## Phi has a unique additive relationship.

The powers of phi have unusual properties in that they are related not only exponentially, but are additive as well. We know that:

Phi^{ 2} = Phi + 1

Which is the same as:

Phi^{ 2} = Phi^{ 1} + Phi^{ 0}

And this leads to the fact that for any n:

Phi^{ n+2} = Phi^{ n+1} + Phi^{ n}

Thus each two successive powers of phi add to the next one!

n | Phi^{n} |

0 | 1.000000 |

1 | 1.618034 |

2 | 2.618034 |

3 | 4.236068 |

4 | 6.854102 |

5 | 11.090170 |

6 | 17.944272 |

## Powers of Phi and its reciprocal:

Another little curiosity involves taking phi to a power and then adding or subtracting its reciprocal:

For any even integer n:

Phi^{ n} + 1 / Phi^{ n} = a whole number

For any odd integer n:

Phi^{ n} – 1 / Phi^{ n} = a whole number

Examples are shown in the tables below:

for n = even integers

n | Phi^{n} | 1/Phi^{ n} | Phi^{ n} + 1/Phi^{ n} |

0 | 1.000000000 | 1.000000000 | 2 |

2 | 2.618033989 | 0.381966011 | 3 |

4 | 6.854101966 | 0.145898034 | 7 |

6 | 17.944271910 | 0.055728090 | 18 |

8 | 46.978713764 | 0.021286236 | 47 |

10 | 122.991869381 | 0.008130619 | 123 |

for n = odd integers

n | Phi^{ n} | 1/ Phi^{ n} | Phi^{ n} – 1/Phi^{ n} |

1 | 1.618033989 | 0.618033989 | 1 |

3 | 4.236067977 | 0.236067977 | 4 |

5 | 11.090169944 | 0.090169944 | 11 |

7 | 29.034441854 | 0.034441854 | 29 |

9 | 76.013155617 | 0.013155617 | 76 |

11 | 199.005024999 | 0.005024999 | 199 |

The whole numbers generated by this have a relationship among themselves, creating an additive series, similar in structure to the Fibonacci series, and which also converges on phi:

Exponent n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |

Result | 2 | 1 | 3 | 4 | 7 | 11 | 18 | 29 | 47 | 76 | 123 | 199 |

Heyhey1D says

Good info but where is phi found?

Yunista says

This also leads to the result, which I’ve alywas remembered from my old college abstract algebra class, that the ratio of terms of the Fibonacci sequence approaches phi. Set up the ratio and take the limit, though one could also say that as the corrective term shrinks, the elements of the sequence approach (phi^n)/sqrt(5), and the sequence approaches a geometric one One could take that further and use the general formula for the sum of a geometric series to at least approximate the sums you began all this with, but it’d take a fairly large n to get good results Just found your blog today, and I’m really enjoying it!

Don says

Phi = [1+5^(1/2)]/2

or phi = (1+sq root of 5)/2

Larry G says

The expression with square root of 5 stems from solving the quadratic equation with A = 1. B = 1 and C = -1

Gary Meisner says

See the Phi Basics category listed in the menu above and https://www.goldennumber.net/what-is-phi/ as a start.

Salahi says

Thank you very much for these clear explanations of phi proeertips. I tried to solve some exercices about it and I was a bit lost I’m in premiere in a French high school and I have to do for the next week a big, big work in a group about a subject with both Maths and Biology (OK, we have to do it for months, but we’re a bit late ).We chose the golden number and how you can find it in nature, but we didn’t find some demonstrations (that’s right, we’re not cleverer than a banana). I’ve been looking for these on the web for long and you’re the first site which I understand !Thank you !

Joe says

Doesn’t the last series mentioned here (φ^n)+/-(φ^-n) for even and odd numbers respectively converge on φ^2, not on φ?

Ahh, my bad. The series generated by either the odd or the even integers converges on φ^2, but the series with both results converges on φ.

Beautiful!

The algebraic proof does pridove certain insights as well. For example, it tells you that the property,F(n+1)^2 – F(n)*F(n+2) = –[F(n)^2 – F(n–1)*F(n+1)] for all n,is independent of the initial values F(1) = 1, F(2) = 1, and therefore that something like Cassini’s identity will hold for any choice of initial conditions.This property can be rewritten F(n+1)^2 + F(n)^2 = F(n)*F(n+2) + F(n–1)*F(n+1), which has a simple geometric interpretation:The area of the two squares* * * * * * * ** * * * * * * ** * * * * * * ** * * * ** * * * *is the same as that of the two rectangles* * * * * * * ** * * * * * * ** * * * * * * ** * * * ** * * * *

new mathematics. look at http://www.phimath.net

The result of the exponent is called Lucas numbers.

all the powers of phi have the same form: http://prntscr.com/45ybe5. whem L is lucas numbers – the result of the exponent and F is fibonacci sequence.

Just messin’ around raising Phi to powers on my calculator.. I discovered for myself the Lucas numbers! which is a series similar to Fibonacci, with one astounding fact :-

Firstly fyi – to get the series the same rule applies where the next term is the sum of the previous two, just let the sequence start with 2 and 1 instead of 1 and 1.

2 1 3 4 7 11 18 29 etc…

The sequence also carries on leftwards where there is a spooky similar sequence with alternating sign:

…. -29 18 -11 7 -4 3 -1 2 1 3 4 7 11 18 29 …

The rule to get next term means, as per such series, that the further to the right we go (positive numbers) the ratio between successive terms approaches Phi.

The astounding fact to me is that Phi raised to higher and higher integer powers directly gives these numbers more and more accurately. Why does Phi raised to a very high number tend towards an integer boundary? I wonder if this is significant or useful.

I also noticed that Phi raised to higher powers tends toward integer results.. (i.e. mod1[phi^x} ==>0 as x==> infinity for all natural x). I wonder if there are any other non integers greater than 1 that have this property?

Interesting question. If you do this in Excel or most calculators, numbers can appear to converge to an integer simply because most calculators are only accurate to 15 significant digits. Accordingly, any number with more than 15 digits is going to round to the nearest integer. I tested this at http://keisan.casio.com/calculator, an online calculator with up to 50 significant digits and the results for the powers of Phi still converged on integers. To do this yourself, enter (5^.5*.5+.5)^x where x is the integer exponent. Then (5^.5*.5+.5)^114 yields 667714778405043259651217.999999999999999999999999 and (5^.5*.5+.5)^118 is the first number to round at 50 places to integer 4576585175559979410668403.

Now the more interesting question: Can anyone prove whether Phi is unique in this property???

Yes, I can. More interesting to me is that almost all of you are just ‘scratching the surface’. The whole area is much more profound than you would imagine from reading these posts.

And that’s why there are more than one hundred other pages on the topic on this site!

The part where you add or subtract the reciprocal of powers of phi are related to the Lucas numbers

Where does the Phi^2=Phi+1 come from?? Would you mind explaining how you got this rule?

See the derivation at https://www.goldennumber.net/math/. It comes from the expression of the relationships that exist when dividing a line A into two segments B and C such that A = B + C, and A / B = B / C. Solving for A on both sides give us B + C = B²/C. Say that C is 1 so we can determine the relative dimensions of the line segments. Now we simply have this: B² = B + 1.

Phi^2=phi+1

Phi^3=2phi+1

Phi^4=3phi+2

Phi^5=5phi+3

Phi^6=8phi+5

….etc….

Here we can notice the Fibonacci sequence

Phi has this amazing property

Phi^2=phi+1

Phi^3=2phi+1

Phi^4=3phi+2

Phi^5=5phi+3

Phi^6=8phi+5

….etc….

Here we can notice the Fibonacci sequence

I know this thread is almost 2 years old, but…

If F(n) is the nth Fibonacci number, and we extend the Fibonacci numbers backwards to cover all n integers, then you can generalize this

Phi^n = F(n)Phi + F(n-1)

Using the Fibonacci rule of F(n) + F(n+1) = F(n+2) we get F(n) = F(n+2) – F(n+1)

This gives, …, -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, …

Using simple induction, you can show that this will hold for all n integers.

For example:

Phi^(-4) = (-3)Phi + 5

Also, take any two consecutive Fibonacci numbers, F(n-1) and F(n), and two corresponding powers of Phi, Phi^(n-1) and Phi^n, then

F(n-1)Phi^(n-1) + F(n)Phi^n = Phi^(2n – 1)

For example,

13Phi^7 + 21Phi^8 = Phi^15

Simply multiply the formula Phi^n = F(n)Phi + F(n-1) by Phi^(n-1) to get this result.

Interested in power of Phi values appearing to approach whole numbers. Would be appreciative of any information regarding this.