Thread: How 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?

Originally Posted by ribbie

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?

this sequence of numbers you listed is not generated by a "quadratic".

my calculator gets a cubic equation to fit rather closely ...

Originally Posted by ribbie

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?

Code:

 1    4     26      137
   3    22     111
     19    89
        70

Watch the left-hand numbers 1,3,19,70:

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

OK?

Amazing! Thanks to both of you!

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.