Activity 2-17: Zeroes of a Recurrence Relation

We all know the Fibonacci sequence, 1, 1, 2, 3, 5, 8, We can write this as u n+1 = u n + u n-1, with u 1 = 1, u 2 = 1. So the next term is the sum of the previous two, and we have some initial conditions. We have here an example of a Linear Recurrence Relation (or LRS).

u n+1 = a n u n + a n-1 u n-1 + a n-2 u n a n-k+1 u n-k+1 is the general LRS of order k. So the next term is a linear combination of the last k terms (we also need initial conditions). Now there is a way to calculate future terms for an LRS using a matrix. Let’s demonstrate this using the Fibonacci sequence. We will only be looking at the cases where every a i is an integer: we will call these ‘integer LRSs’.

So it is clear that If we have some computer help, in taking the powers of a matrix, we have a good way of calculating future terms of an LRS.

Task: use the spreadsheet below to find the 21st term, that’s F 21, in the Fibonacci sequence. Powers of a Matrix spreadsheet The 21st term in the Fibonacci sequence is carom/carom-files/carom-2-18.xls

Task: given the LRS u n+1 = u n + u n-1 + u n-2 with u 1 = 1, u 2 = 1, u 3 = 1, find the 22nd term. Using a 3 x 3 matrix this time: The 22nd term in this order-3 integer LRS sequence is

Now we can note that it makes perfect sense to run the Fibonacci sequence backwards....5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, So u 0 = 0, u -1 = 1, u -2 = -1, and so on: the rule still holds. Now a further question: how many 0s are there in the Fibonacci sequence? It seems clear that there can only be one. In that case, how many 0s can any order-2 integer LRS have? Can it have an infinite number? Or is there a maximum?

It is easy to construct an LRS that does have an infinite number of 0s. Consider u n+2 = 2u n, with u 1 = 0, u 2 = 2. This gives us the sequence 0, 2, 0, 4, 0, 8, 0, This clearly has an infinite number of 0s, BUT we call this type of LRS degenerate. The LRS u n+1 = a n u n + a n-1 u n-1 has associated with it the characteristic equation 2 = a n + a n-1. This has two roots, 1 and 2, and if their ratio is a root of unity, we say the LRS is degenerate.

So for our example u n+2 = 2u n, the characteristic equation is 2 = 2, and so 1, 2 are ±√2, and their ratio is -1. So for our Fibonacci sequence, the characteristic equation is 2   1 = 0, In general, the solution to an LRS is u n = A 1 n + B 2 n. We can find A and B from the initial conditions. which gives solutions 1, 2 = Using u 1 = 1, u 2 = 1, we find A =, B =, and so F n = Task: test this out for various n.

So let’s go back to our question and rephrase it: how many 0s can a non-degenerate order-2 integer LRS have? Theorem: Skolem-Mahler-Lech (proved in 1953). Any non-degenerate integer LRS has a finite number of 0s. It can also be proved that the largest number of 0s an order-2 non-degenerate integer LRS can have is 1, (so the Fibonacci sequence has the maximum). How can we prove this?

The order-2 LRS u n+1 = a n u n + a n-1 u n-1 has associated with it the characteristic equation 2 = a n + a n-1. Suppose the roots are 1, 2. So we have u n = A 1 n + B 2 n. Suppose u n = 0 for n = p and n = q, with p > q. 0 = A 1 p + B 2 p  Similarly, Subtracting, we have, and thus the ratio of the roots is a root of unity, and the LRS is degenerate. So the largest number of 0s an order-2 non-degenerate integer LRS can have is 1.

What is the largest number of 0s an order-3 non-degenerate integer LRS can have? Beukers has proved (1991) that the answer is 6. Can we find an order-3 integer LRS with six 0s ? The mathematician Berstel managed to do exactly that.

Task: using our matrix spreadsheet, see if you can find all six zeroes there are to be found. So we have zeroes for a 0, a 1, a 4, a 6, a 13, and rather surprisingly, for a 52 (!)

With thanks to: Graham Everest and Tom Ward. Carom is written by Jonny Griffiths,