# Phi and Mathematics

Note: n^{x} means n raised to the x power. Some browsers may not display exponents as superscripts or raised characters.

## Deriving Phi mathematically

Phi can be derived by solving the equation:

n^{2} - n^{1} - n^{0} = 0,

which is the same as:

n^{2} - n - 1 = 0

This equation can be rewritten as:

n^{2} = n + 1 and 1 / n = n – 1

Using the Quadratic formula, the solution to the equation is (1 plus or minus the square root of 5) divided by 2:

( 1 + √5 ) / 2 = 1.6180339… = Phi

( 1 – √5 ) / 2 = -0.6180339… = -phi

The reciprocal of Phi (denoted with an upper case P), is known often as by phi (spelled with a lower case p).

This, then, results in two properties unique to Phi:

- If you square Phi, you get a number exactly 1 greater than Phi: 2.61804…

Phi^{2} = Phi + 1

- If you divide Phi into 1, you get a number exactly 1 less than Phi: 0.61804…:

1 / Phi = Phi – 1

Phi, curiously, can also be expressed all in fives as:

5 ^ .5 * .5 + .5 = Phi

This provides a great, simple way to compute phi on a calculator or spreadsheet!

## Determining the nth number of the Fibonacci series

You can use phi to compute the nth number in the Fibonacci series (f_{n}):

f_{n} = Phi^{ n} / 5^{½}

As an example, the 40th number in the Fibonacci series is 102,334,155, which can be computed as:

f_{40} = Phi^{ 40} / 5^{½} = 102,334,155

This method actually provides an estimate which always rounds to the correct Fibonacci number.

You can compute any number of the Fibonacci series (f_{n}) exactly with a little more work:

f_{n} = [ Phi^{ n} - (-Phi)^{-n} ] / (2Phi-1)

Note: 2Phi-1 = 5^{½}= The square root of 5

## Determining Phi with Trigonometry and Limits

Phi can be related to e, the base of natural logs,

through the inverse hyperbolic sine function:

Phi = e ^ asinh(.5)

It can be expressed as a limit:

or

## Other unusual phi relationships

There are many unusual relationships in the Fibonacci series. For example, for any three numbers in the series Phi(n-1), Phi(n) and Phi(n+1), the following relationship exists:

Phi(n-1) * Phi(n+1) = Phi(n)^{2} - (-1)^{n}

( e.g., 3*8 = 5^{2}-1 or 5*13=8^{2}+1 )

Here’s another:

Every nth Fibonacci number is a multiple of Phi(n),

where Phi(n) is the nth number of the Fibonacci sequence.

Given 0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765

(Every 4th number, e.g., 3, 21, 144 and 987, are all multiples of Phi(4), which is 3)

(Every 5th number, e.g., 5, 55, 610, and 6765, are all multiples of Phi(5), which is 5)

And another:

The first perfect square in the Fibonacci series, 144,

is number 12 in the series and its square root is 12!

0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

or, if not starting with 0:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

{ 21 comments… read them below or add one }

We should develop a logical system based on powers of phi to define itself.

if you start with φ^-∞ for zero state. Next φ^-2 would be useful to add to φ and φ^2 to get 2 and 3. With zero power you have 1 together with 0,2,3 whole numbers represented in endless possible combinations of powers of φ. Each combination could represent a different set of logic. I am only barely grasping what kind of operations would be possible in virtual machine code! With a base of φ…

Your logical systems resembles that of Godel on the primes for p>p, where the series is 2,3,5,7 taken in powers representing characters, such that p=0,>=1. The 2^0 * 3^1 * 5^0 = 1*3*1 = 3, which is a prime representing p>p. To do the same on the Fibonacci series 0,1,1,2 …. 0^1 * 1^1 *1^0 = 0*1*1= 0

If you divide phi into 1, you get a number exactly 1 less than phi: 0.61804…:

1 / Phi = Phi – 1 = 0.3 + 1/pi = 1.61804 – 1 = 0.3 + 0.318..

1/Phi – 1/pi = 0.3 For example 34/55 = 0.6181818181818182…, 34 and 35 are Fibonacci numbers; 1/pi = 7/22 = 0.3181818181818182, which difference is 0.3; and 34/55 – 7/22 = 1/11 *(34/5 -7/2) = 1/11 * (68 – 35)/10 = 33/11 * 1/10 = 0.3

Even though I agree that recursion is not the right way to find a Fibonacci number, I have to say using it to teach recursion is a good idea. Because its simple. Easy to get the concept thru this example.The most common real world application for recursion is to walk thru a tree structure. That example will be complicated. Also you will have to have an understanding of trees and stuff before hand.Then again, there may be a simple, but real world example that I’m not yet aware of. If someone knows of one, let me know.

You mean like ‘i’?

It would be very interesting.

First of all i would like to thank and appreciate you for this tremendous work and the content you have used in this site.

but can you please tell me the logic behind the series given below?…………here is the series

2 3 8 21 55

The series you list is part of the Fibonacci series, in which each number is the sum of the two before it. In the case of your particular series the logic appears to be that you ignore every other result, i.e., the 5, 13 and 34. Another way of getting to the same result is to use this logic: 2+3+8*2=21 and 3+8+21*2=55 so the next number in the series should be 8+21+55*2=139.

this is very helpful

How do you know all of this math (golden number)? I’m in the 8 grade i want to learn this. How do I get help?

I want to know what is the englis equivalent word of greek word ‘phi’

Phi is a letter of the Greek alphabet, so its closest English equivalent is the letter F. Phi was chosen to represent the golden ratio to recognize the Greek sculptor Phidias, who lived from about 480 – 430 BC. He is regarded as one of the greatest sculptors of classical Greece. His statue of Zeus at Olympia was one of the Seven Wonders of the Ancient World. He also designed the statues of the goddess Athena on the Athenian Acropolis. Mark Barr, a mathematician, first used the Greek letter phi (Φ) to designate the golden ratio. See the Golden Ratio History page for more.

I learned how the square root of 1 plus the square root of 1 plus square root of 1… etc.

if you let x=squareroot(1+squareroot(1+squareroot(1…))) then square both sides you get

x^2 = 1 + squareroot(1+squareroot(1+square root(1….)))

then substitute x in for the squareroot(1+…)

x^2=1+x

then subtract both sides by (1+x) to get

x^2-x-1 = 0

Then use the quadratic formula

x=(1 +/- squareroot((-1)^2 – 2*(1)*(-1)))/(2*1)

Clean everything up to get:

x= (1 +/- squareroot(5))/2

Take the positive outcome, and get phi!!

Thank you.

Φ^n = F(n)*Φ + F(n-1)

When F(n) is the “n”th Fibonacci number (n≠1, obviously)

Has anyone else noticed this?

This family of equations has other solutions besides phi, (i.e. (1-sqrt5)/2 and complex numbers) but phi is a solution for all of them, and also the only positive solution for any of them.

Hi Peter,

yes, I found this, too.

And you mentioned the second root of eq. n^2 – n – 1 = 0: let’s call it phi (small phi, it is < 0)

Here is another nice recursion for Fibonacci, using Phi and phi:

F[n] = Phi^(n-1) + phi * F[n-1]

Have fun!

I’m in 10th grade and doing my math project on phi. Can anyone give me a few more, preferably simple, interesting things about phi in math? I need to explain it to people who’ve never heard of it before. Please reply by 5th July

Check out the Phi Basics and Geometry pages for some practical and visual applications.

How do you show that [1+√5/2][1+√5/2]= [1+√5/2]+1. I’ve tried it in so many ways but I never make it equal.. Help..

Hi Jessica,

The proof you are looking for is as follows;-

(1 + √5)/2 * (1 + √5)/2 = ((1 + √5) * (1 + √5)) /4

= (1 + 2*√5 +5) /4

= (6 + 2*√5) /4

= (6 + 2*√5)/4 – 4/4 + 4/4

= (6 + 2*√5 – 4)/4 + 1

= (2 + 2*√5) /4 + 1

= (1 + √5) /2 + 1

Cheers,

Ted.

Also if anyone’s interested here is the proof of Φ = 1/Φ + 1 ;-

Since Φ = (1 + √5) / 2

Then, if Φ = 1/Φ + 1, it follows that;-

(1 + √5) / 2 = 2/(1 + √5) +1

= 2/(1 + √5) + (1 + √5)/(1 + √5)

= (2+ 1 + √5)/(1 + √5)

= (3 + √5)/(1 + √5)

= (3 + √5)/(1 + √5) * 1

= (3 + √5)/(1 + √5) * (2/(1 + √5))/(2/(1 + √5))

= ((3 + √5) * (2/(1 + √5))) / ((1 + √5) * (2/(1 + √5)))

= ((6 + 2√5) / (1+√5)) / 2

= ((1 + √5)^2) / (1+√5)) / 2

= (1 + √5) / 2

Therefore Φ = 1/Φ + 1

Regards,

Ted Krapkat

I’ve just started down this path, I’m excited. Carroll’s “curiouser and curiouser” keep popping into my head