NP (Point Based) Calculations (Page 19)
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These formulas have rewritten the function of zero as described in the Riemann hypothesis. This solves/answers more than one millenial question.DescriptionFunctionsLegend
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Completed FormulasT = Total Points Measured For Item Being Defined
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Proof That Zeros Must In All Cases Of Measurement Be A Defined Value; (Riemann Q):
Stationary Ellipses (Limiter/Maximum)
0 = Currently Accepted 0
DeterminationN = Point 2 (b) Traveled From 0 Axis
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Iterations of (( n + in (np) - in ) / np )) ;Stationary Ellipses (Limiter/Maximum)
Partially Equal
DeterminationIN = Total Distance Traveled
From First And Last Point
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defined version:NP = New Point Distance (Final) From First Point
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Proof #1:Determination
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( in(0) - (n(0) + [ in(0) *np(0) /in ] /in ) * ( in / [ in(0) - (n(0) + [ in(0) *np(0) /in ]) * [ (( 0 * in ) * in)) / ( 0 * in )) ] - This proof has been tested into the millions and completed.
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Can be easier understood as:
input = in; 0*in; - ((0*n + 0*in(0*np))
= in - (n + in(np))
-> new function added; /in
= where does the new function go?: in - (((n + ((in(np);/in)))
= output ; {Hold value} waiting for:
= output of ; /(in) ;
Or in other words:
= output ; {Hold value} ... wait for ... 0*in - ((0*n + 0*in(0*np))) ... wait for ... value of new function /in "(in - (((n + ((in(np)/in)))" AKA: (Feedvalue)
= output ; /in * (in /feedvalue) or the output divided by the input, multiplied by (the input divided by the feedvalue) AKA: (Feedback)
= output now equals 1 - Because it has followed a path that allowed a division of the outcome multiplied by the original input value/feedvalue.
= output ; *((0 *in) *in)) / (0 *in))
= input (Now you have found the original input value - found through separate, non-destructive paths
Stationary Ellipses (Limiter/Maximum)
0 = 1/2, Whole Limiter
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There are many applications for this. It is a set within a modular group of functions. There are a lot more functions to be written. This is how our universe operates.
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Proof #2:
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[in - [(( n + in (np) - in ) / np ); * np)) / np)) *np )) /0 )) *in )) *0)) = ans;] / ((ans)] * (in))] - Where 0 = 1/2. Input equals output - Whole Limiter. No Imaginary Information.
Another easier way to understand why it works
Stationary Ellipses (Limiter/Maximum)
0 = 1/2, Whole Limiter
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This is a less defined version than proof 1
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What is this doing?: Picture a reservoir system or electrical current circuit. It has two paths, with one input. There is a gate at the end of both paths, and each path is fed different portions of the input.
Once the path which has received a higher portion is filled, all of its excess is then fed into the second path until both are filled, and nothing is allowed past the gate until it has reached the fill point for both.
Then picture the paths being subdivided by the initial input values in such a way that it allows for two paths to fill at the same rate. Each time you multiply by 0, you have a subdivision, because 0 equals 1/2.
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Please note that the graph I provided does not show both paths, but an overall description of the whole.
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Linear Group Functions (Navier-Stokes) - Where 0 = 1/2
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( in(0) - (n(0) + [ in(0) *np(0) /in ] /in ) * ( in / [ in(0) - (n(0) + [ in(0) *np(0) /in ]) * [ (( 0 * in ) * in)) / ( 0 * in )) ] - This proof has been tested into the millions and completed.
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The same equation which solves the Riemann hypothesis solves Navier-Stokes, as shown in one of my graphs, intervals increase over time across x, y, or z, as you move closer to 0; when you describe each point in the function as having a location in time.
This equation stays the same equation for both X,Y, and Z. This can be refined which is why I am working out a second proof.
If you want to use the Riemann proof for the Navier-Stokes, you just need to do a pole reversal (-, to +). I will provide more later on, but you really should just graph it yourself. It's pretty simple.
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Location of X/Y is determined as: [[(in(0) -n(0) +[in(0)-np(0)]) / [(z/x(0) *t) * 0] / final point y]] /in - this is the start to the second proof if you feel like trying to finish it
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This function uses ratios of X/Y for "in", "n", "np"
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*I will not be working on these for a while as there are other things I need to spend time on. Mainly having to create VST versions of these equations to further show how and why they are operational, along with developing a program to
help you visualize what is actually going on here, to help create your own inputs, and to show you why it works. I really wish more people got these things without me holding your hand, but I know that one day more people will.
My mom was able to understand this within 10 minutes, and I have people with diplomas telling me its nonsense.
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Please See Proof Tables Below
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Incomplete - Calculating Points Based On Group Ratios; (Birch And Swinnerton Q) - Where 0 = 1/2
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in - ( n + np (in) - in) / np * 2 - this is really old and needs to be rewritten entirely
Calculating 3 Points (Reverse/Exact Point)
Lookup
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(( in - ( n + np (in) - in) / np * 2)) / (( n + np (in) - in) / np ) * 2in + ( -2n + np ); * (n(np)) - this is really old and needs to be rewritten entirely
Calculating 4th Point (Reverse/Exact Point)
Lookup
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Need to add a single 4th point value to determine location based on ratio; with possibility for addition to determine location for other points.
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Proof Tables
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[( n + in (np) - in ) / np ) * np) / np)] ; Where 0 = 1/2. 0 is halfway between 1 and -1. There should be a -0 and positive 0. Non Whole
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(( (0 + 0(0) - 0 ) / 0 ) * 0 ) / 0 ) = 0 ))
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(( (1 + 1(0) - 1 ) / 0 ) * 0 ) / 0 ) = 1 ))
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(( (1 + 2(1) - 2 ) / 1 ) * 1 ) / 1 ) = 1 )) < --- Answers will be 1 step down from whole function version. I moved on from here. This was enough to prove it could be done. Defined further below.
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If you are confused on 0, please refer to website, or text document (Letter for Science).
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This sheet contains more than 1 proof of 0 definition (riemann). Proof #1 is so far the most stable; fully defined formula, but has not been tested with all numbers.
Proof #2 is completely stable, but not as defined as Proof #1. I prefer a fully defined function, which is why I have shown proof table #1 first.
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[in - [(( n + in (np) - in ) / np ); * np)) / np)) *np )) /0 )) *in )) *0)) = ans;] / ((ans)] * (in))] - Where 0 = 1/2. Input equals output - Whole Limiter. No Imaginary Information. (Stable) Please See Graph.
Works: Needs to be defined
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Alternate ideas (old)
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[in - [( n + in (np) - in ) / (in (np) -n)) * (( 0 *in ) * in )) / ( 0 *in ))]
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[in - [( n + in (np) - in ) / (in (np) -n)) * in]
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Incomplete versions (old):
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[in - [(( n + in (np) - in ) / np ); * np)) / np)) *np )) /0 )); *in )) *0)); /in (in (np) -n); * (( 0 *in ) * in )) / ( 0 *in ))] - - Where 0 = 1/2. Input equals output .Whole Limiter. Broken - Looking for more paths.
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Written as: [((in - (( n + in(np) - in ) / np ); *np)) / np)) *np )) /0 )); *in )) *0)); /((in *(in(np) - n)); *(( 0 *in ) * in )) / ( 0 *in ))] Broken - Looking for more paths.
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Proof #1
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0*1267 - (+0*922 + 0*1267(0*345)

= 633.5 - (461 + (((633.5(172.5))/in))) = 258.75

= 258.75 / 1267 *(in / 258.75)

= 1 * (( 0 *in ) * in )) / ( 0 *in ))

= 1267
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0*0; - (0*0 + 0*0(0*0)

= .25 - (.25 + (.25(.25);.../.50) = .125

= .125 / 0 *(in / .125)

= 1 * (( 0 *in ) * in )) / ( 0 *in ))

= 0
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0*1; - (0*1 + 0*1(0*0)

= .5 - (.5 + (.5(.25);.../1) = .375

= .375 / 1 *(in / .375)

= 1 * (( 0 *in ) * in )) / ( 0 *in ))

= 1
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0*2; - (0*1 + 0*2(0*1)

= 1 - (.5 + (1(.5);.../2) = .75

= .75 / 2 *(2 / .75)

= 1 * (( 0 *in ) * in )) / ( 0 *in ))

= 2
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0*3; - (0*1 + 0*3(0*2)

= 1.5 - (.5 + (1.5(1);.../3) = -.5

= .5 / 3 *(3 / .5)

= 1 * (( 0 *in ) * in )) / ( 0 *in ))

= 3
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0*4; - (0*1 + 0*4(0*3)

= 2 - (.5 + (2(1.5);.../4) = -1.5

= -1.5 / 4 * (4 / -1.5)

= 1 * (( 0 *in ) * in )) / ( 0 *in ))

= 4
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0*3376547; - (0*1278314 + 0*3376547(0*2098233)

= 1688273.5 - (639157 + (1688273.5(1049116.5);.../3376547) = 524558.25
= 524558.25 / 3376547 * (3376547 / 524558.25)

= 1 * (( 0 *3376547 ) * 3376547 )) / ( 0 *3376547 ))

= 3376547
Please note at this point some calculators struggle.
You can subdivide all numbers by 1/2 (0) at any point in the process
to ensure accurate results
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0*934; - (0*105 + 0*934(0*829)

= 467 - (52.5 + (467(414.5);.../934) = 621.75

= 621.75 / 934 * (934 / 621.75)

= 1 * (( 0 *in ) * in )) / ( 0 *in ))

= 934
reminder: 0 = .5
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I don't need to continue to calculate proofs for Riemann, which is what this is. You are welcome to try and break it. If you do, let me know and I will update.
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Notepad:
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ip + in (np) - in / np
Attempted new definition - Incomplete. Stick to 3
Old - Misc.
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n + (np-2n)
FFT rewrite (Old) - Incomplete
Old - Balancing
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(( n + np (in) - in) / np ) *2) * 2in + ( -2n + np )) * 1/3, Or IN/N/NP, Or the ratio of X to Y, Z to X, Y to Z etc
Needs compressor (idea) - come back to this
Alteration
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/0)) / ( n + np ))
Re-evaluating none whole numbers
Balancing
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/0
Doubling numbers
Rectifier or bias
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*0
Halving numbers
Rectifier
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(( n + in (np) - in ) / np ) * np)) / np)) ;Before an input - partial limiterLimiter
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Turning 7 into 6: 2in + ( -2n + np )) and /np - Do not use multiples of two like this. Saved idea because it would be good practice to rework.Old form of adjustment (Rerouting)Old - Balancing
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* 1/3 Do not use factors of 3 like this. The idea is the ratio.Just an idea not to be used literallyIdea
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0.85714Noteable number (Consistent)Return
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* (n(np))Very helpful in raising ratios to larger numbersBalancing
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* np)) / np))Very helpful in raising ratios to larger numbersBalancing
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(( n + np (in) - in) / np ) * 2in + ( -2n + np )) * (1 1/3) - 2) / 6IdeaRatio
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(((( n + np (in) - in) / np ) * 2in + ( -2n + np )) /np + ( -n + np ))) *2) * 3 * 2 * 10IdeaRatio
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(( n + in (np) - in ) / np ) * np)) / np)) *np /0 ))IdeaRatio
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(( n + in (np) - in ) / np ) * np)) / np)) *np /0 ))IdeaRatio
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(( (1 + 2(1) - 2 ) / 1 ) * 1 ) / 1 )) * 1 ) / 0 ) = 1Idea (Old)Ratio
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-in (( n + in (np) - in ) / np ) * np)) / np)) *np /0 )) +nBrokenMisc
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-in (( n + in (np) - in ) / np ) * np)) / np)) *np /0 )) *in ) *0BrokenMisc
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in - (( n + in (np) - in ) / np ); * np)) / np)) *np )) /0 )) *in )) *0)); / (( 0 *in ) * in )) * (in)) - Where 0 = 1/2. Input equals output - Whole Limiter. No Imaginary Information. Please See Graph.BrokenMisc
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(( n + np (in) - in) / np ) * 2in + ( -2n + np ))BrokenMisc
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(( -in + ( n + np (in) - in) / np * 2)) / (( n + np (in) - in) / np ) * 2in + ( -2n + np )* (n(np))BrokenMisc
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(( n + np (in) - in) / np ) * 2in + ( -2n + np )) / ((-in + ( n + np (in) - in) / np * 2in - n))BrokenMisc
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-in + ( n + np (in) - in) / np * 2in - n ( n + np (in) - in) / np ) * 2in + ( -2n + np ))BrokenMisc
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(( n + in (np) - in ) / np ) /2in * (2)(1)(in(np) //// 0 = 1/2BrokenMisc
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