Question

How do you find the pattern for a list?

Answer 1

A few thoughts...

Explanation:

I think you are referring to some kind of sequence, but it is not clear what sort of problems you want to address.

Let's look at a few sequence types and their associated rules.

Arithmetic sequence

An arithmetic sequence has a common difference between terms. So to get from one term to the next you need to add the common difference.

We can write a recursive rule:

`a_(n+1) = a_n + d`

where `d` is the common difference.

We also need to specify the starting point, the first term `a`:

`a_1 = a`

We can write the explicit rule for a general term as:

`a_n = a + d(n-1)`

Geometric sequence

A geometric sequence has a common ratio between terms. So to get from one term to the next you need to multiply by the common ratio.

We can write a recursive rule:

`a_(n+1) = a_n * r`

where `r` is the common ratio.

We also need to specify the starting point:

`a_1 = a`

We can write the explicit rule for a general term as:

`a_n = ar^(n-1)`

Linear recursion

A sequence may also be defined using a linear recursive rule, where each successive term is based on the two (or more) previous terms.

The classic example of such a sequence is the Fibonacci sequence, definable recursively by:

`F_0 = 0`

`F_1 = 1`

`F_(n+2) = F_(n+1)+F_n`

It starts:

`0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,...`

How do you get an explicit rule for a linear recursion like this?

Consider a geometric sequence:

`1, x, x^2, x^3,...`

If it satisfies the linear recursive rule for `1, x, x^2`, then it will continue to satisfy it.

So given a rule:

`a_(n+2) = p a_(n+1) + q a_n`

we can associate the quadratic equation:

`x^2 = px+q`

Calling the two roots of this equation `alpha` and `beta` note that any sequence given by:

`a_n = A alpha^n + B beta ^n`

will satisfy the recursive rule.

We then just have to choose `A` and `B` so that the first two terms match the initial two terms of the sequence.

In the case of the Fibonacci sequence, we find:

`F_n = 1/sqrt(5)(varphi^n - varphi^(-n))`

where `varphi = 1/2+1/2sqrt(5)`