1. Activity 2-17: Zeroes of a Recurrence Relation www.carom-maths.co.uk

2. We all know the Fibonacci sequence, 1, 1, 2, 3, 5, 8, 13... 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).

3. u n+1 = a n u n + a n-1 u n-1 + a n-2 u n-2 +......+ 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’.

4. 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.

5. 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 10 946. http://www.s253053503.websitehome.co.uk/ carom/carom-files/carom-2-18.xls

6. 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 157 305.

7. 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, 13... 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?

8. 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, 16... 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.

9. 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.

10. 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?

11. 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.

12. 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.

13. 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 (!)

14. With thanks to: Graham Everest and Tom Ward. Carom is written by Jonny Griffiths, hello@jonny-griffiths.nethello@jonny-griffiths.net