Saturday, October 08, 2011

Riemann Hypothesis & Strings

Riemann's Hypothesis is that all the non-trivial zeros lie on the line ( y = ½ ) and these zeros have something to do with prime numbers. The calculations done up to the present time have shown that the zeros are on the line ( y = ½ ) and there are yet more calculations to go. Let us suppose that each of Riemann's zeros are part of separate strings of numbers and each time the Riemann equation crosses the line ( y = ½ ) it does so where a zero is located. To make it simple, suppose that the zeros are in the middle of the number and the number 1 is on both ends ( 101, 1001, 10001 etc. ). If we add up the digits in these numbers we find the total is always 2 ( 1 + 0 + 1 = 2 ), ( 1 + 0 + 0 + 1 = 2 ), ( 1 + 0 + 0 + 0 +1 = 2 ). To take it one step further we could say these numbers are on the line ( y = 2 ) since all the numbers total 2. Riemann's Hypothesis is that the non- trivial zeros lie on the line ( y = ½ ). If we flip our analogy, the numbers on the line ( y = ½ ) become ( 1 / 101, 1/1001, 1/ 10001, etc. ). Riemann's equation still crosses the line ( y = ½ ) where the zeros are located. Riemann's equation isn't exact in terms of the positions of the zeros. It has been calculated that the first position of the zeros is around 14 and the second calculated position around 21. Our analogy is exact because we can create these numbers to infinity and the principle is the same. Riemann said that the zeros have a real value of ½ or .5 . If we add ( ½ + 1/101 ) we obtain ( .50990099 ). Prime number ( 11 ) is the 6th prime ( 1, 2, 3, 5, 7, 11 ). ( 11 X .50990099 = 5.6089108911 ) or 6 rounded to one digit. The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's and can be adjusted to get a more accurate estimate. As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .50990099 ) X ( .50990099 )) you get ( 25.22969903 )which is very close to 26. If you adjust the Riemann zeros to form the number ( .509999 ) and do the multiplication ( 97 x (.509999 X .509999) ) you get ( 25.22960106 ) which is further away from 26. You can see from these calculations that if the Riemann zeros are manipulated as Riemann has suggested to obtain the location of the prime ( or calculate the number of primes up to that point ) you can vary the distance to the true position of the prime. As a matter of interest Riemann's equation first cuts the line ( y = ½ ) at a position just over 14. The 14th prime is 41. ( 41 X ( ½ + 1/ 101) X ( ½ + 1 / 101 ) is ( 10.664099959 ) which is just off 14 by the value of Pi ( 3.141592554 ). The 21st prime is 71. It's position is calculated at 18 using the same method ( 71 X ( ½ + 1/ 101 ) X ( ½ + 1 / 101 )) which is very close to 21. The calculation of the prime numbers and their positions are a little bit more complicated than this illustration but this is the basics.

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