1. Take any number which is a multiple of 9 starting at zero ( 0 ) ( 0, 9, 18, 27, etc. ) .
2. Add 1, 2, 4, 5. 7, 8 to the number so that the first column of the number ends in ( 1, 3, 7, 9 ) . For example ( 11, 13, 17, 19 ) .
3. Test to see if it is a prime . A prime number is defined as being only evenly divisible by itself and 1.
Saturday, December 31, 2011
Sunday, December 25, 2011
Time Is Gravity, Entropy, Chaos In Disguise
Gravity appears to us as a weak force, but it is really the effects of time in disguise. Einstein said that mass bends space around the mass and that mass and energy are the same thing in two forms. The more that the mass / energy flexes the space string, the greater the gravity which is seen by us as a force . Time follows the flexed space string at differing velocities depending on the amount of flex in the space string near the mass / energy . Time has a numerical value of nine ( 9 ) and can be broken up into 9 segments ( 1/9 ). The 9 segments can have different velocities and can effect what happens. We can see this phenomena in our own life. Each of us, have adventures in life which may appear spontaneous or random . The things that happen to you or I aren't always at the same time. This is because the energy around us is flexed arbitrarily by either outside , random , or self induced actions that flexes the energy string . If you and I experience the same thing, the length of time may be different because the time segments travel at different speeds. In addition, if something bad happens, it usually follows in units of three ( 3 ), because the universe runs on the number three ( 3 ) and tries to stabilize around the number three ( 3 ) when things get out of whack.
In summary:
1. Mass
2. Weight
3. Time
are all related to space strings. If the space strings flex, time follows the flexed string at different velocities and we see the effects of this flexing as gravity if a mass is involved or entropy / chaos if a mass isn't involved but the results are mild or severe.
In summary:
1. Mass
2. Weight
3. Time
are all related to space strings. If the space strings flex, time follows the flexed string at different velocities and we see the effects of this flexing as gravity if a mass is involved or entropy / chaos if a mass isn't involved but the results are mild or severe.
Friday, December 09, 2011
Being Good
I got to thinking about being good. For most people, the connotation of being good is equivalent to having a dull and boring life. Most religions stress the need to be good in life and to help others. I also got to thinking about God or in other words the person who designed this universe so you and I would have a reasonable chance of survival. Looking at it from a science perspective it suddenly occurred to me, that being good is necessary to survive. Our universe is filled with energy and mass. Einstein said that energy and mass are equivalent which means that energy and mass can be converted from one to the other and back again. String theory says that all the energy is basically in the format of a string. A vibrating string in a loop gives the mass including us its' properties and behavior. A flexed string around a mass, such as the earth gives the earth its' gravity. The wind or other force is simply strings of energy or mass in motion. Velocity or acceleration is just a form of rolled up energy string ( think car here ) disturbing the energy fields as the mass ( car ) moves along. You and I are essentially a mass doing its' own thing by moving in a space occupied by strings. If we design something such as a car we have to round out its' rough edges so it won't cause any disturbance to the energy strings commonly known as turbulence. Similarly we want to round out our own rough edges in order to be acceptable to other people. Time in our universe is just energy that is still creating strings of space from dark energy and dark matter as it moves in a forward direction. Sometimes dark energy and dark matter break off from time before it gets converted into space. This broken off dark energy and dark matter sometimes brings disaster to ordinarily good people. This is the chance that keeps our world from getting too dull.
Wednesday, November 30, 2011
The Quantum World
The quantum world is made up of individual particles that don't have to travel through space to go somewhere. This world looks like foam continuously bubbling away. Travel in the quantum world is through the action of cloning, superposition and entanglement. Cloning is the method by which the individual particle reproduces itself at new locations. Superposition means that more than one particle can occupy the same space at the same time because space hasn't a physical reality in the quantum world. Quantum particles don't have to travel through space to get from A to B because space doesn't exist. Entanglement is the method by which a particle transfers information from one point to another point . A quantum particle can bubble many copies of itself at the same approximate location. This means that the bubbling quantum particle at the new location forms a cloud. This cloud may also appear as a wave which we can measure. We can also measure the position of the quanta, but we can't do both measurements at the same time. Different quanta may come together to form units of something else. If these quanta can't be separated out then we say a strong nuclear ( cloud ) force exists between them. This strong nuclear ( cloud ) force can also be considered as an unbreakable wave. If these quanta can be separated out in a reaction then we can say it has a weak nuclear force and a breakable wave that forms relationships with other quanta. Sometimes quanta form a rotational cloud with each other and these clouds carry plus and minus charges. These are your electromagnet quanta forces.
Tuesday, November 22, 2011
Time
Time is the basis of everything. When the universe was initially created, time separated out into dark matter, dark energy and string space. Dark matter acts like a thick honey which tends to keep everything in its' place like inertia. Dark Energy acts as an accelerator for the universe. String space when forming circles creates things and the string's circular vibration creates different characteristics which we see as that thing's properties. Unattached curved strings, when vibrating, appear to us a waves which when modulated transmit information. Radios use that technology. A non-vibrating curved string is gravity or mass. Inertia is related to the mass of the object. Spaceless time from another universe that existed before the creation of our universe is now our quantum world. In the quantum world there is time but not space. Einstein spoke of spooky action at a distance which is true in the abstract but not existing physically. If space doesn't exist in the quantum world, you don't have any delays. The absence of space means you can have superposition because more than one thing can exist literally on / in the same spot. You can also have entanglement which means that if something happens in one location the something immediately happens / clones in the other location. This happens because the something doesn't have to travel through physical space. The speed of change is infinite ( timeless ), because the limitation on the speed of light only applies when it is moving through space and not time. To take it one step further, suppose that something went wrong and spaceless time in another universe cloned space creating an expanding bubble resulting in our universe. Since our space includes classical time, and at the time of the creation of our universe everything wasn't immediately converted there would be more space than we first calculated from our perspective of logical events. Also not all the time would be converted which meant dark time, dark matter and dark energy still exists. In fact there is much more dark time, dark matter and dark energy in comparison to matter and anti-matter.
Friday, November 18, 2011
Holographic Projections, Black Holes & Event Horizons
The whole universe is organized and run by time which has a mathematical value of nine ( 9 ). Space is composed of a series of strings whose basic value is one to nine ( 1 – 9 ). The zero ( 0 ) string holds imaginary numbers. Everything that isn't time, is associated with a string that is equal to the one digit sum of its' digits . For instance ( 97 ) contains the digits 9 and 7 which total 16 ( 9 + 7 = 16 ) 16 contains the digits 1 and 6 which total 7 ( 1 + 6 = 7 ). Therefore something with the value of 97 rests on the 7th string. If you take the numbers from 1 to 99, and add the digits until one digit is left you will find that the sequence of the one digit numbers go from 1 to 9 and then repeat themselves. These strings can be thought of as strings of space. These space strings rest on a complex woven background string whose value is zero ( 0 ) and is equivalent to imaginary numbers since you can't show any values on string zero ( 0 ) except zero ( 0 ). The world around us ,including you and I, appears to be solid because our vision can't see finer detail on its' own. You and I including the quantum world is really like a swarm of flies. If you look at a swarm of flies, and someone asks you its' location, you tend to answer in terms of where the densest part of the fly swarm is located. In essence when someone asks you where someone or something is located, you answer in terms where the densest part of someone or something is located. Obviously in this instance the someone or something appears solid because that is where our sight begins and ends. If everything is basically similar to a fly swarm, you can also say it is a holograph of what exists on the average at that location. Dark matter in which the universe rests is really sticky time acting as inertia and gravity while dark energy is the force of time or entropy pulling / accelerating the expansion of the universe. Dark matter as disguised time also appears to us as time or inertia which tends to keep things from falling apart. Dark energy or expanding time appears to us as entropy which tends to pull things apart which we also see as aging, force or deterioration. If you think of the universe as having direction, west to east is the expansion of the strings ( 1 to 9 ) into larger numbers totaling the value of that particular string . For instance the east to west expansion of string seven could create the value of 250 ( 2 + 5 + 0 = 7 ). The value of 250 would be on string seven ( 7 ) since the digits of ( 250 ) would total 7 ( 2 + 5 + 0 = 7 ). This east to west expansion would form a holographic pattern of information or processes which time would then project north and south which we would see as an image, process or new facts depending on our location. Time only moves outward in our universe which we see as entropy, otherwise called deterioration, force and aging. If a holographic projection reaches a black hole, the first thing it sees is an event horizon. At this point the holographic projection is changed because only time and location enter the black hole while space information stays at the event horizon. If time and location leave the holographic projection, the only thing left is the space information string 7 since ( 2 + 5 + 0 = 7 ). Thus the information at the event horizon is proportional to the area of the event horizon rather than the volume of the black hole since the black hole doesn't contain space except from our perspective. The time and location which enters the black hole from the holographic projection is a non – physical distorted quantum representation of the holographic projection without the space string which represents physical space in our seen world. Thus in a spaceless quantum world represented by time and location you get superposition, entanglement and cloning which is instantaneous. If the black hole contracts in size, time and location reverses itself and rejoins its' string which appears to us as Hawking radiation due to the creation of , to us , a new holographic projection. Time and location for the old particle is destroyed in the black hole so conservation of energy is preserved.
Wednesday, November 16, 2011
Holographic Universe
Let's suppose the whole universe is organized and run by time which has a value of nine ( 9 ). Everything that isn't time, is associated with a string that is equal to the one digit sum of its' digits . For instance ( 97 ) contains the digits 9 and 7 which total 16 ( 9 + 7 = 16 ) 16 contains the digits 1 and 6 which total 7 ( 1 + 6 = 7 ). Therefore something with the value of 97 rests on the 7th string. If you take the numbers from 1 to 99, and add the digits until one digit is left you will find that the sequence of the one digit numbers go from 1 to 9 and then repeat themselves. These strings can be thought of as strings of space. These space strings rest on a complex woven background string whose value is zero ( 0 ) and is equivalent to imaginary numbers since you can't show any values on string zero ( 0 ) except zero ( 0 ). The world around us ,including you and I, appears to be solid because our vision can't see finer detail on its' own. You and I including the quantum world is really like a swarm of flies. If you look at a swarm of flies, and someone asks you its' location, you tend to answer in terms of where the densest part of the fly swarm is located. In essence when someone asks you where someone or something is located, you answer in terms where the densest part of someone or something is located. Obviously in this instance the someone or something appears solid because that is where our sight begins and ends. If everything is basically similar to a fly swarm, you can also say it is a holograph of what exists on the average at that location. Dark matter in which the universe rests is really sticky time acting as inertia and gravity while dark energy is the force of time or entropy pulling / accelerating the expansion of the universe. Dark matter as disguised time also appears to us as time or inertia which tends to keep things from falling apart. Dark energy or expanding time appears to us as entropy which tends to pull things apart which we also see as aging, force or deterioration. If you think of the universe as having direction, west to east is the expansion of the strings ( 1 to 9 ) into larger numbers totaling the value of that particular string . For instance the east to west expansion of string seven could create the value of 250 ( 2 + 5 + 0 = 7 ). The value of 250 would be on string seven ( 7 ) since the digits of ( 250 ) would total 7 ( 2 + 5 + 0 = 7 ). This east to west expansion would form a holographic pattern of information or processes which time would then project north and south which we would see as an image, process or new facts depending on our location. Time only moves outward in our universe which we see as entropy, otherwise called deterioration, force and aging.
Sunday, November 06, 2011
Riemann's Hypothesis – Calculating Zeros On ( y = ½ )
Sometimes in life you stumble upon something that is basically true and it's up to other people to prove it. 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. Riemann provided a formula to prove his belief which was ( s = ½ + it ). The current method of calculating the position of the zeros is by plugging numbers into the formula and calculating. I don't have any quarrel with this method except that there isn't any guarantee that you haven't missed a zero which is supposed to be on the line ( y = ½ ) and is actually off the line ( y = ½ ) . The rub in the calculation is that if you miss a “t” or an “i”, it might be the number that proves Riemann's Hypothesis is incorrect. Personally, I don't think it matters because Riemann accidentally stumbled onto something that is correct so missed numbers in “t” or “i” are academic. I don't know what numbers they are using but the available literature says the first few zeroes were calculated around 14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. Even if you proved that all the zeros lay on the line ( y = ½ ), that proof still doesn't tell you how all these zeros relate to the position of the primes on the line ( y = ½ ) in useable terms.
Here's my proof for what it is worth:
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 + it ). 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.
Here's my proof for what it is worth:
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 + it ). 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
The Riemann Hypothesis Resolved ( Hopefully finally)
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.
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.
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.
Tuesday, September 27, 2011
The Complicated Fractal Nature Of Prime Numbers
Mandelbrot, the inventor of fractals, when he worked for IBM was presented with a problem involving interference in the transmission of information. Every so often parts of the transmitted information would seem to randomly drop off / scramble which, needless to say, caused problems with the transmitted informational message. The problem had to be fixed, but as the information drop off / scramble appeared to be random , everyone was flummoxed because of the lack of a recognizable pattern. Mandelbrot thought about it and started experimenting. One time he decided to take a straight line and divide it into 1/3rd . He discarded the middle 1/3rd and kept the other 2/3rd separated by a space ( ------ ------ ). He continued on and discovered that the pattern produced by this method matched the pattern of the informational message drop off / scramble. This discovery proved that the informational drop off / scramble wasn't random but followed a fractal pattern.
Prime Numbers also follow a 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 ).
The second thing you will notice is that if the sum of the digits of any number ending ending in 1, 3, 7, 9 total a multiple of 3 ( divide by 1/3rd and discard the potential prime just like Mandelbrot discarded his string sections ), except for 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 ).
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 ( for instance 97 = ( 9 + 7 = 16 ) ( 1 + 6 = 7 ) you will find the column zero or far right column one digit totals are ( 1, 2, 4, 5, 7, 8 ). All the rest of the columns are zero ( 01, 02, 04, 05, 07, 08 ).
In summary:
1. Prime numbers, if they are prime numbers, have the 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 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number. 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).
3. Except for the one digit prime number 3 in the one digit prime number series ( 1, 2, 3, 5, 7 ) if you add the digits of a prime number ( for instance 97 = ( 9 + 7 = 16 ) ( 1 + 6 = 7 ) the column zero or far right column one digit totals are ( 1, 2, 4, 5, 7, 8 ).
Prime Numbers also follow a 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 ).
The second thing you will notice is that if the sum of the digits of any number ending ending in 1, 3, 7, 9 total a multiple of 3 ( divide by 1/3rd and discard the potential prime just like Mandelbrot discarded his string sections ), except for 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 ).
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 ( for instance 97 = ( 9 + 7 = 16 ) ( 1 + 6 = 7 ) you will find the column zero or far right column one digit totals are ( 1, 2, 4, 5, 7, 8 ). All the rest of the columns are zero ( 01, 02, 04, 05, 07, 08 ).
In summary:
1. Prime numbers, if they are prime numbers, have the 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 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number. 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).
3. Except for the one digit prime number 3 in the one digit prime number series ( 1, 2, 3, 5, 7 ) if you add the digits of a prime number ( for instance 97 = ( 9 + 7 = 16 ) ( 1 + 6 = 7 ) the column zero or far right column one digit totals are ( 1, 2, 4, 5, 7, 8 ).
Thursday, September 22, 2011
One Small Step For The Riemann Hypothesis
The Riemann Hypothesis says that all the non-trivial zeros ( 0 ) are on the line ( y = ½ ) and that this hypothesis has something to do with prime numbers. Another way of expressing it, is 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 error term in the prime number theorem is closely related to the position of the zeros.
In summary:
1. We have a line ( y = ½ ) .
2. We have some zeros that are real and we can use them since they touch / cross the line ( y = ½ )
3. The error term in the prime number theorem is related to the position of the zeros.
The Prime Number Theorem is concerned with the number of primes preceding a number. Due to the nature of mathematics when you multiply, it is probably better to have the number as a prime number for considerations of accuracy. In other words, if you can calculate the location of a prime number in the prime number series you automatically know how many prime numbers precede it.
We have the line ( y = ½ ) . Therefore let the first digit of the multiplier be ½ or .5. We have some zeros that are real and we can use them since they touch / cross the line ( y = ½ ) . Therefore let the middle digits be zeros ( 0 ). We now have the number .50. To finish our multiplier add the digit 9 to the end forming the number ( .5099999999 ).
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 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 in 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. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range. The present method of proving the Riemann Hypothesis consists of calculating whether or not zeros cross or touch the line ( y = ½ ) . While I appreciate the effort, it seems to me that I have proved the Riemann Hypothesis because my zeros can be infinitely added between the digits ½ and 9 ( .5----9 ). It can be seen that the calculation of the position of a prime and the number of primes preceding that prime all depend on the Riemann Hypothesis' real zeros. Thus the Riemann Hypothesis has been proved by illustrating its’ real relationship to the position of the primes and the number of primes preceding it.
In summary:
1. We have a line ( y = ½ ) .
2. We have some zeros that are real and we can use them since they touch / cross the line ( y = ½ )
3. The error term in the prime number theorem is related to the position of the zeros.
The Prime Number Theorem is concerned with the number of primes preceding a number. Due to the nature of mathematics when you multiply, it is probably better to have the number as a prime number for considerations of accuracy. In other words, if you can calculate the location of a prime number in the prime number series you automatically know how many prime numbers precede it.
We have the line ( y = ½ ) . Therefore let the first digit of the multiplier be ½ or .5. We have some zeros that are real and we can use them since they touch / cross the line ( y = ½ ) . Therefore let the middle digits be zeros ( 0 ). We now have the number .50. To finish our multiplier add the digit 9 to the end forming the number ( .5099999999 ).
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 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 in 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. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range. The present method of proving the Riemann Hypothesis consists of calculating whether or not zeros cross or touch the line ( y = ½ ) . While I appreciate the effort, it seems to me that I have proved the Riemann Hypothesis because my zeros can be infinitely added between the digits ½ and 9 ( .5----9 ). It can be seen that the calculation of the position of a prime and the number of primes preceding that prime all depend on the Riemann Hypothesis' real zeros. Thus the Riemann Hypothesis has been proved by illustrating its’ real relationship to the position of the primes and the number of primes preceding it.
Monday, September 19, 2011
Theory Of Everything
Here's The Theory Of Everything in a nutshell without the mathematics. The classical world that you and I see every day consists of:
1. Time
2. Energy / Weight / Mass
3. Space
Einstein in his equation E = MC^2 said that energy and mass were equivalent. Weight is the pull of gravity on an object. That is why Newton's apple fell from the tree since the pull of the earth ( gravity ) is greater than the pull of the apple ( gravity ) on the earth. Mass is the same thing as weight without gravity pulling on you and I. As far as space is concerned, Einstein said that space doesn't exist unless something extends into it. If you look at something you are extending your sight ( observation ) into space. If you drive a car you are extending it into space ( distance ). If we didn't have space there wouldn't be velocity, acceleration or force ( Force = Mass X Acceleration ( into space) ). Time is a marker for something happening ( The apple fell from the tree at 2:00 a.m.. ). Time is also a measure of distance ( I traveled 30 kilometers / miles in 30 minutes ).
In the quantum world there is time but not space. Einstein spoke of spooky action at a distance which is true in the abstract but not existing physically. If space doesn't exist in the quantum world, you don't have any delays. The absence of space means you can have superposition because more than one thing can exist literally on / in the same spot. You can also have entanglement which means that if something happens in one location the something immediately happens / clones in the other location. This happens because the something doesn't have to travel through physical space. The speed of change is infinite ( timeless ), because the limitation on the speed of light only applies when it is moving through space and not time. Information also travels using the same principle. If we attempt to measure the quantum / spaceless world from our classical / space world, by using particles ( light / x-ray etc. ) we disturb the something we are measuring by adding quanta / energy ( particles ) to it. This energy addition changes the original quantum something so that we aren't measuring the original.
In summary quantum time world without classical space is:
1. Cloning
2. Superposition
3. Entanglement
In summary our classical world is:
1. Time
2. Space
3. Energy / Weight / Mass
1. Time
2. Energy / Weight / Mass
3. Space
Einstein in his equation E = MC^2 said that energy and mass were equivalent. Weight is the pull of gravity on an object. That is why Newton's apple fell from the tree since the pull of the earth ( gravity ) is greater than the pull of the apple ( gravity ) on the earth. Mass is the same thing as weight without gravity pulling on you and I. As far as space is concerned, Einstein said that space doesn't exist unless something extends into it. If you look at something you are extending your sight ( observation ) into space. If you drive a car you are extending it into space ( distance ). If we didn't have space there wouldn't be velocity, acceleration or force ( Force = Mass X Acceleration ( into space) ). Time is a marker for something happening ( The apple fell from the tree at 2:00 a.m.. ). Time is also a measure of distance ( I traveled 30 kilometers / miles in 30 minutes ).
In the quantum world there is time but not space. Einstein spoke of spooky action at a distance which is true in the abstract but not existing physically. If space doesn't exist in the quantum world, you don't have any delays. The absence of space means you can have superposition because more than one thing can exist literally on / in the same spot. You can also have entanglement which means that if something happens in one location the something immediately happens / clones in the other location. This happens because the something doesn't have to travel through physical space. The speed of change is infinite ( timeless ), because the limitation on the speed of light only applies when it is moving through space and not time. Information also travels using the same principle. If we attempt to measure the quantum / spaceless world from our classical / space world, by using particles ( light / x-ray etc. ) we disturb the something we are measuring by adding quanta / energy ( particles ) to it. This energy addition changes the original quantum something so that we aren't measuring the original.
In summary quantum time world without classical space is:
1. Cloning
2. Superposition
3. Entanglement
In summary our classical world is:
1. Time
2. Space
3. Energy / Weight / Mass
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