Phi's Fascinating Figures

On this page we meet some more marvellous maths about the number Phi itself, its multiples and powers.

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Powers of Phi

We return to the definition of Phi, as the positive valued solution to

P(P-1)=1

P here is either of 2 values, Phi and 1-Phi.
So P has the property that P²=P+1.
Let's look at Phi² first.

Phi squared

We can read the equation above as to find P², just add 1 to P.

You do the maths...

Use your calculator to evaluate Phi=(1+5)/2. If you have a Memory on your calculator, store this value in it.
Square it and check that it is just Phi+1 (i.e. subtract one and compare with the Memory value).
Use your calculator to evaluate the other value (1-Phi) or (1-Memory) = (1-5)/2
Again square the value just found and check that it is just the same as adding 1 to the value you squared.

Now let's look at Phi³.

Phi cubed

Is there another way to calculate Phi³ apart from just Phi x Phi x Phi?

Yes - let's see how to compute it in two more ways. We use the basic P(P-1)=1 formula or, in another form, P²=P+1.
P³ is just P.P² = P(P+1) by our "basic formula", which expands to P²+P

...now that's interesting!... 1 + P = P² AND P + P² = P³ ...hmmm! Is there anything in this do you think?
Question: How could this be generalized? We'll use this result in the next sub-section about a Phi-bonacci Brick... but, for now, let's get back to the original equation...

and P² + P = (P+1) + P after using the "basic formula" again and this is just 2P + 1. So

Phi³ = 1 + 2 Phi

Notice that this needs just one multiplication rather than two if we evaluated Phi x Phi x Phi. That's the first quick way.

The second answer was spotted by Scott Beach but it is also in the table at the foot of this page: since Phi=(√5+1)/2 then 2Phi is √5+1 and 1+ 2 Phi = &radic 5 + 2:

Phi³ = 2 + √5

Interesting Facts about a Phi-bonacci Brick

[This variation was suggested by Hud Nordin of Sunnyvale, California.]
phiBrick

If we have a brick with sides of lengths 1, Phi=1·61803... and phi=1/Phi=0·61803... then:-

the longest side is the sum of the other two lengths since 1 + phi = Phi
the largest face (area C=1 x Phi) is the sum of the other two face's areas (area A =1 x phi and area B=phi x Phi) since
Area A + Area B
= 1 x phi + phi x Phi
= phi + 1
= Phi
= Area C

The next three interesting facts and figures on the Phibonacci brick were first pointed out by Donald Seitz in The Mathematics Teacher, 1986, pages 340-341 in an article entitled A Geometric Figure Relating the Golden Ratio to Pi.

What is the surface area S of the brick?
Above we saw that the sum of the 2 smaller face's areas equals the largest face's area, and that this is Phi.
Since there are 2 faces with smallest area, 2 of middle-sized area (which total 2 times the largest face area, that is 2 Phi) and we also have two other faces of the largest area (Phi) , then:
The surface area of the brick is 4 Phi
How long is the diagonal across the brick?
Another surprise awaits us when we calculate the length of the diagonal across the brick.
The formula is a 3-dimensional analogue of Pythagoras Theorem.
For a rectangle of sides x and y, its diagonal is (x² + y²).
For a 3-D brick with sides x,y and z, its diagonal has length (x² + y² + z²).
So how long is the diagonal of our Phi-bonacci brick? Since its sides (x,y and z) are 1, Phi and phi, the length of its diagonal is: (1² + Phi² + phi²). I'll leave you to check the algebra but the surprisingly simple answer is
The diagonal of the brick has length 2
A relationship between Phi and Pi
Even more surprising is that the brick shows a simple relationship between Pi and Phi, using the values for the diagonal D that we have just found and the surface area S.
If we imagine the brick tightly packed into a sphere, the centre of the sphere will be half way along diagonal D.
So the radius of the sphere will be 1 and its surface area will be 4 Pi.
We showed above that the surface area of the brick, S, is 4 Phi.
Putting these two together we have:
The ratio of the surface areas of
the Phibonacci brick and its surrounding sphere
is Phi : Pi

You do the maths...

Suppose we have a brick with sides a, b and c. Suppose also that the sum of the smaller sides equals the longest side (i.e. that a+b=c). Also suppose we have the "Area sum property" too, that the smaller two areas add up to the larger, or that ab + ac = bc. The questions here show that the only way to have BOTH properties is when a and b as well as b and c are in the golden ratio (Phi):- Show that, if we let k stand for a/b then we must also have k = b/c .
Show that k must also satisfy k² = 1 + k
... and hence show that k must be Phi.

Golden Triangles, Rectangles and Cuboids by M Bicknell and V E Hoggatt Jr in Fibonacci Quarterly vol 7 (1969), pages 73-91 is a very readable account of some generalizations of these results.

The Golden rectangle and powers of phi

On a golden rectangle with sides of length phi and 1, dividing at the golden ratio point gives two overlapping squares whose sides are phi.
The gap between a square and the longer side has length 1-phi which is also phi².
So the phi by phi squares are themselves divided at their own golden section ratios by the side of the other overlapping square. The size of the overlap is phi-phi² which is just phi³ (because the equation 1-phi=phi² can be multiplied by phi to give the relation phi-phi² = phi³).

As an exercise, how would you easily construct a line of length phi⁴ = phi²-phi³ on the diagram?

Higher powers of phi

Here is another geometrical illustration of phi², phi³ and phi⁴:

If we take a square with sides of length 1 and divide two sides at the golden section point, we form two new squares (yellow and blue) and two (red) rectangles. Notice that the diagonal is also divided at its golden section point too.
The large yellow square has sides of length phi, so its area is phi².
The sides of a red rectangle are of length phi and 1-phi=phi² so each has an area of phi³!
The small blue square has sides of length 1-phi=phi² so its area is phi⁴!!
So, the squares plus the rectangles area must add up to 1, the area of the big square and we have yellow + 2 blue + red = 1:

1 = phi² + 2 phi³ + phi⁴

You do the maths...

Evaluate Phi and raise it to the power 4 (hint: square it twice) on your calculator. Save this value in a memory (or write it down).
Compare the value you have just found with (3 Phi+2).
What is Phi⁵ in the form a Phi + b (ie find the integers a and b) ?
Write Phi⁶ in the same form.
Can you spot a general pattern for Phiⁿ? Hint: spot the well-known series for the values of a and b in the successive powers of Phi?
Here is a cube with sides divided at the golden section points to split it into eight pieces. The eight pieces are of 4 different shapes:

There are 8 blocks in the cube: how many of each shape are there?
What is the volume of each? ( Each volume will be a power of phi!)
If the total volume is 1 (since the original cube has sides of length 1), what relationship does this suggest between your multiples of powers of phi and 1?
Expand (1+phi)³. Since 1+phi=phi^-1 your answer will also equal (phi^-1)³=phi^-3.
How is this answer related to the relationship you found in the last question?

A relation on Powers of Phi

There's an even more intriguing relationship between the powers of Phi than the one you discovered for yourself in the last section. Here's how you can find it:

You do the maths...

What do you notice about the difference between Phi³ and Phi²?
and how does Phi⁴-Phi³ compare with another power of Phi?
Can you guess what we must add to Phi⁴ to get Phi⁵.
What is the general pattern:
Phiⁿ = Phi^n-1 + ?
Can you justify this using your answer to the last section? that is, using your pattern for
Phiⁿ = a Phi + b

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Multiples of Phi

We've just seen that we can use multiples of Phi to calculate its powers easily and you might think that there's not much else we can discover about multiples of Phi.

Here we show another relationship and also explain how the Rabbits family Tree diagram was made, originally seen in the Fibonacci and Nature. It's all to do with the multiples of Phi, or, rather, the fractional parts of the multiples.

You do the maths...

On a piece of paper, draw a line 10cm long near the top, or use graph paper. This is going to be used to plot the fractional parts of the multiples of Phi=1·618.. so it will represent values from 0 to 1 (0·5 being at 5cm, 1 at 10cm etc).
Draw the first point at the fractional part of Phi itself=0.618, ie at 6·18 cm along the line. Mark it 1, since we used 1 times Phi.
Draw another line underneath the first. On it put a point for the fractional part of 2Phi=3.236 ie at 2·36cm. Mark the point as 2.
Draw another line underneath the last. On it put a point for the fractional part of 3Phi=4.854, ie at 8·54cm. Mark it as 3.
Draw a new line and this time mark TWO points labelled 4 and 5,
so that this line will have the next 2 multiples of Phi: 4Phi and 5Phi.
The next line will have the next THREE multiples of Phi: 6Phi, 7Phi and 8Phi.
The next line will have the next FIVE multiples of Phi.
The next line will have the next EIGHT multiples of Phi.
..you can guess the next one!
When you're tired of drawing lines and plotting points, add some thick vertical lines down from each point.
Where each vertical line begins, draw a thick horizontal line to the nearest vertical line.

We can interpret the thick lines of the Rabbit Family Tree as follows:

The 10 cm lines across the page each represent one new generation of rabbits, or each new month in Fibonacci's original problem.
Each thick vertical line represents one of our rabbit pairs and..
.. each pair is "born" when it appears for the first time as a branch from an thick vertical line that started higher up the page (its "parents").
Notice how each vertical line goes through two levels (generations or months) before it starts to branch and ..
.. then it branches on every line after that. This shows where our rabbits must be 2 months old to start producing their own young and then do so every month after that.
Notice how on each generation level, the new rabbits appear equally and optimally spaced across the page so that there is no overcrowding at one point in the tree...
.. and also that each horizontal branching line is the same length on each generation and ...
.. all are on the same side of their "parents" line in each generation and...
..the side they appear on swops each time, from left side to right side to left.

You can probably spot some more relationships in this Tree (for instance, that the lengths of the horizontal branches are each 0·618 times shorter than those of the previous generation).

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Continued Fraction representations

A continued fraction is a fraction of the form:

    a +            1            = a + 1/( b + 1/( c + 1/( d + ... )))
         b +        1     
              c +    1    
                   d + ...

where a, b, c etc are integers (often only positive).
Phi is the simplest form of continued fraction since all the integers a, b, c, ... are 1.
Writing such fractions in the usual form, as above, takes a lot of space on the page, so an abbreviation often used is to just give the integers as a list: [a,b,c,...].

    a +            1            = [ a, b, c, d, ... ]
         b +        1     
              c +    1    
                   d + ...

In this notation then Phi=[1,1,1,1...] which we can further abbreviate to [1] where the numbers shown like this indicate that they are repeated in the same cycle for ever. In some books, these periodic parts may have dots over them in the same way that we write infinite decimal fractions such as

	.		.		.		.
1/3 =0·3333.. =0·	3	; 1/7 = 0·	1	4285	7	= 0·142857142857142...; 1/12 = 0·0833333... = 0·08	3

and other authors use an overline like this so that 1/7=0·142867.
If we have more than one integer in the repeating part, such as [6,1,5,1,5,1,5,...] we write [6,1,5].

For more see Introduction to Continued Fractions at this site.

1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ..More..

Formulae for Phi

Phi denotes (

5+1)/2 =1·6180339... and
phi denotes (

5-1)/2 =0·6180339... .

The basic formulae are

Phi = 1/phi = phi + 1
Phi² = Phi + 1 and therefore
phi² = 1 − phi.

These can be generalized and give an important and remarkable pair of results for all integers n (positive, zero and negative):

Phi^n-2 + Phi^n-1 =	Phiⁿ
	phiⁿ	= phiⁿ⁺¹ + phiⁿ⁺²

Other relationships between the powers of Phi and phi and the Fibonacci numbers that also hold for all integer values of n are:

Main	= Other	= X + Y Phi	= A + B phi	= C + D √5
`Phiⁿ`	`= phi⁻ⁿ`	`= F(n−1) + F(n) Phi`	`= F(n+1) + F(n) phi`	`L(n) + F(n)√5 2`
`(−phi)ⁿ`	`= (−Phi)⁻ⁿ`	`= F(n+1) − F(n) Phi`	`= F(n-1) − F(n) phi`	`L(n) − F(n)√5 2`

where L(n) is the n^th Lucas number, defined as L(n) = F(n+1) + F(n-1).

With thanks to John McNamara. There is a comprehensive list of these and other kinds of formlae on my Fibonacci, Phi and Lucas numbers Formulae page at this site.

Powers and Multiples of Phi Calculator

C A L C U L A T O R

Numerical Relationships between Phi and its powers

Summarising the above, we can make a table where we relate the powers of Phi to its multiples - and we note the Fibonacci numbers appearing again! We can also expand the multiples of Phi using its definition as (

5+1)/2 and find Fibonacci again. Finally, we give the continued fraction representations of these powers with the repeating period part (if any) shown like this. In both the

5 forms and the continued fractions forms we note that a new series of numbers appears - - the Lucas numbers 2, 1, 3, 4, 7, 11, 18, 29, ... .

Phi power	phi power	A + B Phi	C + D phi	Ksqrt(5)+L	decimal value	continued fraction
...
Phi⁹	phi⁻⁹	21 + 34 Phi	55 + 34 phi	38 + 17√5	76·0131556174..	[76]
Phi⁸	phi⁻⁸	13 + 21 Phi	34 + 21 phi	(47 + 21√5)/2	46·9787137637..	[46, 1, 45]
Phi⁷	phi⁻⁷	8 + 13 Phi	21 + 13 phi	(29 + 13√5)/2	29·0344418537..	[29]
Phi⁶	phi⁻⁶	5 + 8 Phi	13 + 8 phi	9 + 4√5	17·9442719099..	[17, 1, 16]
Phi⁵	phi⁻⁵	3 + 5 Phi	8 + 5 phi	(11 + 5√5)/2	11·0901699437..	[11]
Phi⁴	phi⁻⁴	2 + 3 Phi	5 + 3 phi	(7 + 3√5)/2	6·8541019662..	[6, 1, 5]
Phi³	phi⁻³	1 + 2 Phi	3 + 2 phi	2 + √5	4·2360679774..	[4]
Phi²	phi⁻²	1 + 1 Phi	2 + 1 phi	(3 + √5)/2	2·6180339887..	[2, 1]
Phi¹	phi⁻¹	0 + 1 Phi	1 + 1 phi	(1 + √5)/2	1·6180339887..	[1]
Phi⁰	phi⁰	1	1	1	1·0000000000..	[1]
Phi⁻¹	phi¹	−1 + 1 Phi	0 + 1 phi	(−1 + √5)/2	0·6180339887..	[0, 1]
Phi⁻²	phi²	2 − 1 Phi	1 − 1 phi	(3 − √5)/2	0·3819660112..	[0, 2, 1]
Phi⁻³	phi³	−3 + 2 Phi	−1 + 2 phi	−2 + √5	0·2360679774..	[0, 4]
Phi⁻⁴	phi⁴	5 − 3 Phi	2 − 3 phi	(7 − 3√5)/2	0·1458980337..	[0, 6, 1, 5]
Phi⁻⁵	phi⁵	−8 + 5 Phi	−3 + 5 phi	(−11 + 5√5)/2	0·0901699437..	[0, 11]
Phi⁻⁶	phi⁶	13 − 8 Phi	5 − 8 phi	9 − 4√5	0·0557280900..	[0, 17, 1, 16]
Phi⁻⁷	phi⁷	−21 + 13 Phi	−8 + 13 phi	(−29 + 13√5)/2	0·0344418537..	[0, 29]
Phi⁻⁸	phi⁸	34 − 21 Phi	13 − 21 phi	(47 − 21√5)/2	0·0212862362..	[0, 46, 1, 45]
Phi⁻⁹	phi⁹	-55 + 34 Phi	-21 + 34 phi	-38 + 17√5	0·0131556174..	[0, 76]
...

phi = 1/Phi so the powers of phi are the negative powers of Phi.
Note that [0,a,b,c,...] is 0 + 1/(a + 1/(b + 1/(c + ...) = 1/[a,b,c, ...].
The negative powers of Phi are the positive powers of phi and are obtained by putting a 0 at the head of the continued fraction lists above: eg phi = 1/Phi = [0,1]

You do the maths...

Phi³ looks as if it is just phi³+4. Perhaps this is just correct to a few decimal places but is it really true and accurate? What other such "coincidences" can you spot in the Table above? Almost all of them will be accurate and true relationships, in fact.
Try to prove some of your "coincidences" using the formulae above the table.
What about those integers in the continued fraction forms of the powers? It looks as if the series is
1, 3, 4, 7, 11, ..
If the next two are 18 and 29, can you spot how they are generated?
These are not the Fibonacci numbers but the pattern of generating them is similar. The numbers are, in fact, those of a series very closely related to the Fibonacci numbers called the Lucas numbers.
From the table above, what do you notice about the third and fourth columns for the following pairs of rows?
Phi and Phi^-1,
Phi² and Phi^-2,
Phi³ and Phi^-3,
and so on?

What if we added some of these pairs to get rid of the Phi's and subtracted the others? What series of integers results?

The Mathematical Magic of the Fibonacci Numbers
Fibonacci Home Page
Phi in 3 dimensions

Where to now?
(Optional) Introduction to Continued Fractions
Phigits and Base Phi Representations

The next Topic is...
The Golden String

updated 25 September 2016