Comments on: Number Five (5) and Phi

By: Ad

Ad — Mon, 10 Mar 2025 23:19:38 +0000

In reply to Gary B Meisner. For any integer n; (n+√n) x (n-√n) = n x n-1

By: Jeffrey

Jeffrey — Fri, 21 Oct 2022 17:36:58 +0000

In reply to Matty. 5 = five = PHIve which is phi + v + e.. The letter V is the roman numeral for five and the letter E is the fifth letter of the roman alphabet. Coincidence?!

By: Gary B Meisner

Gary B Meisner — Sun, 24 Oct 2021 19:25:20 +0000

In reply to Dude.

Very true! The unique mathematical property in this is that the cube of Phi is the square root of 5 plus 2:

Root 5 = 2.236…
2.236… + 2 = 4.236…
Phi cubed = 4.236

By: Dude

Dude — Fri, 22 Oct 2021 16:28:36 +0000

Phi can also be ((5^.5)+2)^(1/3)

By: Matty

Matty — Fri, 04 Jun 2021 03:38:15 +0000

In reply to Sergio Viana. 5 = PHIve.

By: Tea

Tea — Tue, 25 Jun 2019 09:39:12 +0000

In reply to Rhuben. It depends where the starting point is as most people start with 1 but the 1st Fibonacci number is 0. It's up for interpretation really but makes for sense to start 0 1 ... 0+1=1 1+1=2 and so on. Either way I'm just being pedantic and you are still right haha!

By: Jack Brunner

Jack Brunner — Thu, 06 Jun 2019 05:10:16 +0000

take a pentagon, then put all the lines in from each corner, that makes a new pentagon in the middle.
use that pentagon and mirror of each side of the pentagon and your big pentagon is almost full
repeat the same with every pentagon you have, then do it again and again and again, well I wnet to 6 levels, but it is infinite.
you still have the big pentagon but inside that a very beautiful pattern with all the gaps develops.
guess what in these gaps you can fill up with more pentagons of sizes you already have.
keep going and going till it is almost full. you have tiny diamonds left in the pattern.

By: randy selner

randy selner — Wed, 12 Dec 2018 05:52:01 +0000

Powers of phi can be expressed as the midpoint formula between two separate functions, both of which follow the Fibonacci sequence rule of adding the two previous values to get the next value.
Phi^3 = (4 + 2xsqrt(5))/2 ; Phi^2 = (3 + 1xsqrt(5))/2 ; Phi^1 = (1 + 1xsqrt(5))/2 ; Phi^0 = (2 + 0xsqrt(5))/2 ;
Phi^-1 = (-1 + 1xsqrt(5))/2 ; Phi^-2 = (3 – 1xsqrt(5))/2 ; Phi^-3 = (-4 + 2xsqrt(5))/2…
The two patterns can be seen to be whole numbers (of the Lucas number set, -11, 7, -4, 3, -1, 2, 1, 3, 4, 7, 11…) and the square root of 5 multiple numbers (of the recognizable Fibonacci sequence), of which each set contains alternating members of two specific functions relating to Phi.

Y = (0.5sqrt(5)+0.5)^X+(0,5sqrt(5)+0.5)^-X ; Y = (0.5sqrt(5)+0.5)^X-(0.5sqrt(5)+0.5)^-X

These two functions are as similar to e as hyperbolic cosine and sine functions (respectively), in that it is a power function split into a function similar to y=x^2, and a function similar to x^3, both of which add to make the function itself.
It is important to note, however, hyperbolic cosine and sine are 1/2 the difference between their power difference functions, so adding sinh(x) to cosh(x) gives e^x instead of 2e^x.
So the functions represented by the parts given relate exactly to the values including the 2 in the denominator, if functions similar to hyperbolic cosine and sine but relating to phi^x instead, the functions as given above would need to be halved.

Y =0.5 ((0.5sqrt(5)+0.5)^X+(0,5sqrt(5)+0.5)^-X) ; Y =0.5 ((0.5sqrt(5)+0.5)^X-(0,5sqrt(5)+0.5)^-X) or

Y = (Phi^x+Phi^-x)/2 and Y = (Phi^x-Phi^-x)/2

This gives legitimacy as to why the fibonacci numbers in negative powers alternate between positive and negative values.

By: Gary B Meisner

Gary B Meisner — Tue, 01 Mar 2016 02:35:35 +0000

In reply to Simon Smallwood. It's just another way to express four fives in a relationship, although admittedly not one that represents Phi.

By: Simon Smallwood

Simon Smallwood — Mon, 29 Feb 2016 18:48:20 +0000

In reply to Fred Whelch.

Try √((A2+A3)/(A2-A3)) where A2=5 AND A3=√(5).

THIS WORKS FOR ME!

BUT I couldn’t make any sense of

(5 + √5) x (5 – √5) = 5 + 5 + 5 + 5 so what?