## Saturday, October 27, 2012

### The Riemann Hypothesis Solved Using Calculus Limits

Mathematics basically breaks down into the following three concepts:

1. Numbers.
2. Symbols.
3. Lengths
The symbols such as x or y are usually equal to something which is up to you to discover. These symbols may be in a formula expressing relationships which are familiar to anyone that has ever studied algebra. These relationships in algebra are generally a static value as only one numerical answer is needed. Calculus was the next major step. Calculus is the study of limits which involve fractions. Limits are concerned with how close you can get to something I call a brick wall ( limit ) without actually touching it. How close is actually the calculation of length before you hit the wall or limit. Here you are dealing with fractions. Calculus, using symbols, is the study of the change in the length of a formula containing symbols that change in length. If you have symbols in a formula, what happens to the formula when that symbol or symbols change in length ???? To make life simpler, the initial formula is simplified to lessen the potential confusion. In any language, as in mathematics, there are many ways to express an idea .
Riemann expressed his idea using the following formulas:

Riemann said that the “s” in the above equation was ( s = ½ + it ). The “i” in it represented an imaginary number and the “t” represented a real number. He also said that all his non – trivial zeros lie on the line ( y = ½ ) . Innumerable calculations up to the present time have proven that Riemann's conjecture is true but there also may be an exception yet to be discovered. The Riemann Hypothesis also implies that if the zeros consistently cross the line ( y = ½ ) then these zeros have something to do with the distribution of primes. There is an implicit idea here that every crossing of zeros on the line ( y = ½ ) is related to the position of a relative prime beginning at the beginning of the line ( y = ½ ) and the end of the line ( y= ½ ) at infinity. Theoretically, if the values of Riemann's formula were chosen correctly the
calculation of the zeros crossing the line in terms of distance of an unknown prime from the beginning of ( y = ½ ) would be accurate. The only variable in the equation is in the one ( s = ½ + it ) or ( it ). Riemann's formula is really a disguised calculus problem using zeros ( 0 ) to adjust the placement of a prime on the line ( y = ½ ) . In other words, if you adjust the zero in a mathematical expression how close can you get the relevant unknown prime to its' actual position on the line ( y = ½ ) ???? The Riemann function backs up this concept by saying that the magnitude of the oscillations of primes around their expected position is controlled by the real parts of the zeros of the zeta function. In particular the amount of error is closely related to the position of the zeros . This means that the unknown value of ( it ) has adjustable zeros as well as a real number which is used to fine tune the position or location of a particular prime on the line ( y = ½ ) . Primes are numbers that can only be divided evenly by themselves and one ( 1 ). The single digit prime numbers are 1, 2, 3, 5, 7 . If you draw a straight line and put some or all the primes on it, you will soon see that there isn't a standard distance between the primes. You can also count the primes from the beginning of the line but there isn't any way of accurately saying how many primes precede a prime without counting all the primes from the beginning up to that particular prime. Let's assume that the non-trivial zeros lie on the line ( y = ½ ) . For the sake of convenience we can also put all the primes from one ( 1 ) to infinity on the line ( y = ½ ) and indicate their position on the line ( y = ½ ) from the beginning of the line ( y = ½ ) . The single digit primes are five ( 5 ) in number consisting of 1, 2, 3,. 5, 7. If you arbitrarily multiply the primes by ( ½ ) you will soon see that your answer has usually very little relationship to the primes true position on the line ( y = ½ ) . In other words, by multiplying by ( ½ ) you will soon see that if all the numbers are written down from zero ( 0 ) to infinity that ( ½ ) of them aren't primes. The Riemann Hypothesis also implies that if the zeros consistently cross the line ( y = ½ ) then these zeros have something to do with the distribution of primes. If the zeros consistently cross the line ( y = ½ ) than these zeros ( 0 ) can be used in the calculation . We now have the value of the line ( ½ ) plus the zeros to use in some way to see if we can calculate the position of the primes on the line ( y = ½ ) from the beginning. Multiplying by ( ½ ) or ( .5 ) or ( .50 ) doesn't change your original answer. Using ( .05 ) also doesn't improve things to any great extent. Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. This manipulation changes the physical distance between the primes. To accomplish this manipulation, place some digits after the zero ( 0 ) in ( .50 ) such as forming ( .509999 ) . Therefore:

1. 5 X ( .509999 ) = 2.549995 (4)
2. 7 X ( .509999 ) = 3.569993 ( 5 )
3. 11 X ( .509999 ) = 5.609989 ( 6 )

This basic principle calculates the position of the primes fairly accurately up to prime 31.

Using the same multipliers on the next prime ( 37 ) we get ( 37 X .509999 = 18.869963. Prime ( 37 ) is actually the 13th prime. Riemann said that the zeros can be manipulated ( positions changed or zeros added ). We still need the ( ½ ) in Riemann's line ( y = ½ ) and the ( 9's ). If we take ( .509999 ) and multiply it by itself ( ( .509999 X .509999 = .26009898 ).

1. ( 37 X .509999 X ,509999 = 9.62366226 ( 13 )

This calculation shows that the calculated position is short of the true position by about ( 13 – 9.62366226 = 3.37633774 ). The difference is approximately equal to Pi ( 3.141592654 ). In some calculations the difference is about the natural number ( e ) ( 2.71828`828 ).

In summary, Riemann's intuition told him that the zeros had a real value of ( ½ ) which was true since the digit ( ½ ) is used in the calculation. Riemann's intuition also told him that the zeros can be manipulated ( position changed or zeros added ) which is also true. This manipulation of zeros ( 0 ) is really using the principles of limits in calculus to see how close you can find a factor when multiplied by a prime that will give you the position of that prime on the line ( y = ½ ). What Riemann missed was that the calculation involved the number 9 and that Pi ( 3.141592654 ) and the natural number ( e ) ( 2.718281828 ) might have to be added or subtracted from the final answer. Riemann also missed that powers would also have to be used.

In general for calculating the location of any prime you:

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 which equals 995.2299524. 7919 is the 1000th prime. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).

The Clay Mathematics Institute is offering a \$1,000,000 prize for the solution to the Riemann Hypothesis. From my reading, it seems to involve proving whether or not all the zeros lie on the line ( y = ½ ) . Since I have shown how the Riemann Hypothesis relates to the location of the primes and by extension how many primes precede that prime ( counting 1, 2, 3, 4 ), it seems to me it is largely academic as to whether all the zeros lie on the line ( y = ½ ) . I've also proved that the manipulation of the zeros as implied by Riemann can bring a prime closer to its' position using the principles of Calculus.