING   20%          6%            2%
PERCENT EXPECTED IN PREFERENTIALLY SELECTED STRING   23.2%        9.76%         5.92%  —Preceding unsigned comment added by 92.20.205.185 (talk) 10:09, 12 April 2011 (UTC)[reply] 

When 6 out of 51 values are observed to be within 0.03 away from an integer, then the probability for this to happen is beta distributed with mean value ${\displaystyle \mu ={\frac {1+6}{2+51))=0.132}$ and standard deviation ${\displaystyle \sigma ={\sqrt {\frac {\mu (1-\mu )}{3+51))}=0.0461}$, and so the deviation of the theoretical value 0.06 from the mean value is ${\displaystyle {\frac {0.06-\mu }{\sigma ))=-1.56}$ standard deviations, which is not significantly different from zero. So the null-hypothesis, that the fractional parts of the imaginary parts of the zeroes of the zeta-function have uniform distribution, is not rejected by the observations. Bo Jacoby (talk) 18:36, 12 April 2011 (UTC).[reply]

It's not significant when looking at the whole list, but I'm still suspicious about the first few entries. A more spectacular list of numbers, where it is clear that the real numbers are close to integers is given by the function f(n) = exp[pi sqrt(n)] for integer n. Some are very close to integers, but I think that this is only true for a finite number of values for n. So, if you make the list arbitrarly large, you can make the statistical significance arbitrarily low. You can imagine similar cases where the effect is less spectacular, where a deep mathematcal reason does exist, but where the effect never rises above statistical significance. Count Iblis (talk) 22:09, 16 April 2011 (UTC)[reply]

I think it's just a case of you having looked through so many lists of numbers that eventually you were bound to find some pattern. A discussion of a similar effect (a correlation of the location of Uranus with quakes on Earth), is now on the Science Desk: Wikipedia:Reference_desk/Science#Help.21_I_have_some_questions_about_Venus.2C_Uranus_and_the_moon. StuRat (talk) 06:06, 17 April 2011 (UTC)[reply]

Yes that could be. However, one does have to consider if it is a priori likely if there could be an effect. Then math is different in this repect, as the exp[pi sqrt(n)] example clearly shows (here you do have real numbers that are very close to integers, but the reason involves quite advanced math). Count Iblis (talk) 15:09, 17 April 2011 (UTC)[reply]

In relation to the above sequence; If a have a random number generating function which produces a number from a set of 'interesting numbers' with probability p, and an 'uninteresting number' with probability 1-p, and analyse the produced sequence by deliberately selecting from the total sequence a continuous string that starts and finishes with one of the interesting numbers, what is the expected increase in the expectation of 'interesting numbers' in my string compared to a randomly selected string? —Preceding unsigned comment added by 92.20.205.185 (talk) 12:08, 11 April 2011 (UTC)[reply]

(I have no idea what your first six words mean, so am ignoring them.) Just thinking aloud, nothing rigorous here. You know the first and last terms are interesting; otherwise, the string is randomly selected. Assuming the string has length ${\displaystyle n>2}$, the expected number of interesting numbers will be ${\displaystyle 2+p(n-2)=pn+2(1-p)}$, whereas the the expected number in a randomly chosen string will be ${\displaystyle pn}$. Thus, you can expect ${\displaystyle 2(1-p)}$ more interesting numbers in your string, regardless of length.—msh210℠ 17:59, 11 April 2011 (UTC)[reply]

So for a random string it will have a fraction ${\displaystyle p}$ of interesting numbers, and for a preferentially selected string ${\displaystyle p+{\frac {2(1-p)}{n))}$. I.e.the preferential selection causes an expected increase of ${\displaystyle {\frac {2(1-p)}{n))}$. Thanks.

Hello. I've recently got interested in game theory. One thing that confuses me is Nash equilibrium - the concept sounds rather similar to saddle point. What's the difference between the two, or am I way off in thinking that these two have something in common? Thank you! 212.68.15.66 (talk) 12:46, 11 April 2011 (UTC)[reply]

Well they are both examples of Equilibrium points. Indeed you might find a saddle point as the Nash equilibrium for a two player zero-sum game, where you can use the vertical axis for the payoff function for one player and minus the payoff for the second. If the game is not zero-sum you would need more dimensions.--Salix (talk): 13:46, 11 April 2011 (UTC)[reply]