Riemann said that all his non – trivial zeros lie on the line ( y = ½ ) . These zeros are at the location that the primes form a right angle line to ( y = ½ ) or in other words, are vertical to the line ( y = ½ ) . Prime numbers are defined as numbers that can only be divided by themselves and ( 1 ). An example of one digit primes are ( 1, 2, 3, 5, 7 ). You can see from these numbers that primes aren't linear ( 1, 2, 3, 4 ) but form harmonics. If all the primes were played musically, you would get a tune. The physical distance between the primes would be similar to the value of the rests in a musical piece. Calculations have been done on the location that some of Riemann's zeros lie on the line ( y = ½ ) using Riemann's formula ( ½ + bi ). The first few zeroes were calculated around 14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. These numbers are not close to the position of the beginning one digit primes ( 1, 2, 3, 5, 7 ). You will see that the distance between ( 1, 2, 3, ) is ( 1 ) and the distance between 3 & 5, 5 & 7, is 2. Riemann, in his Riemann's Hypothesis, said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . 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 which is equivalent to altering the value of the rests in a musical piece.
1. 1 X .99 = .99 ( 1 )
2. 2 X .99 = 1.98 ( 2 )
3. 3 X .99 = 2.97 ( 3 )
If we add the digit ( ½ ) or ( .5 ) in Riemann's line ( y = ½ ) and Riemann's zero ( 0 ) to .99, we form the number ( .509999 ).
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. 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 have also found the missing pieces to Riemann's Hypothesis which was correct as far as it went.
Monday, October 31, 2011
Sunday, October 30, 2011
The Universe Is Built On Strings
There is an old joke which said that all the things I needed to know were covered in kindergarten. Strings appear to be in that category. If you try to measure the definitive length of a string you will find that the task is impossible. First of all the string isn't rigid so you don't know where it begins and ends. Secondly, the length of the string depends on your measuring instrument. Different measuring instruments give different results. To add to the fun, the measuring instrument is not always accurate. Lastly the measurer may need glasses or alternately disturbs the measurement when he observes it ( quantum measurement ). If you tie the ends of a string so it forms a loop and vibrate it, you have illustrated that vibrating strings have different properties when they vibrate and by extension represent different things. For instance, metals, bricks, sand, etc. all look different to each of us because we can't see all their detail. If you hold a string straight it represents quantum time which doesn't have space. Therefore, without space to travel through, in the quantum world, you have cloning, entanglement and superposition .Everything is instantaneous and reproducable without space,having to be traveled and superposition because everything can be in the same location because there is no space to interfere with location. If you hold a string in an arc without moving it, you have gravity which is otherwise inertia. The closer the ends of the string are together in a loop the stronger the gravity ( inertia ). If a string is held in a loop and moved while in a loop, the movement of the string represents entropy, force, acceleration and velocity. The best demonstration of this phenomena is watching a car accelerate from a standing start. First of all the back end dips until the force of gravity or inertia is broken. Then entropy, force, acceleration and velocity happen. You can see this phenomena as soon as the front end of the car comes down and entropy, acceleration and velocity begin. Entropy is the deterioration of the car from position to position as it goes down the track. You and I can't see it because we can't see fine detail.
In summary:
1. A straight string represents spaceless quantum time leading to cloned copies, entanglement and superposition.
2. A tied vibrating looped string represents various properties which we see as things ( bricks, buildings, people ).
3. A static untied looped string represents gravity and inertia.
4. A dynamic, moving untied looped string represents entropy, force, velocity and acceleration.
In summary:
1. A straight string represents spaceless quantum time leading to cloned copies, entanglement and superposition.
2. A tied vibrating looped string represents various properties which we see as things ( bricks, buildings, people ).
3. A static untied looped string represents gravity and inertia.
4. A dynamic, moving untied looped string represents entropy, force, velocity and acceleration.
Wednesday, October 26, 2011
Patterns, Harmonics, Disorder, Chaos
Our universe is based on fundamental patterns. The use of fundamental patterns in our universe is necessary because they help us to organize our surroundings into something that is understood. An example of patterns is language, mathematics and speech. We may have variations on the fundamental pattern such as language, accents and spelling. Mathematics is broken down into adding, subtracting, multiplying and dividing as well as algebra, geometry, calculus and trigonometry. Speech is broken down into sounds that convey a message. Wild animals use speech in the form of sound to communicate. These variations on the fundamental patterns could be called harmonics, similar to harmonics in music as a common example. We also have the opposite to patterns which could be called disorder or in the extreme, chaos. Disorder can be as innocuous as designing a different type of car, building, or inventing a new word. Chaos is basically extreme disorder. An example of this is death or other forms of extensive destruction. Common examples of disorder leading to eventual chaos is aging in ourselves, our possessions or the environment ( read climate change here ). The problem with chaos is that you don't always know it is present unless you can see it or are in the midst of it. Our universe is based on mathematics. Theoretically if you mapped all the incoming data, you could see when chaos is about to start or has started. That is a great theory but it breaks down in its' applicability when the incoming data is overwhelming. A simple example is counting from one ( 1 ) to infinity. If you graph all the numbers you will see it climbs up in one direction like climbing a hill. A simple method is to add all the digits in a number until you get a one column number ( for instance 476, ( 4 + 7 + 6 = 17, ( 1 + 7 =8 ) ). You will find that all the numbers from one to infinity total from ( 1 to 9 ) and then repeat themselves in a pattern. The resulting pattern looks like a saw tooth. You could call this a fractal pattern since it repeats itself indefinitely. Prime numbers are an example of a seemingly patternless chaos in our world, but they have an underlying complicated fractal pattern . Prime Numbers are defined as numbers that can be only divided by themselves and one ( 1 ). Prime Numbers follow a complicated fractal pattern. First of all, if you look at a list of prime numbers you will find that they always have the numbers 1, 3, 7, 9 in column zero or otherwise known as the far right column ( 11, 13, 17, 19 ). The second thing you will notice is that all numbers ending in ( 1, 3, 7, 9 ) aren't prime numbers ( 21, 33, 27, 39 ). This is our first pattern. The second thing you will notice is that if the sum of the digits of any number ending in 1, 3, 7, 9 total a multiple of 3 , except for prime number 3, ( for instance 6, 9, 12, etc. ) it isn't a prime number. If a number ending in 1, 3, 7, 9 in column zero ( far right column ) isn't a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column zero ( far right column ). These are more patterns. Lastly, except for the one digit prime number 3 in the one digit prime number series ( 1, 2, 3, 5, 7 ), you will find if you continuously add the digits of a prime number you will find the column zero or far right column one digit totals for prime numbers are ( 1, 2, 4, 5, 7, 8 ). These numbers ( 1, 2, 4, 5, 7, 8 ) are the strange attractors of the prime numbers.
In summary:
1. Prime numbers, if they are prime numbers, have the pattern numbers 1, 3, 7, 9 in column 0 ( farthest right column ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for prime number 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number.
3. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0).
4. If you add the digits and there is more than one total ( for instance 97 has a first digit total of 16 ( 9 + 7 = 16 )) then 16 is a harmonic of ( 7 ), since ( 1 + 6 = 7 ).
5. The column zero or far right column one digit totals for primes are ( 1, 2, 4, 5, 7, 8 ).
6. The one digit totals ( 1, 2, 4, 5, 7, 8 ) are all strange attractors of the complex prime number fractals.
The adding of these digits is a neat way of quantizing energy levels or Riemann numbers, which really is only a series of numbers in which we simplify the pattern in order to study it. Any chaotic system can be quantized in this manner similar to any harmonic oscillator or anything else for that matter.
In summary:
1. Prime numbers, if they are prime numbers, have the pattern numbers 1, 3, 7, 9 in column 0 ( farthest right column ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for prime number 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number.
3. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0).
4. If you add the digits and there is more than one total ( for instance 97 has a first digit total of 16 ( 9 + 7 = 16 )) then 16 is a harmonic of ( 7 ), since ( 1 + 6 = 7 ).
5. The column zero or far right column one digit totals for primes are ( 1, 2, 4, 5, 7, 8 ).
6. The one digit totals ( 1, 2, 4, 5, 7, 8 ) are all strange attractors of the complex prime number fractals.
The adding of these digits is a neat way of quantizing energy levels or Riemann numbers, which really is only a series of numbers in which we simplify the pattern in order to study it. Any chaotic system can be quantized in this manner similar to any harmonic oscillator or anything else for that matter.
Thursday, October 20, 2011
The Riemann Digits ½ , Zero ( 0 ) & My 9.
Prime numbers and the Riemann Hypothesis digits are all intertwined. A prime number is defined as a number that can only be divided by itself and one ( 1 ) ( 1, 2, 3, 5, 7 ).
Here's the skinny on prime numbers:
1. Prime numbers, if they are prime numbers, have the digits 1, 3, 7, 9 in column 0 ( farthest right column ) ( 11, 13, 17, 19 ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number.
3. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0). If not try other numbers.
The calculation of the location of the prime numbers on a line or calculating the number of previous primes on a line involve the digits ( ½, 0 & 9 ). Riemann in his Riemann's Hypothesis said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . 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. I have separately discovered that the location of the prime numbers on the line ( y = ½ ) or any other line for that matter involves the digit 9.
The primary number for the location of a prime on any line from the Riemann Hypothesis is ( .509999 ). For the primes ( 1, 2, 3, ) you multiply them by ( .99 ) since their actual positions are ( 1, 2, 3 ). For the primes (5 to 23 ) you multiply them by ( .509999 ). For the primes 29 & 31 you multiply them by ( .509999 ) and subtract Pi ( 3.141592654 ) to bring them closer to their true position.
For the rest of the primes up to 97 you raise ( .509999 ) to the power of 2 and adjust using either Pi or the natural number e ( 2.718281828 ).
In summary, Riemann anticipated the number ( ½ ) & ( zero ( 0 ) and zero adjustment, but missed my discovery of the number 9 and the necessity of Pi and ( e ).
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).
It would seem from the above, that Riemann anticipated the zeros ( 0 ) and the value of those zeros being ( ½ ) since they were on the line ( y = ½ ). Riemann didn't anticipate the number 9 or the option of adjusting the calculation by adding 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 ) .
Here's the skinny on prime numbers:
1. Prime numbers, if they are prime numbers, have the digits 1, 3, 7, 9 in column 0 ( farthest right column ) ( 11, 13, 17, 19 ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number.
3. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0). If not try other numbers.
The calculation of the location of the prime numbers on a line or calculating the number of previous primes on a line involve the digits ( ½, 0 & 9 ). Riemann in his Riemann's Hypothesis said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . 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. I have separately discovered that the location of the prime numbers on the line ( y = ½ ) or any other line for that matter involves the digit 9.
The primary number for the location of a prime on any line from the Riemann Hypothesis is ( .509999 ). For the primes ( 1, 2, 3, ) you multiply them by ( .99 ) since their actual positions are ( 1, 2, 3 ). For the primes (5 to 23 ) you multiply them by ( .509999 ). For the primes 29 & 31 you multiply them by ( .509999 ) and subtract Pi ( 3.141592654 ) to bring them closer to their true position.
For the rest of the primes up to 97 you raise ( .509999 ) to the power of 2 and adjust using either Pi or the natural number e ( 2.718281828 ).
In summary, Riemann anticipated the number ( ½ ) & ( zero ( 0 ) and zero adjustment, but missed my discovery of the number 9 and the necessity of Pi and ( e ).
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).
It would seem from the above, that Riemann anticipated the zeros ( 0 ) and the value of those zeros being ( ½ ) since they were on the line ( y = ½ ). Riemann didn't anticipate the number 9 or the option of adjusting the calculation by adding 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 ) .
Tuesday, October 18, 2011
The Riemann Hypothesis Resolved Using ½ , 0 and 9
Like a lot of people, I've been fooling around trying to prove Riemann's Hypothesis without much success. Let's suppose for the sake of argument that all the zeros lie on the line ( y = ½ ) . If they do, the solution for the Riemann Hypothesis brings us no closer to showing what the deep connection is between the zeta function and the distribution of the primes. My reading of the literature on the Riemann Hypothesis indicates that the zeros lie on the line ( y = ½ ) and that if the zeroes are manipulated in some fashion you can calculate the location of the primes to a greater and greater accuracy. If you know the location of one prime you know how many primes precede it because you are counting ( 1, 2, 3, 4 etc. ). Riemann's genius or intuition was that the distribution of the primes involve the number zero ( 0 ) and ½. The line is important because if you count something, that something is being placed on a line. If the something is a prime it lies on a line. Riemann's Hypothesis says that all the zeros lie on a line so for the sake of argument you have primes on a line and zeros ( 0 ) on a line. Riemann said that if you manipulate the zeros ( 0 ), you can bring the prime sitting on the line closer and closer to its' true position. What the line ( y = ½ ) means is a mystery, although the number ( ½ ) may prove useful. The Riemann Hypothesis equation brings the zeros ( 0 ) to the line ( y = ½ ) depending on what you are using for ( t ) or the other variables. Let's suppose the primes lie on a straight line which also has zeros ( 0 ) on it.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891). You will see from these calculations that the best multiplier is .5099999 since the multiplication ( answer ) is the closest to 6.
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
I calculate that the best usable number for the location of a prime is ( .5099999 ). Riemann said that the proportion of primes is about ( 1 / ln (x) ) where ( ln(x) is the natural logarithm of x to base ( e ) or ( 2.718281828 ) . Using Riemann's formula the proportion of primes below prime number 11 is ( 1 / ln(11) or ( .417032391 ). The number of primes below ( 11 ) using this formula is ( 11 X .4107032391 = 4.587356306) or in round numbers ( 5 ). My calculation using ( .5099999 ) is ( 6 ) in round numbers which is the exact location of prime number ( 11 ). Riemann worked out that if the zeros really do lie on the critical line, then the primes stray from the ( 1/ln(x) ) distribution exactly as much as a bunch of coin tosses stray from the 50:50 distribution law. I have proved that the non-trivial zeros have to lie on the line, because my calculation for the location of the primes uses Riemann's non-trivial zeros that have to be on the line ( y = ½ ) or a line since you are calculating the location of the primes in a sequence ( or the number of preceding primes ).
The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only conjectures the zeros are on the line ( y = ½ ) . The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. I believe that I have proved that the Riemann zeros are all on the line whether it be ( y = ½ ) or some other line, because those zeros have to be used in a calculation while being on a line to prove that the location of the primes are on a straight line and not somewhere else in space.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891). You will see from these calculations that the best multiplier is .5099999 since the multiplication ( answer ) is the closest to 6.
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
I calculate that the best usable number for the location of a prime is ( .5099999 ). Riemann said that the proportion of primes is about ( 1 / ln (x) ) where ( ln(x) is the natural logarithm of x to base ( e ) or ( 2.718281828 ) . Using Riemann's formula the proportion of primes below prime number 11 is ( 1 / ln(11) or ( .417032391 ). The number of primes below ( 11 ) using this formula is ( 11 X .4107032391 = 4.587356306) or in round numbers ( 5 ). My calculation using ( .5099999 ) is ( 6 ) in round numbers which is the exact location of prime number ( 11 ). Riemann worked out that if the zeros really do lie on the critical line, then the primes stray from the ( 1/ln(x) ) distribution exactly as much as a bunch of coin tosses stray from the 50:50 distribution law. I have proved that the non-trivial zeros have to lie on the line, because my calculation for the location of the primes uses Riemann's non-trivial zeros that have to be on the line ( y = ½ ) or a line since you are calculating the location of the primes in a sequence ( or the number of preceding primes ).
The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only conjectures the zeros are on the line ( y = ½ ) . The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. I believe that I have proved that the Riemann zeros are all on the line whether it be ( y = ½ ) or some other line, because those zeros have to be used in a calculation while being on a line to prove that the location of the primes are on a straight line and not somewhere else in space.
Sunday, October 16, 2011
The Position Of Primes Using The Riemann Hypothesis Numbers & The One That Was Missed
Like a lot of people, I've been fooling around trying to prove Riemann's Hypothesis without much success. Let's suppose for the sake of argument that all the zeros lie on the line ( y = ½ ) . If they do, the solution for the Riemann Hypothesis brings us no closer to showing what the deep connection is between the zeta function and the distribution of the primes. My reading of the literature on the Riemann Hypothesis indicates that the zeros lie on the line ( y = ½ ) and that if the zeroes are manipulated in some fashion you can calculate the location of the primes to a greater and greater accuracy. If you know the location of one prime you know how many primes precede it because you are counting ( 1, 2, 3, 4 etc. ). Riemann's genius or intuition was that the distribution of the primes involve the number zero ( 0 ) and ½. The line is important because if you count something, that something is being placed on a line. If the something is a prime it lies on a line. Riemann's Hypothesis says that all the zeros lie on a line so for the sake of argument you have primes on a line and zeros ( 0 ) on a line. Riemann said that if you manipulate the zeros ( 0 ), you can bring the prime sitting on the line closer and closer to its' true position. What the line ( y = ½ ) means is a mystery, although the number ( ½ ) may prove useful. The Riemann Hypothesis equation brings the zeros ( 0 ) to the line ( y = ½ ) depending on what you are using for ( t ) or the other variables. Let's suppose the primes lie on a straight line which also has zeros ( 0 ) on it.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891).
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only proves the zeros either cross the line or are on the line ( y = ½ ). The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. The most fascinating part is that the tweaking involves Pi and the natural number ( e ) but that is another story.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891).
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
The Clay Mathematics Institute is offering a $1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only proves the zeros either cross the line or are on the line ( y = ½ ). The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. The most fascinating part is that the tweaking involves Pi and the natural number ( e ) but that is another story.
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.
Wednesday, October 05, 2011
Riemann Hypothesis Solved Using A Quantum Mechanical System ( Revised )
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. Since Riemann's Hypothesis includes the line ( y = ½ ), let's chose a quantum energy system in which the energy levels are 2. Riemann's equation also included complex numbers. In a quantum energy system Riemann's complex numbers could be represented by zero's ( 0 ) since ( 2 + 0 = 2 ). ( 1 + 1= 2 ) also equals 2. Riemann also went on to say that the prime number locations were influenced by the position of the zeros. To extend our ( 1 + 1 = 2 ) analogy, we now have ( 1 + 0 + 1 = 2 ). We can convert these additions to numbers ( 11, 101, etc. ). We can now say that we have a real axis ( y = 2 ) with infinite numbers with zeros between their ones ( 11, 101, 1001, 10001, etc. ) on an infinite real axis line ( y = 2 ). We can also say that each of the zero's in ( 101, 1001, 10001, etc. ) are on the line ( y = 2) since the numbers ( 101, 1001, 10001, etc. are also on this line. Riemann also said that the values of the zeros ( 0's ) on the line are equal to ( ½ ) which is also true when the line ( y = 2 ) is flipped. We could also extend this argument to one which says that the value of the ones ( 1's ) are also ( ½ ) since the reciprocal of one ( 1 ) is ( 1 ) and the fraction ( 1 / 1 ) when added ( 1 / 101 ) has the value ½ on the line ( y = ½ ) . These numbers would also be evenly spaced on the real axis line ( y = 2 ) since each infinite number would total energy level 2 which means the distance between the numbers would be ( 2 + 2 = 4 ) or a spacing of 4. Flipped the distance between the numbers would be one ( 1 ) since ( ½ + ½ = 1 ). Riemann's Hypothesis says that all his formula's non-trivial zero's are on the line ( y = ½ ) and this is what we've proved up to now since the non-trivial zeros are in the numbers ( 101, 1001, 10001, etc. ). If we flip our infinite real axis line ( y = 2 ), we create a real axis line ( y = ½ ) with flipped infinite real numbers ( 1/11, 1/ 101, 1/1001, etc. ) all the way to infinity. 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. The calculation of the prime numbers and their positions are a little bit more complicated than this illustration but this is the basics.
Sunday, October 02, 2011
Riemann Hypothesis Solved Using A Quantum Mechanical System
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. Since Riemann's Hypothesis includes the line ( y = ½ ), let's chose a quantum energy system in which the energy levels are 2. Riemann's equation also included complex numbers. In a quantum energy system Riemann's complex numbers could be represented by zero's ( 0 ) since ( 2 + 0 = 2 ). ( 1 + 1= 2 ) also equals 2. Riemann also went on to say that the prime number locations were influenced by the position of the zeros. To extend our ( 1 + 1 = 2 ) analogy, we now have ( 1 + 0 + 1 = 2 ). We can convert these additions to numbers ( 11, 101, etc. ). We can now say that we have a real axis ( y = 2 ) with infinite numbers with zeros between their ones ( 11, 101, 1001, 10001, etc. ) on an infinite real axis line ( y = 2 ). These numbers would also be evenly spaced on the real axis line ( y = 2 ) since each infinite number would total energy level 2 which means the distance between the numbers would be ( 2 + 2 = 4 ) or a spacing of 4. Riemann's Hypothesis says that all his formula's non-trivial zero's are on the line ( y = ½ ). If we flip our infinite real axis line ( y = 2 ), we create a real axis line ( y = ½ ) with flipped infinite real numbers ( 1/11, 1/ 101, 1/1001, etc. ) all the way to infinity. If we add ( ½ + 1/11+ 1/ 101 + 1001 + etc. ) we eventually start to get numbers resembling ( .509999. .500999, etc. ) The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's. If you raise these numbers to the power of 2, you get somewhere over ¼ . As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .5099999 ) X ( .5099999 ) you get very close to 26. The calculation of the prime numbers are a little bit more complicated than this illustration but this is the basics.
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