If I have a series of numbers such as 1, 4, 26, 137, how do I work out an equation that will define the function that produces it?

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- December 5th 2010, 04:49 AMribbieHow to define a quadratic series algebraically?
If I have a series of numbers such as 1, 4, 26, 137, how do I work out an equation that will define the function that produces it?

- December 5th 2010, 05:35 AMskeeter
- December 5th 2010, 06:26 AMWilmerCode:
`1 4 26 137`

3 22 111

19 89

70

1 + 3(n-1) + 19(n-1)(n-2) / 2! + 70(n-1)(n-2)(n-3) / 3! : see the pattern?

Simplifies to:

(70n^3 - 363n^2 + 617n - 324) / 6

You can now get next term (407) and see what it "gives":

Code:`1 4 26 137 407`

3 22 111 270

19 89 159

70 70

0

- December 5th 2010, 07:06 AMribbie
Amazing! Thanks to both of you!

- December 5th 2010, 07:14 AMrtblue
I do agree that this function is more suited to a cubic, but if you wanted to derive a quadratic equation from a set of data, you could use Wilmer's method, or you could simply substitute points into the standard form of a quadratic, and solve for a, b, and c.