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f – The Golden Ratio

A brief teaching paper by Adrian Larner

What is The Golden Ratio?

The Golden Ratio is a number, f: it is the ratio of two lengths, say AC and the shorter, co-linear AB, with this property:

AB and BC (i.e. AC - AB) are in the same ratio.

This ratio, f, has long been considered aesthetically pleasing. Its numerical value is easily calculated. Let AB be arbitrarily set to 1. Then AC is f, and so is AB/BC, i.e. 1 / (f-1) Thus we have merely to solve:

f = 1 / (f-1)

Multiplying through to remove the denominator:

f² - f = 1

Now completing the square:

f² - f + 1/4 = 1 + 1/4 = 5/4 = (f - 1/2)²

We obtain the solutions:

f = ½(1 ± Ö5)

Of the two solutions, which are approximately 1.618 and -0.618, the desired value, f, is the former (and the latter is clearly 1-f). It is not known why this ratio should be perceived as particularly pleasing. It has been used by a number of artists (for example Whistler, in his Nocturnes and other works).

The pages in this workspace (unless shrunk until the absolute sizes demanded by the gif files are unobtainable) are divided horizontally into three areas, say (from the left), of widths a, b, c, such that a is approximately in golden ratio to b, c to a + b, and (inevitably if more subtly) a + ½b + ½c to ½b + ½c, so that the midline of the combined b and c, area divides the page in golden ratio.

Fibonacci Numbers

The Fibonacci Numbers (u_k ), which – like f – occur frequently in nature, are defined as follows:

u₀ = 0

u₁ = 1

For k > 1, u_k = u_k-1 + u_k-2

Pascal's Triangle

You may be acquainted with Pascal's Triangle, which is an exhibition of the values of the binomial function, mFn, each value appearing at row n, position m. mFn is the number of combinations of n things taken m at a time (“m From n”):

            1

          1   1

        1   2   1

      1   3   3   1

    1   4   6   4   1

  1   5  10  10   5   1

1   6  15  20  15   6   1

Notice that each number in the triangle is the sum of the two numbers above it:

mFn = (m-1)F(n-1) + mF(n-1)

The formula for mFn is well known. It uses the factorial function, j! = 1×2× ... ×j. (0! = 1):

mFn = n!/m!(n-m!)

The recursive formula given above follows simply:

(m-1)F(n-1) + mF(n-1) = (n-1)!(1/(m-1)!(n-m)! + 1/m!(n-m-1)!)

= (n-1)!(m + (n-m))/m!(n-m)! = n(n-1)!/m!(n-m)! = n!/m!(m-n)!

If we now rotate Pascal's Triangle somewhat to the right, we get:

                  1

                1

              1   1

            1   2

          1   3   1

        1   4   3  

      1   5   6   1

    1   6  10   4  

  1   7  15  10   1

Summing across the (formerly diagonal) rows we get:

1, 1, 2, 3, 5, 8, 13, 21, 34, ...

... the Fibonacci Series.

So the n'th Fibonacci Number is ...

One question that arises for number series like the Fibonacci is: what is the formula for the n'th member? (u_k in terms of k)

The association with Pascal's Triangle gives one approach to a solution:

u_k = 0F(k-1) + 1F(k-2) + 2F(k-3) + ...

... terminating before m exceeds n (because mFn = 0 when m>n). Such a sum of a series is, however, not particularly convenient, and its reduction to a single expression is by no means obvious.

Let us return to f, the golden ratio. Because the ratio of successive Fibonacci Numbers approaches f, we can obviously obtain an approximation to u_k in the form C.f^k for some constant, C. (It turns out that C is approximately 0.4472)

But let us attack the problem less directly by considering first the powers of f under the simplifying reduction (the very definition of f, see above):

f² = 1 + f

This gives us:

f⁰ = 1

f¹ = f

f² = 1 + f

f³ = f(1 + f

) = f + 1 + f = 1 + 2f

f⁴ = f(1 + 2f

) = f + 2 + 2f = 2 + 3f

f⁵ = f(2 + 3f

) = 2f + 3 + 3f = 3 + 5f

... and so on. In general:

f^k = u_k-1 + u_kf

But can we get from here to a formula for u_k in terms of k? We now consider the negative powers of f, i.e. the powers of 1/f. Recall that 1/f = f - 1. Accordingly, we get:

1/f⁰ = 1

1/f¹ = -1 + f

1/f² = (f-1)(-1 + f) = -f + 1 + f + 1 -f = 2 - f

1/f³ = (f-1)(2 - f) = 2f - 2 -f - 1 + f = -3 + 2f

1/f⁴ = (f-1)(-3 + 2f) = -3f + 3 +2f + 2 -2f = 5 - 3f

1/f⁵ = (f-1)(5 - 3f) = 5f - 5 -3f - 3 + 3f = -8 + 5f

... and so on. Again we see the Fibonacci Numbers appearing, but with an inconvenient flip-flop of signs. This can be corrected by replacing 1/f by -1/f, which, it will be recalled is 1 - f, the other solution to the quadratic equation by which we calculated f. We get the general expression:

(1-f)^k = u_k+1 - u_kf

Subtracting from the previously derived f^k = u_k-1 + u_kf we get:

f^k - (1-f)^k = u_k-1 - u_k+1 + 2u_kf = 2u_kf - u_k = u_k(2f - 1)

Rearranging, and substituting f - (1 - f) for 2f - 1 in order to exhibit the symmetry of f and -(1 - f) in the final expression:

u_k = (f^k - (1-f)^k)/(f - (1-f))

How strange that this function uses the irrational parameter, f, and yet invariably returns an integer value. So much for the golden ratio.

Links

As you would imagine, the Web has plenty on the Golden Ratio and the Fibonacci Numbers. Try some of these (or search on “Golden Ratio” or “Fibonacci”):

www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib
 

www.geom.umn.edu/~demo5337/s97b/
 

A rather mathematical but very interesting site:
    http://www.friesian.com/golden.htm
 

http://www.mcn.net/~jimloy/golden.html

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