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> Averaging Phi
PJS
post Oct 07, 2010, 09:52 AM
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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.

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Jupiter
post Jul 23, 2011, 10:21 AM
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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 .
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P.j.S
post Jul 26, 2011, 04:00 AM
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QUOTE(Jupiter @ Jul 23, 2011, 10:21 AM) *

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 non-intelligently 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
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PJS
post Oct 25, 2011, 05:20 PM
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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
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Phi
post Oct 27, 2011, 11:03 PM
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so they say...
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PJS
post Nov 07, 2011, 11:11 AM
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i thought that i found a new whole number but a calculator with a longer display proved me wrong.

phi^35 = 20633239.00000004846548813785357

QUOTE(PJS @ Oct 25, 2011, 05:20 PM) *

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.
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PJS
post Jan 03, 2012, 04:11 AM
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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.
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PJS
post Jan 03, 2012, 06:09 AM
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QUOTE(PJS @ Oct 25, 2011, 05:20 PM) *

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.
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PJS
post Jan 03, 2012, 06:11 AM
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QUOTE(PJS @ Oct 25, 2011, 05:20 PM) *

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.
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P. Jay
post Jan 03, 2012, 04:42 PM
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QUOTE(PJS @ Jan 03, 2012, 06:11 AM) *

QUOTE(PJS @ Oct 25, 2011, 05:20 PM) *

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
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P JayS
post Apr 19, 2012, 02:46 PM
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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.
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P JayS
post Apr 21, 2012, 06:27 AM
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QUOTE(P JayS @ Apr 19, 2012, 02:46 PM) *

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
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P JayS
post May 12, 2012, 07:02 AM
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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
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P JayS
post May 12, 2012, 07:57 AM
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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
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Flex
post May 12, 2012, 08:12 AM
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QUOTE(P JayS @ May 12, 2012, 08:57 AM) *

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! smile.gif
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P JayS
post May 12, 2012, 08:23 AM
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QUOTE(Flex @ May 12, 2012, 08:12 AM) *

QUOTE(P JayS @ May 12, 2012, 08:57 AM) *

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! smile.gif

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!
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P JayS
post May 12, 2012, 08:36 AM
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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
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Flex
post May 12, 2012, 08:52 AM
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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/1-r

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/(e-1) = 1.5819..... Pretty damn close to phi eh?

Now if we take this and square it, we get (e/e-1)^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 smile.gif

*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
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P JayS
post May 12, 2012, 08:55 AM
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QUOTE(Flex @ May 12, 2012, 08:52 AM) *

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/1-r

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/(e-1) = 1.5819..... Pretty damn close to phi eh?

Now if we take this and square it, we get (e/e-1)^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 smile.gif

Good work!
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Flex
post May 12, 2012, 01:17 PM
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Beautiful man... Those A-holes who made fun of you had no idea that you held the secret to the interplay between space and time itself (potentially smile.gif ) . You are a good man! Don't ever let them tell you otherwise.
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P JayS
post May 14, 2012, 03:07 PM
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QUOTE(Flex @ May 12, 2012, 08:52 AM) *

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/1-r

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/(e-1) = 1.5819..... Pretty damn close to phi eh?

Now if we take this and square it, we get (e/e-1)^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 smile.gif

*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/e-1) approximately pi?
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Flex
post May 15, 2012, 03:14 AM
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QUOTE(P JayS @ May 14, 2012, 04:07 PM) *

QUOTE(Flex @ May 12, 2012, 08:52 AM) *

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/1-r

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/(e-1) = 1.5819..... Pretty damn close to phi eh?

Now if we take this and square it, we get (e/e-1)^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 smile.gif

*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/e-1) 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.
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P JayS
post May 15, 2012, 10:36 AM
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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.
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post Nov 15, 2012, 10:23 AM
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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.
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P JayS
post Jan 08, 2013, 02:05 PM
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QUOTE(PJS @ Oct 07, 2010, 09:52 AM) *

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!!!!!:.::!???!!.>::.
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P JayS
post Jan 19, 2013, 03:31 PM
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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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