Maths investigations for children at home 2: Fibonacci numbers

The Fibbonaci sequence starts 0,1,1,2,3,5,8,13,21…; each term is the sum of the two preceding terms. Using Excel (or another tool if you prefer, but I think Excel will be good for this) work first 40 terms of the sequence (you can do this by entering a formula that says “this cell is the sum of the cell above it and the cell two above it”, and copy pasting it down a column.

Now, using another column on the same spreadsheet, work out the ratios of consecutive terms. 1/1 = 1; 2/1 = 2; 3/2 = 1.5; 5/3 = 1.6666…; 8/5 = 1.6. What does the ratio do as the terms get larger?

You should see it converging to a value that mathematicians call  (pronounced “phi” – it’s a Greek letter). . Work out the first 11 decimal places of φ (φ is what is called an irrational number, which means that annoyingly the decimal expansion never stops or recurs, but 11 places is good enough for government work).

Multiply φ by itself (we call this “squaring”, and the result is written φ2). Compare φ2 to φ – what do you notice?

Translate your observation into a quadratic equation that φ satisfies. (Don’t be put off by the technical jargon. This just means “find a whole number multiple of φ2 which is equat to the sum of a whole number multiple of φ and a whole number that sum to zero” – for example, 32 = 9 and 2*9 = 7*3 – 3 = 0, so x=3 satisfies the quadratic equation 2x2 = 7x- 3). There are lots of quadratic equations that φ satisfies, but there’s one simple, obvious one, where no number larger than one divides all the three whole numbers you’ve found, that you should be able to find easily, and that’s the one we’re after.

Now, let’s try this again. We started the Fibonacci series with 0 and 1, but there was nothing special about those numbers. The Lucas sequence starts 1,3,4,7, and continues using the same rule as the Fibonacci sequence. Compute the first 40 terms of it, and the ratios of consecutive terms. What do you notice?

We can do this for any two starting values. Pick some pairs of numbers, build the sequences from them by adding each pair of terms together to get the next, and see what the ratio of consecutive terms converges to.

At this point, you should see that the value of φ is coming from the rule we’ve chosen, and doesn’t really care where we start from. So let’s try some different rules.

For example, what happens to the ratios of consecutive terms if you build a sequence where each term is twice the term before it, minus the term before that? Or if it is the term before it, plus twice the term before that? For each pair of multipliers (the fancy word is “coefficients”), try a bunch of sets of starting values, and work out the constant that is equivalent to φ for that rule.

In all cases, the constant should satisfy a simple quadratic equation. Can you figure out what the relation between the coefficients defining a rule and the quadratic equation that the limit of the ratios of consecutive terms satisfies is?

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