Averaging Phi 
Averaging Phi 
PJS 
Oct 07, 2010, 09:52 AM
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#1

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
x^3 + 2x^2 = 1
2x^2 = 1  x^3 x^2 = (1  x^3)/2 x = sqrt (1  x^3)/2 1/x = averaged phi For example: x = 0.6 x = sqrt (1  0.216)/2 x = sqrt (0.784 / 2) = 0.392 x = sqrt 0.392 = 0.62609903369994111499456862724476 1/x = 1.5971914124998497831494097633795 next enter x = 0.62609903369994111499456862724476 Continue to enter each subsequent x into sqrt (1  x^3)/2 then invert the answer 1/x for averaged phi. After a few entries the number phi will start to be produced. P.j.S 
Jupiter 
Jul 23, 2011, 10:21 AM
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#2

Newbie Group: Basic Member Posts: 13 Joined: Jul 19, 2011 Member No.: 33460 
This is a non intelligent way to compute Phi ,the best would be to use one of the infinite series of this constant .
When one ,wants to compute a special constant it is more effective to use its properties ,an intuitive approach is to say that the more separate information is used the more effective and precise the method of computation is . 
P.j.S 
Jul 26, 2011, 04:00 AM
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#3

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
This is a non intelligent way to compute Phi ,the best would be to use one of the infinite series of this constant . When one ,wants to compute a special constant it is more effective to use its properties ,an intuitive approach is to say that the more separate information is used the more effective and precise the method of computation is . x^3 + 2x^2 = 1 is an equation used to confirm the digits of phi. To say that I have originated this equation nonintelligently is an insult. I doubt that another equation can be found to duplicate what I have done. Can you write a new equation to approach the digits of phi in some manner and confirm phi by solving for the unknown quantity "x"? I doubt it very much. Quickly, I think that I can. (x^3 + 2x^2 + sqrt 5) / 2 = averaged phi = 1/x. For example start at 0.6 again then subsequently enter the new "x" value found from inverting averaged phi. Then enter the new x value at the start of the equation to get a nearer value to phi. PJS 
PJS 
Oct 25, 2011, 05:20 PM
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#4

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
New Discovery about Phi.
(sqrt 5 + 1) / 2 = phi phi^35 = 20633239.0 17.5th root of phi^35 = 1 + phi = 2 + 1/phi As a diameter phi^35 * Creative Pi = 64967633.0 a perfect circle for rational pi. 64967633 / phi^35 = 1080/7^3 From the School of P.j.S Original Mathematics & Related Sciences 
Phi 
Oct 27, 2011, 11:03 PM
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#5

God Group: Basic Member Posts: 1351 Joined: Jul 11, 2008 From: Las Vegas, NV Member No.: 25755 
so they say...

PJS 
Nov 07, 2011, 11:11 AM
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#6

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
i thought that i found a new whole number but a calculator with a longer display proved me wrong.
phi^35 = 20633239.00000004846548813785357 New Discovery about Phi. (sqrt 5 + 1) / 2 = phi phi^35 = 20633239.0 17.5th root of phi^35 = 1 + phi = 2 + 1/phi As a diameter phi^35 * Creative Pi = 64967633.0 a perfect circle for rational pi. 64967633 / phi^35 = 1080/7^3 From the School of P.j.S Original Mathematics & Related Sciences phi^35 / Creative Pi = a whole number to 7 decimal places. The diameter = 6552963.8675926079848726215590504... A 0 in the 8th decimal place and a 1 in the next 9th decimal position yields the binary logic of 0 and 1. What else could the digits of phi reveal in the next 10th decimal position from the first 1? Just an interesting observation. PJS standing by. 
PJS 
Jan 03, 2012, 04:11 AM
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#7

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
x = 1/phi
x = (1 + sqrt 5) / 2 = 1/x = 1/phi 1/phi = x = ((x^3  2x^2) + sqrt 5) / 2 = 1/phi = x Note: x^3  2x^2 = 1 when x^2 + 1 = 0 in this expression. x, x^2 and x^3 = 1. (1 + sqrt 5) / 2 = 1/phi. 1/phi = x = (x^3 + 2x^2 = 1). (1 + sqrt 5) / 2 = phi. PJS standing by. 
PJS 
Jan 03, 2012, 06:09 AM
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#8

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
New Discovery about Phi. (sqrt 5 + 1) / 2 = phi phi^35 = 20633239.0 17.5th root of phi^35 = 1 + phi = 2 + 1/phi As a diameter phi^35 * Creative Pi = 64967633.0 a perfect circle for rational pi. 64967633 / phi^35 = 1080/7^3 From the School of P.j.S Original Mathematics & Related Sciences phi^35 / Creative Pi = a whole number to 7 decimal places. The diameter = 6552963.8675926079848726215590504... A 0 in the 8th decimal place and a 1 in the next 9th decimal position yields the binary logic of 0 and 1. What else could the digits of phi reveal in the next 10th decimal position from the first 1? Just an interesting observation. PJS standing by. 
PJS 
Jan 03, 2012, 06:11 AM
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#9

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
New Discovery about Phi. (sqrt 5 + 1) / 2 = phi phi^35 = 20633239.0 17.5th root of phi^35 = 1 + phi = 2 + 1/phi As a diameter phi^35 * Creative Pi = 64967633.0 a perfect circle for rational pi. 64967633 / phi^35 = 1080/7^3 From the School of P.j.S Original Mathematics & Related Sciences phi^35 / Creative Pi = a whole number to 7 decimal places. The diameter = 6552963.8675926079848726215590504... A 0 in the 8th decimal place and a 1 in the next 9th decimal position yields the binary logic of 0 and 1. What else could the digits of phi reveal in the next 10th decimal position from the first 1? Just an interesting observation. PJS standing by. 
P. Jay 
Jan 03, 2012, 04:42 PM
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#10

Aspiring Group: Basic Member Posts: 79 Joined: Dec 30, 2011 Member No.: 33864 
New Discovery about Phi. (sqrt 5 + 1) / 2 = phi phi^35 = 20633239.0 17.5th root of phi^35 = 1 + phi = 2 + 1/phi As a diameter phi^35 * Creative Pi = 64967633.0 a perfect circle for rational pi. 64967633 / phi^35 = 1080/7^3 From the School of P.j.S Original Mathematics & Related Sciences phi^35 / Creative Pi = a whole number to 7 decimal places. The diameter = 6552963.8675926079848726215590504... A 0 in the 8th decimal place and a 1 in the next 9th decimal position yields the binary logic of 0 and 1. What else could the digits of phi reveal in the next 10th decimal position from the first 1? Just an interesting observation. PJS standing by. With a new whole number to 7 decimal places would not the binary language start at the 8th decimal with 0 then next 9 decimal places later with 1 then to the next 0 or 1? With the weighted decimals in between for opposing purposes? 0.........1...0.0 and so on with the new Perfect Circle in the diameter? 0000000........1 in the circumference 1..1.0...............0 with phi 0...0...............0 with 1/phi 0...............0.1.......1 with sqrt 5 
P JayS 
Apr 19, 2012, 02:46 PM
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#11

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
operators: * / +  o
phi oppose 1/phi phi o 1/phi = 1.0 1.6180339887 o 0.6180339887 = 1.0 averaged phi, 1.0 * averaged phi = averaged phi sqrt 5 o sqrt 5 = 0 2.2360679774 o 2.2360679774 = 0.0 sqrt 5 o sqrt 2 2.2360679774 o 1.4142135623 = a. 1.2222544151 sqrt 5 0 1/phi 2.2360679774 o 0.6180339887 = b. 2.4220340113 a. o b. 1.2222544151 o 2.4220340113 = 1.2002204042 = 1..00..0.0.. if the weighted binary calculations are correct. 
P JayS 
Apr 21, 2012, 06:27 AM
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#12

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
operators: * / +  o phi oppose 1/phi phi o 1/phi = 1.0 1.6180339887 o 0.6180339887 = 1.0 averaged phi, 1.0 * averaged phi = averaged phi sqrt 5 o sqrt 5 = 0 2.2360679774 o 2.2360679774 = 0.0 sqrt 5 o sqrt 2 2.2360679774 o 1.4142135623 = a. 1.2222544151 sqrt 5 0 1/phi 2.2360679774 o 0.6180339887 = b. 2.4220340113 a. o b. 1.2222544151 o 2.4220340113 = 1.2002204042 = 1..00..0.0.. if the weighted binary calculations are correct. 2 random decimals opposed: a. 0.1793462469 o b. 0.5623798547 = 0.4170336122 = 0..1.0...1.. 0.4170336122 Digit Weight a. 0.0170000022 Digit Weight b. 0.4000336100 0..1.0...1.. if calculated correctly 
P JayS 
May 12, 2012, 07:02 AM
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#13

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
2x^2 + 3y + z = 1
z = 1  (2x^2 + 3y) 2x^2 + 3y = 1 1) z = 0 3y = x^3 x^3 + 2x^2 + 3y + 2x^2 = 2 x^3 + 3y = 2  4x^2 6y = 2  4x^2 6y  2 = 4x^2 (6y  2) / 4 = x^2 1.5y  0.5 = x^2 sqrt (1.5y  0.5) = x note x = averaged 1/phi from x = sqrt (1  x^3) / 2 2) x = 1/phi (0.6180339887)^3 + 4(0.6180339887)^2 + 3y = 2 0.236067977 + 1.527864045  2 = 3y 1.763932022  2 = 3y x^3 = 3y x^3 / 3 = y 3) y = 0.078689325 x = 1/phi y = x^3 / 3 z = 0 
P JayS 
May 12, 2012, 07:57 AM
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#14

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
x = 0.5
2x^2 + 3y + z = 1 .5 + 0.125 + z = 1 0.625 + z = 1 z = 1  0.625 z = 0.375 3y = 1  (2x^2 + z) 3y = 0.125 y = 0.041667 
Flex 
May 12, 2012, 08:12 AM
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#15

God Group: Basic Member Posts: 1954 Joined: Oct 17, 2006 From: Bay area CA Member No.: 5877 
x = 0.5 2x^2 + 3y + z = 1 .5 + 0.125 + z = 1 0.625 + z = 1 z = 1  0.625 z = 0.375 OR 1 + 1 + z = 1 z = 1  2 z = 1 Very interesting indeed. You know P Jay, Jupiter may be right. You are no doubt finding some very unique connections that seem crazy, but are really pretty sound. The fact that you lack the classical training, makes it so that you are presenting new connections that are very difficult to see in the confines of academia. The key is balance. I have my own math to contribute! I am working on it now, be back in like 30 when I figure it out! 
P JayS 
May 12, 2012, 08:23 AM
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#16

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
x = 0.5 2x^2 + 3y + z = 1 .5 + 0.125 + z = 1 0.625 + z = 1 z = 1  0.625 z = 0.375 OR 1 + 1 + z = 1 z = 1  2 z = 1 Very interesting indeed. You know P Jay, Jupiter may be right. You are no doubt finding some very unique connections that seem crazy, but are really pretty sound. The fact that you lack the classical training, makes it so that you are presenting new connections that are very difficult to see in the confines of academia. The key is balance. I have my own math to contribute! I am working on it now, be back in like 30 when I figure it out! I made an error in the "B" part of my comment. 1 + 0.125 + z = 1 instead of 1 + 1 + z = 1. I accidentally cubed 2x^2 = 1 as 3y. Sorry! 
P JayS 
May 12, 2012, 08:36 AM
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#17

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
1 + 1 + z = 1
(x^3 + 2x^2) + (3y + 2x^2) + z = 1 x = 1/phi y = x^3/3 z = 1  2 z = 1 x^3  2z = 1 z = 1 z = x^2 x^3 + 2 = 1 x^3 = 1  2 x^3 = 1 x = 1 
Flex 
May 12, 2012, 08:52 AM
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#18

God Group: Basic Member Posts: 1954 Joined: Oct 17, 2006 From: Bay area CA Member No.: 5877 
Sum (n=0 to infinity) tr^n =t+tr+tr^2...tr^n
Ok so this is the series expansion whose sum converges to t/1r Let now look at the solution we get to this series expansion when we look at the sum where n=i and the function itself is e^ui/KbT and ui=ixKbT Reduces to sum n=i of e^i and when t=1 and r=1/e we get the ratio of e/(e1) = 1.5819..... Pretty damn close to phi eh? Now if we take this and square it, we get (e/e1)^2 = 3.16395 Notice that the first number is less than phi, but close, and the squared term is more than pi, but close. If we carry this on forever, I believe we get close and closer to true values of both numbers. The reason we never get a real solution is that the answer is relating two irrational numbers, resulting in an irrational solution. Looks like a fractal to me *I apologize for the chemistry variables I used, I came the this conclusion in class the other day lol. Basically the idea is that if you take something like (1/e^0)+(1/e^1)+(1/e^n) you near phi, and when we square phi, we near pi. I do not think it any coincidence the random ass solution I came up with in class happens to be very similar to the time constant. In fact, to me it almost seems like two sides of the same coin. http://en.wikipedia.org/wiki/Time_constant 
P JayS 
May 12, 2012, 08:55 AM
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#19

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
Sum (n=0 to infinity) tr^n =t+tr+tr^2...tr^n Ok so this is the series expansion whose sum converges to t/1r Let now look at the solution we get to this series expansion when we look at the sum where n=i and the function itself is e^ui/KbT and ui=ixKbT Reduces to sum n=i of e^i and when t=1 and r=1/e we get the ratio of e/(e1) = 1.5819..... Pretty damn close to phi eh? Now if we take this and square it, we get (e/e1)^2 = 3.16395 Notice that the first number is less than phi, but close, and the squared term is more than pi, but close. If we carry this on forever, I believe we get close and closer to true values of both numbers. The reason we never get a real solution is that the answer is relating two irrational numbers, resulting in an irrational solution. Looks like a fractal to me Good work! 
Flex 
May 12, 2012, 01:17 PM
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#20

God Group: Basic Member Posts: 1954 Joined: Oct 17, 2006 From: Bay area CA Member No.: 5877 
Beautiful man... Those Aholes who made fun of you had no idea that you held the secret to the interplay between space and time itself (potentially ) . You are a good man! Don't ever let them tell you otherwise.

P JayS 
May 14, 2012, 03:07 PM
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#21

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
Sum (n=0 to infinity) tr^n =t+tr+tr^2...tr^n Ok so this is the series expansion whose sum converges to t/1r Let now look at the solution we get to this series expansion when we look at the sum where n=i and the function itself is e^ui/KbT and ui=ixKbT Reduces to sum n=i of e^i and when t=1 and r=1/e we get the ratio of e/(e1) = 1.5819..... Pretty damn close to phi eh? Now if we take this and square it, we get (e/e1)^2 = 3.16395 Notice that the first number is less than phi, but close, and the squared term is more than pi, but close. If we carry this on forever, I believe we get close and closer to true values of both numbers. The reason we never get a real solution is that the answer is relating two irrational numbers, resulting in an irrational solution. Looks like a fractal to me *I apologize for the chemistry variables I used, I came the this conclusion in class the other day lol. Basically the idea is that if you take something like (1/e^0)+(1/e^1)+(1/e^n) you near phi, and when we square phi, we near pi. I do not think it any coincidence the random ass solution I came up with in class happens to be very similar to the time constant. In fact, to me it almost seems like two sides of the same coin. http://en.wikipedia.org/wiki/Time_constant When you square phi you get 1 + phi. But when you multiply phi by 2 you get a figure near pi. 2(phi)  1  sqrt 5 = 0 2(phi) = 1 + sqrt 5 How does multiplying your answer by 2 instead of squaring affect your answer 2(e/e1) approximately pi? 
Flex 
May 15, 2012, 03:14 AM
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#22

God Group: Basic Member Posts: 1954 Joined: Oct 17, 2006 From: Bay area CA Member No.: 5877 
Sum (n=0 to infinity) tr^n =t+tr+tr^2...tr^n Ok so this is the series expansion whose sum converges to t/1r Let now look at the solution we get to this series expansion when we look at the sum where n=i and the function itself is e^ui/KbT and ui=ixKbT Reduces to sum n=i of e^i and when t=1 and r=1/e we get the ratio of e/(e1) = 1.5819..... Pretty damn close to phi eh? Now if we take this and square it, we get (e/e1)^2 = 3.16395 Notice that the first number is less than phi, but close, and the squared term is more than pi, but close. If we carry this on forever, I believe we get close and closer to true values of both numbers. The reason we never get a real solution is that the answer is relating two irrational numbers, resulting in an irrational solution. Looks like a fractal to me *I apologize for the chemistry variables I used, I came the this conclusion in class the other day lol. Basically the idea is that if you take something like (1/e^0)+(1/e^1)+(1/e^n) you near phi, and when we square phi, we near pi. I do not think it any coincidence the random ass solution I came up with in class happens to be very similar to the time constant. In fact, to me it almost seems like two sides of the same coin. http://en.wikipedia.org/wiki/Time_constant When you square phi you get 1 + phi. But when you multiply phi by 2 you get a figure near pi. 2(phi)  1  sqrt 5 = 0 2(phi) = 1 + sqrt 5 How does multiplying your answer by 2 instead of squaring affect your answer 2(e/e1) approximately pi? PJ! Brilliant! You are hitting that damn nail on the head. Thank you for lending this insight! Thought I intentionally chose to square the term as this is symbolic in the quantum chemistry world of creating a probability distribution. 
P JayS 
May 15, 2012, 10:36 AM
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#23

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
a. x^3 + 2x^2  3y  2x^2 = 0
x^3  3y = 0 3y = x^3 y = x^3 / 3 y = x^3 / 3 x^3 = 1 y = 1/3 3y = 1 x^3 = 3y x^3  3y = 0 1  1 = 0 b. x^3 + 2x^2 + 3y + 2x^2 = 2 x^3 + 4x^2 + 3y = 2 x^3 + 3y = 2  4x^2 x^3 + 3y = x x / 2 = 3y x^3 = 3y 3y + 3y  2 = 4x^2 (6y  2) / 4 = x^2 1.5y + 0.5 = x^2 (y = 0.5 /  1.5) = x^2 (y = 1/3) = ((1.5 * 1/3) + 0.5) = 0 = x^2 x^2 = 0 x = sqrt 0 x = 0 y = 1/3 y = x^3/3 x^3 = 1 3y = 1 x = 1 x^3 + 3y + 4x^2 = 2 1 + 1 + 0 = 2 Here x = 1 and x = 0 at the same time are proved in this equation. This makes the Subtraction Law for energy totally possible. an  bn  cn = 0 where energy = 1an and the void = 0 are present at the same time. 1 and 0 are at opposite ends of time at the same time that they are both either 1 and 0 at the same time. This means that 0 is nothing the same time that 1 is everything instead of an either/or situation from a much slower cycle. When invisible energy penetrates the void then energy slows down from c^2 speed and collects to become visible matter. We know that x^3 + 3y + 4x^2 = 2 is correct by x = 1/phi and y = x^3/3. 
balika 
Nov 15, 2012, 10:23 AM
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#24

Newbie Group: Basic Member Posts: 15 Joined: Oct 26, 2012 Member No.: 34648 
This is a good idea to all the people in the world which make some thing new. hiprospeed.com . I know the silicone hose which make some new and different to all the people in the world.

P JayS 
Jan 08, 2013, 02:05 PM
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#25

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
x^3 + 2x^2 = 1 2x^2 = 1  x^3 x^2 = (1  x^3)/2 x = sqrt (1  x^3)/2 1/x = averaged phi For example: x = 0.6 x = sqrt (1  0.216)/2 x = sqrt (0.784 / 2) = 0.392 x = sqrt 0.392 = 0.62609903369994111499456862724476 1/x = 1.5971914124998497831494097633795 next enter x = 0.62609903369994111499456862724476 Continue to enter each subsequent x into sqrt (1  x^3)/2 then invert the answer 1/x for averaged phi. After a few entries the number phi will start to be produced. P.j.S Averaged Phi space on a mathematical exercise board in the room of objectivity in a moment of time before the most courageous leader our Heavenly Father Jehovah God would ever bring any destruction upon the wicked. In stead please save everyone!. 16 / 61 = 7^3 = 1 / 0 = 1, 2, 3 in a three fold chord = 3.0 = 4: Love, Joy, Peace and Happiness. P.j.S standing by Lord. Father knows why!!!!!:.::!???!!.>::. 
P JayS 
Jan 19, 2013, 03:31 PM
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#26

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
x = ?
x = 1 1 = ? 0 = ?  x 0 = ?  1 ? = 1 x = ? = 1 x + ? = 2 = 1 x + x = 2 = 1 2x = 2 = 1 + x x = x * ? (x = [x) * (x] = x) x * x = ? x + ? = x * x x + x = x^2 (2x = x^2) 2x / x^2 = ? ? = (1, 2) x = (1, 2) 2x = x^2 2 *1 = 1 * 1 (2 * 1) / (1 * 1) = 2 2x / x^2 = 2 2x = 2x^2 2x / 2 = x^2 x = x^2 = 1 2x^2 = 2 * 1 * 1^2 = 2 * x = 2 * 1 = 2 2x^2 = 2 x^2 = 2/2 x = 1 2x^2 = 4 = x 2 = 4 = x * x^2 4 / 2 = x^3 2 = x^3 = 1 x = ( 1, 2, 4, 8, 16, 32, 64, 128...) 2 = 2x^2 2x^2 = x^3 = 64 ? = x^3  2x^2 ? = 1 x^3  2x^2 = 1 x^2 = 1 x^3  2(1) = 1 x^3  (2) = 1 x^3 + 2 = 1 2 = 2x^2 x^3 + 2x^2 = 1 2x^2 = 1  x^3 x^2 = (1  x^3) / 2 x = sqrt (1  x^3) / 2 1 / x = averaged phi; the mind twisting mess of x = ? P.j.S . 
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