I illustrated this approach with respect to the aggregate sum of factors showing how it is compiled for the composite numbers up to 10.
Once again 4 has 2, 6 has 3, 8 has 3, 9 has 2 and 10 has 3 factors. So we obtain therefore the sum (4 * 2) + (6 * 3) + (8 * 3) + (9 * 2) + (10 * 3) = 8 + 18 + 24 + 18 + 30 = 98.
I will now likewise illustrate up to n = 10, with respect to the corresponding aggregate of the Riemann zeros.
We have to remember that these zeros are measured on the imaginary scale up to t where n = t/2π. Thus we add the zeros up to t = 62.83, i.e. 14.13 + 20.02 + 25.01 + 30.42 + 32.94 + 37.59 + 40.92 + 43.33 + 48.01 + 49.77 + 52.97 + 56.45 + 59.35 + 60.83 (correct to 2 decimal places) = 571.74.
Then to express this result with respect to n, we divide by 2π to obtain 90.96 = 91 (to nearest unit).
Up to n

Acc. sum of factors
(1)

Acc. sum of Riemann
zeros (2)

Formula
Est. (3)

(2)/(1) as %

(3)/(2) as %

10

98

91

72

92.86

79.12

20

499

493

419

98.80

84.99

30

1355

1234

1117

91.07

90.52

40

2620

2677

2205

102.17

82.37

50

4277

4221

3713

98.69

87.96

60

6459

6370

5663

98.62

88.90

70

9038

8767

8073

97.00

92.08

80

12073

12200

10956

101.05

89.80

90

15947

15858

14322

99.44

90.31

100

20367

20133

18206

98.85

90.43

110

24608

24958

22591

101.42

90.52

In the above table, I show the results (in 10's) up to n = 110, for both the aggregate sum of factors of the composite nos. (col. 2) and the corresponding sum of Riemann zeros (col. 3) with values rescaled to n.
Then in col. 4, I show the estimated aggregate sum using the formula, i.e.{n(n + 1)(log n – 1)}/2.
Then in col 5, I show the % accuracy of the actual sum of factor values and the corresponding sum of zeros.
These values compare vary well indeed. Sometimes the factor value sum exceeds that of the sum of zeros (and likewise the sum of zeros may exceed the sum of factor values). However even though we are still at a very early point on the n scale, the two sets of values already have converged very close to each other (in the region of 99% accuracy) as indicated in col. 5..
The formula estimate is not quite so accurate (with respect to both factor sums and zero aggregate sums). However it gives a consistent estimate (with % accuracy gradually improving) that already  as can be seen from the final column (col. 6)  is about 90% accurate.
Of course  as was the case for so long with the prime number theorem  I have not provided actual proofs of what is indicated by the data.
However I would confidently assert that the actual sums (with respect to both the factors and zero sum aggregates) would eventually approach 100% accuracy (in relative terms).
Likewise with only slightly less confidence, I would expect that the formula estimate (at high n) would also approach 100% accuracy (in relative terms) with respect to the prediction of both aggregates.
No comments:
Post a Comment