# How to define a quadratic series algebraically?

• Dec 5th 2010, 04:49 AM
ribbie
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?
• Dec 5th 2010, 05:35 AM
skeeter
Quote:

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 ...

$y = \frac{35}{3} x^3 - \frac{121}{2} x^2 + \frac{617}{6} x - 53$
• Dec 5th 2010, 06:26 AM
Wilmer
Quote:

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?
• Dec 5th 2010, 07:06 AM
ribbie
Amazing! Thanks to both of you!
• Dec 5th 2010, 07:14 AM
rtblue
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.