## Phi, Φ, 1.618…, has two properties that make it unique among all numbers.

- If you square Phi, you get a number exactly 1 greater than itself: 2.618…, or

Φ² = Φ + 1.

- If you divide Phi into 1 to get its reciprocal, you get a number exactly 1 less than itself: 0.618…, or

1 / Φ = Φ – 1.

These relationships are derived from the dividing a line at its golden section point, the point at which the ratio of the line (A) to the larger section (B) is the same as the ratio of the larger section (B) to the smaller section (C).

This relationship is expressed mathematically as:

A = B + C, and

A / B = B / C.

Solving for A, which on both sides give us this:

B + C = B²/C

Let’s say that C is 1 so we can determine the relative dimensions of the line segments. Now we simply have this:

B + 1 = B²

This can be rearranged as:

B² – B – 1 = 0

Note the various ways that this equation can be rearranged to express the relationship of the line segments, and also Phi’s unique properties:

B^{2} = B + 1

1 / B = B – 1

B^{2} – B^{1} – B^{0} = 0

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

Now we have a formula that can be solved 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… = Φ

( 1 – √5 ) / 2 = -0.6180339… = -Φ

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

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

5 ^ .5 * .5 + .5 = Φ

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} = Φ^{ n} / 5^{½}

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

f_{40} = Φ^{ 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} = [ Φ^{ n} – (-Φ)^{-n} ] / (2Φ-1)

Note: 2Φ-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:

Φ = 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 Φ(n-1), Φ(n) and Φ(n+1), the following relationship exists:

Φ(n-1) * Φ(n+1) = Φ(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

Daniel Hazelton Waters says

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…

professiona says

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

professiona says

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

Hyunjung says

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.

Vruta says

You mean like ‘i’?

It would be very interesting.

gaurav says

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

Gary Meisner says

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.

Andy Ginthum says

I believe, though I could be wrong, that you made a mistake in the sequence

3+8+21*2=55 therefore 3 + 8 + (21 x 2) = 11 + 42 = 53 not 55

this sequence that guarav has listed includes all the numbers not just the last three.

so you need to include the 2 of the first number to get 55 so the next number would be

2 + 3 + 8 +21 + 2(55) = 144

2 + 3 + 8 + 21 + 55 + 2(144) = 377

2 + 3 + 8 + 21 + 55 + 2(377) = 843

Andy Ginthum says

oops the last line should include 144

2 + 3 + 8 + 21 + 55 + 144 + 2(377) = 987

sorry got ahead of myself

matt says

Next in the series is 144.

fsipsv says

this is very helpful

juan corrales says

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?

Ashok Shah says

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

Gary Meisner says

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.

Kilogram says

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!!

zeb says

Thank you.

Peter Clothespinsstein says

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.

Uli Egg says

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!

Gayatri says

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

Gary Meisner says

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

jessica garica says

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..

Ted says

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.

Ted Krapkat says

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

William J Adamo says

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

jack says

I have become very familiar with this number and its uses sense I stumbled across it a year ago, in fact, it has literally changed my entire out look of the bible, and I wish to include the information found in this web site in a book I am writing about the bible and its patterns, and this numbers and its variations appear so often that its quite spooky.