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  1. #31
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    Dec 2004
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    184

    Some new equations to think about

    New equations:

    2\[Pi] * (remainder (17 / 2\[Pi])) = 4.42

    4.42 * 5 = 22.1

    22.1 * (5 / (17* 2\[Pi])) = 0.884

    0.884 * 5 = 4.42

    4.42 = 4.42
    It's all about ideas.

    constructorscorner.com

  2. #32
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    Dec 2004
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    184
    I know no one believes you can find the product knowing only N, where N is the product of 2 Prime numbers. If so I would have just rendered RSA obsolete. Something that hasn’t happened since it was published in 1978.

    The fact is I believe this is proof that elementary equations can in fact do this. My programming experience is not able to do heavy calculations on 256 bit numbers. I posted this here so someone would prove me right. (Or wrong) I hope someone has taken interest in this and has tried to do something with my work.

    However there is something if it did work. What if x was known and N was unknown. You would be able to find the end to irrational numbers. Say Pi was y and x was the diameter and N was unknown. You could test the unknown N to be any number of digits. You would only need the zero for placement of the decimal position.

    So a diameter of 19 would equal an end of N = 10…100…1000.10000….100000.1000000…etc. And we know the quotient is 3.14159… But we do not need to know y because N is determined alone in the equation by x, which is 19.

    This is because N divided by 19 is relatively Prime until it reaches the Product of 2 Primes.

    Obviously, this is only a theory. It may hurt the reputation of my previous work. But I see something here. It may be easy to refute, but if Pi even ends you need a computational solution and not just a recursive algorithm that crunches numbers.

    Thanks for any input.
    It's all about ideas.

    constructorscorner.com

  3. #33
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    Dec 2004
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    184

    A new approach; The area of a triangle with y/x radians and x radius.

    This is a continuativion of finding a vector as it relates to the Product of 2 Prime numbers. If you have a radius x which is the smallest of 2 Prime products you can rotate it around the x and y axis at an angle of y/x radians; where y is the larger of the Prime products. You have 2 sides of the triangle; both of which equal to x. The only question is how many times around the coordinate plane a radius of x while revolve around a 2 Pi circular revolution. Find the area of this triangle (and include all the times it revolves around the circle; 3 and 2347 for example) and you have N, the product of 2 Prime numbers. The only downfall is that you must isolate x and it has trignometric functions such as Sine, Cosine, and Tangent. So it is not solved, but it should be easier to find equations and geometry to tinker with a solution.

    the area of a triangle = 1/2 (base * height)

    1/2 *[[cos (N/x^2) * x + x] * [sin (N/x^2) * x] = N

    But this equation will not work until you know what quadrant y/
    x resides in.

    test x = 5 ; N = 85

    1/2 *[[cos (85/5^2) * 5 + 5] * [ Pi + sin ((85/5^2)) * 5] = 85

    1/2 *[[cos (85/5^2) * 5 + 5] * [ Pi + sin ((85/5^2)) * 5] approximately = 83.5184





    I will expand this outline at a later time. For now this is an idea worth considering.
    It's all about ideas.

    constructorscorner.com

  4. #34
    Join Date
    Dec 2004
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    184

    More on trig functions

    We will use the definition of sine: Opposite / Hypotenuse and a trigonometric Pythagorean Identity (sin^2 theta) + (cos^2 theta) = 1.


    There is a remainder problem here. We can divide the angle into sectors of Pi but mathematically finding the remainder of a number not between -1 < 0 < 1.

    (y/x) / PI = angle of last quadrant in radians

    [(y/x) / Pi] is the angle in radians. It is the number of quadrants that y encircles a circle of radius x. The area of this circle covered is x * y or N. (Yes Prime numbers again. But the relation is by multiplication and not a pattern of Prime numbers.

    sin(theta) = Opp / Hyp = ( ( (y/x) / Pi) / x)

    Using the equation:

    the area of a triangle = 1/2 (base * height)

    1/2 *[[cos (N/x^2) * x + x] * [sin (N/x^2) * x] = N


    (sin^2 theta) + (cos^2 theta) = 1 That is the Pythagorean identity.

    transforms the equation to


    1/2 * [ [( ( 1- ( (y/x) / Pi) / x)^2) * x + x] * [ ( ( (y/x) / Pi) / x) * (N/x^2) * x] ] = N

    knowing y = N/x or y/x = N / x^2


    1/2 * [ [( ( 1- ( ( (N / x^2) /x) / Pi) / x)^2) * x + x] * [ ( ( ( (N / x^2) /x) / Pi) / x) * (N/x^2) * x] ] = N



    I must test this and write it without all the confusing parenthesize. I will update in the following weeks.
    It's all about ideas.

    constructorscorner.com

  5. #35
    Join Date
    Dec 2004
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    184

    Correction in copy and past error. Now simpler

    We will use the definition of sine: Opposite / Hypotenuse and a trigonometric Pythagorean Identity (sin^2 theta) + (cos^2 theta) = 1.


    There is a remainder problem here. We can divide the angle into sectors of Pi but mathematically finding the remainder of a number not between -1 < 0 < 1.

    (y/x) / PI = angle of last quadrant in radians

    [(y/x) / Pi] is the angle in radians. It is the number of quadrants that y encircles a circle of radius x. The area of this circle covered is x * y or N. (Yes Prime numbers again. But the relation is by multiplication and not a pattern of Prime numbers.

    sin(theta) = Opp / Hyp = ( ( (y/x) / Pi) / x)

    Using the equation:

    the area of a triangle = 1/2 (base * height)

    1/2 *[[cos (N/x^2) * x + x] * [sin (N/x^2) * x] = N


    (sin^2 theta) + (cos^2 theta) = 1 That is the Pythagorean identity.

    transforms the equation to


    1/2 * [ [( ( 1- ( (y/x) / Pi) / x)^2) * x + x] * [ ( ( (y/x) / Pi) / x) * x] ] = N

    knowing y = N/x or y/x = N / x^2


    1/2 * [ [( ( 1- ( ( (N / x^2) /x) / Pi) / x)^2) * x + x] * [ ( ( ( (N / x^2) /x) / Pi) / x) * x] ] = N

    Correction in above equation

    where ((N / x^2) / x) = N / (x^3)

    1/2 * [ [( ( 1- ( ( N / (x^3) / Pi) / x)^2) * x + x] * [ ( ( ( N / (x^3) / Pi) / x) * x] ] = N

    Remember the entire revolution of y/x radians must be added to the sine and cosine. For example 23 and 3 would is in the 1st quadrant after making one revolution.

  6. #36
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    Dec 2004
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    184

    The most clearly explained work so far!

    It's all about ideas.

    constructorscorner.com

  7. #37
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    Dec 2004
    Posts
    184

    Prime Switch

    If you are a WWII Buff or just want to see an encryption machine see the link:


    http://www.constructorscorner.net/id...imeSwitch.html
    It's all about ideas.

    constructorscorner.com

  8. #38
    Join Date
    Dec 2004
    Posts
    184

    Some very important sentences

    The work runs into a problem because we can't mathematically calculate the remaider of N / (PI / 2) and y / (Pi / 2).

    If we knew the remainder of y / (Pi / 2) it would be easier to find the area of the radius versus distanced revolved. Which is the area or N.


    But an interesting thing to take note is that the segment scribed by the angle of y/x radians is proportional to an angle N / 1 on a unit circle. So the difference [(N / (Pi / 2)) - ((N / x^2) / (Pi / 2)) ]
    is very important. It is the propotion from a circle of radius x and a unit circle of radius one.


    This is all I have for now. This is confusing yet these sentences are powerful.
    It's all about ideas.

    constructorscorner.com

  9. #39
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    Dec 2004
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    184

    A look at the smaller to predict the larger.

    I have had this idea for quite a time and never really developed it because of its difficulty. So I post it here that it might inspire someone.

    Idea:

    The problem when working with Prime numbers is that we are working with whole, positive integers. But computers work in decimals. If we were to take a decimal number and find when it equals a whole number integer, this whole number would possibly be Prime.

    This would eliminate the multiples of the Prime number since the decimal would also equal all the multiples. So to find a Prime number instead of factoring a large impossible calculation, patterns in the decimal could be found to see the smallest number that that decimal equals a whole number.

    The reason this works is due to proportions. (Remember how I related the proportions on a unit circle to a value of larger circles?) This is no easy task: to find where a decimal will equal a whole number. If I had a use for a supercomputer this would be it.

    Yes I realize that many decimals could equal the Prime number 3 or 5 or larger, but as the number of digits is in the hundreds the separation of decimals is more unique or specifically matched to a number. Also if you limit the decimals between 0 and 1, you increase computation but find specific values.

    I know this is confusing but the goal is to make it easier to find Prime numbers other than factoring. So if 23 * 0.24 doesn’t equal a whole number at least it presents a clue to what is going on.

    So basically the goal is to find where a decimal equals a whole number, if it ever does. If it doesn’t the multiple that will make it equal a whole number approaches infinity. This could make it easier to find large Prime numbers. (As the Prime number approaches infinity!) I believe that you could eliminate potential decimals faster than dividing numbers when factoring that are becoming larger with more values to divide when the number gets larger. I am a where that numbers between 0 and 1 are infinite, but even though I currently don’t have a solution I am saying it is worthwhile to pursue such an idea, if it hasn’t been done already.

    This may be fluff, but please post to this thread if you get inspired to do some math.
    It's all about ideas.

    constructorscorner.com

  10. #40
    Join Date
    Dec 2004
    Posts
    184

    An important remainder proportion; Work in progress; Just an idea yet

    This is the continuation of the decimal solution to Prime numbers.

    Imagine a circle with 0.26 units in circumference. Now imagine a circle in the same units with a circumference of 1.

    That is 1 / (2*Pi)

    Imagine a value of 144 units wrapping around the circle 144 times until it stops at the first quadrant.
    The value 144 mod 1.

    Now do the same for the circle of 0.26 units. 144 / 0.26. It is going to wrap around many more times according to how much smaller the decimal 0.26 is than the circumference of the first circle with circumference of 1.

    This is all very simple. But what if the difference between the circle of circumference 1 and the 0.26 units circle is not comparable until you take 2Pi radians and subtract the remainder of 144 / 0.26 ; this is the amount of space of the 0.26 units circle that is needed to make one more whole unit.

    So that’s 0.26 – (the remainder of (144 / 0.26)

    The number that should make a whole number is the proportion of (1 / 0.26) * (the remainder of 144 / 0.26)

    This is the lines of how I am thinking. I can’t solve this but wanted to show I am seeing a pattern.

    If you used a reference of say a circle of circumference 8 than it would be ((the remainder of (144/8)) / 0.26) * (the remainder of 144 / 0.26); This proportion would show where a value of 8 would be reached. It would also show a pattern of where 0.26 is reaching a whole number.

    This would still be computationally intense as to narrow the value to the one’s place in a hundred digit number. But it may be an approach to take.

    Do not get me wrong. I am not saying this works. I only want to share an idea as to why I thought it would be valuable to find were small decimal numbers equal who numbers. Finding this might be even more difficult than factoring. But if you find one decimal that would equal a whole number it would be easy to disprove it was Prime if it has several non-proportional decimals that all equal the given number.

    So this is not correct! I just want to explain why I put effort in such a difficult problem.
    It's all about ideas.

    constructorscorner.com

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