Math Is Fun Forum

Bob

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Registered: 2010-06-20

Posts: 10,626

Pascal, Fibonacci and the Golden Ratio

Following a Help Me post, I got inspired to post about these topics in maths.  For me it shows a beautiful connection between geometry, number and algebra.

(1) Start with Pascal's triangle:

1
1   1
1   2   1
1   3   3   1
1   4   6   4   1
etc.

Each row is determined by adding pairs of numbers in the row above.  It's useful for the binomial theorem.

(2) If you draw diagonal lines starting at each right hand 1 and also at the number directly below each 1 ,and tracking down and left, and add these numbers up, you get:

1, 1, 1+1, 2 + 1, 1 + 3 + 1, 3 + 4 + 1, etc which is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, where each number is made by adding the previous two numbers.

That sequence crops up in some unlikely places, such as the number of snowdrop bulbs formed in a clump, year by year.  More on that later.

(3)  Now form a new sequence by dividing each term in Fibonacci, by the one before it:

1/1 = 1   2/1 = 2   3/2 = 1.5   5/3 = 1.666   8/5 = 1.6   13/8 = 1.625   21/13 = 1.61538

These numbers appear to be converging on a number, with one term above and then the next below that number.  Using a MS Excel spreadsheet I got 1.618033998521 after a number of iterations.

This is a number that mathematicians call the Golden Ratio.  It is sometimes given the symbol Greek letter phi.  If you make a rectangle with length to width ratio phi, and cut off a square, the rectangle that is left has length to width ratio phi once more.

(4) If you reverse the division, 1/1, 1/2, 2/3, 3/5, 5/8 and so on, the results also converge on a number which appears to be phi - 1= 0.61803399852

This property is actually true for any Fibonacci sequence, ie the starting values need not be 1 and 1.  eg.  4, 7, 11, 18, 29, ……

What follows is a proof of this.

(5)  Let's say that three terms in a Fibonacci sequence are a, b, a+b where all are positive integers.

Then the next few numbers are a + 2b, 2a + 3b, 3a + 5b and so on.

(6) The ratio sequence is

If I subtract term 2 from term 1

and subtract term 2 from term 3

There are two things to notice about this algebra; firstly the second subtraction is done in the opposite order to the first; and secondly

as (a+b) is greater than a

So the terms are getting closer together and are oscillating either side of the limit.

(7) So, how do we find that limit?  If we carry on the ratio sequence, the values will get closer and closer to the limit.  So equating a pair of terms will reveal the limit:

Using the quadratic formula and discarding the negative values gives

(7) repeating this process with the ratios reversed leads to

and note that

And another interesting result:

I'll stop there for now.  Maybe more later.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob