# FIND THE NEXT NUMBER IN THE SEQUENCE

Given a sequence of n real numbers f(x1), f(x2), f(x3), ... , f(xn), there is always a mathematical procedure to find the next number f(x n+1) of the sequence. The resulting solution may not appear to be satisfying to students, but it is mathematically logical.

The solution involves finding the polynomial f(x) that passes through the points (1, f(1)), (2, f(2)), (3, f(3)), ... , (n, f(xn)). Then f(x n+1) is the next number of the sequence.

There are several ways to calculate this polynomial. One way is to solve n linear equations in n unknowns. Another way is to use Newton's Forward Difference Formula:

f(x) = f(x1) + Δf(x1)(x - x1)/1! + Δ2 f(x1)(x - x1)(x - x2)/2! + Δ3 f(x1)(x - x1)(x - x2)(x - x3)/3! + ... .

Here the forward difference operator Δ is defined: Δf(xn) = (xn+1- xn), Δk f(xn) = Δk-1(Δf(xn)).

For example, to find the next number of the sequence 1, 2, 3, 6, 5 we set up a table of successive forward differences:

 xi f(xi) Δf(xi) Δ2f(xi) Δ3f(xi) Δ4f(xi) 1 1 1 2 2 0 1 2 3 3 2 -8 3 -6 4 6 -4 -1 5 5

Substitute the values from the table into Newton's Forward Difference Formula:
f(x) = f(x1) + Δf(x1)(x - x1)/1! + Δ2 f(x1)(x - x1)(x - x2)/2! + Δ3 f(x1)(x - x1)(x - x2)(x - x3)/3! + Δ4 f(x1)(x - x1)(x - x2)(x - x3) (x - x4)/4!.

f(x) = 1 + 1(x - 1)/1! + 0(x - 1)(x - 2)/2! + 2(x - 1)(x - 2)(x - 3)/3! -8(x - 1)(x - 2)(x - 3)(x - 4)/4!

The polynomial above reproduces f(x) = f(xi), for i = 1, 2, ... , 5.

Substituting the next term, x6 = 6, gives f(6) = -14. This is the next number of the sequence.

However, f(6) could have been any number. If f(6) had been -99, we could have found a fifth degree polynomial fitting all six points.

Here are a few references:
http://mathworld.wolfram.com/FiniteDifference.html
http://en.wikipedia.org/wiki/Difference_operator
Burden, Richard L. and J. Douglas Faires. (1985). Numerical Analysis, Third Edition, Prindle, Weber & Schmidt. ISBN 0-87150-857-5.

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