Given a sequence of n real numbers f(x

The solution involves finding the polynomial f(x) that passes through the points (1, f(1)), (2, f(2)), (3, f(3)), ... , (n, f(x

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(x

Here the forward difference operator Δ is defined: Δf(x

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

x_{i} |
f(x_{i}) |
Δf(x_{i}) |
Δ^{2}f(x_{i}) |
Δ^{3}f(x_{i}) |
Δ^{4}f(x_{i}) |

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(x

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(x

Substituting the next term, x

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).

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