## Sunday, January 08, 2012

### Riemann Hypothesis Hopefully, Finally, Resolved

Let us assume that all the primes to infinity lie on a horizontal line . While it is acknowledged that the primes lie on this horizontal line to infinity and are in order, we don't know precisely where the primes lie on this horizontal line in terms of their distance from each other. Let us assume that wherever the infinite primes are on the horizontal line at varying distances from each other, we have a vertical line ( y = ½ ) off the horizontal line which forms an upside down “T” ( ! ! ! ! etc. ) . The infinite primes are at the intersection of a vertical line ( y = ½ ) and an infinite line holding all the prime numbers . Riemann said all his zeros are on the line ( y = ½ ) and have a value of ( ½ ) when they are on this line ( y = ½ ) . At the present time we are calculating to prove that all of Riemann's non-trivial zeros line on the line ( y = ½ ) and so far so good . If we can establish that the calculation of the position of these primes use ( ½ ) and Riemann's zero's ( 0 ) , then we have established that Riemann's zeros are all on the line ( y = ½ ) because ( y = ½ ) intersects with the horizontal line holding the infinite primes . Riemann also said his zeros can be added or subtracted or placed in different positions in order to adjust the position of the primes from the beginning of the horizontal line . This adjustment as an integer also establishes the number of primes preceding that prime number .The first prime numbers from 1 to 12 in order are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. If we multiply these numbers by ( .509999999 ) our answer is very close to their actual position ( 23 X .509999999 = 11.72999999 ). The actual position of prime number 23 is 10. There are 9 prime numbers preceding 23 ( 1, 2, 3, 5,7,11,13, 17, 19 ). The prime numbers from position 13 to 26 in order are ( 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ). If you multiply these primes by ( .509999999 ) you will find that the prime number positions are hugely incorrect. You will see from the “ In summary “ list that I indicated under ( 3. ) that the error term in the prime number theorem is related to the position of the zeros. You can increase the number of zeros by either adding them between ½ and 9 (for instance .5000999999) or by raising ( .509999999 ) to the power of 2 ( .509999999 X.509999999 = ( .2600999 )). You will see from ( .2600999 ) that we have adjusted the error term in the prime number theorem by adjusting the zeros ( 0 ) from one to two. Multiply the prime numbers by ( .2600999 ) to obtain the prime number location.

If you do the multiplication, you will find that some of the prime number locations are still out. For instance, ( 37 X .2600999 = 9.623699981). The actual location is 13. For some inexplicable reason if you add Pi ( 3.141592654 ) to this number you get ( 12.76529263 ). This trick of either adding or subtracting Pi works in the majority of cases. In some cases adding or subtracting the natural number ( 2.718281828 ) also works.

Here’s how the system works for numbers in general.

1. Count the number of digits in a prime number. For instance 7919 has 4 digits. Subtract 1 from the number of digits ( 4 - 1 = 3 ) for 7919. Form another number equal to the number of digits in 7919 ( 4 ) by putting ( .5 ) in the far left column and 9 in the far right column. ( .5—9 ). Fill the middle with Riemann Hypothesis zeros ( 0 ) forming a four digit number ( .5009 ). Raise ( .5009 ) to the power of 3 ( which is the number of digits in 7919 ( 4 ) minus 1 ( 4 - 1 = 3 ). ( .5009 ) ^ 3 = .125676215. Multiply 7919 X .125676215 = 995.2299524. 7919 is the 1000th prime number. The calculation is short by approximately the value of Pi ( 3.141592654 ). Pi + 995.2299524 is 998.3715451 which is very close to 1000.

It can be seen from these calculations that the magnitude of the oscillations of the primes around their expected position is controlled by the zeros ( 0’s) in the multiplier. The error term is closely related to the position of the zeros in the number ( .509999999 ). The error term can be controlled by either adding zeros ( .500999999, .50009999 ) or by raising these numbers to a power ( multiply the numbers by themselves ) thereby increasing the zeros. The power zeros can also be adjusted. For instance (( .5099999999 ) ^2 = (..260099999 )). If we change the digit 6 to zero ( 0 ) creating ( .200099999 ) and multiplying it by the prime number 347 we get ( 69.43469965 ) Prime number 347 is the 69th prime. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range if the calculated answer is close but too far out.

In Summary:

1. ( .509999999 ) may be adjusted using Riemann zeros.

2. ( .509999999 ) may be raised to a power to increase the zeros.

3. ( .509999999 ) raised to a power may have its' power zeros adjusted.

4. Tweaking can be done by adding or subtracting Pi or “e”.