Finite Difference Method for Determining Coefficients of a Quadratic Equation

Another post gives a geometric solution to solving the quadratic equation. But how does one solve for the coefficients of the quadratic equation if you are given the first few terms of the sequence? 

Suppose you have a sequence of numbers you believe is generated by a quadratic equation and you wish to determine the equation's coefficients. The finite difference method can be used. Consider the series beginning 3, 12, 25. You wish to find the coefficients a, b, and c so that: 

will result in the answers of 3, 12, and 25 when one substitutes 0, 1, and 2 for x in the equation.

The general form for the method is let the first terms of the series be designated t₁, t₂, and t₃. Then the coefficients can be found by the following equations:

a = ½ (t₃ - 2t₂ + t₁)

b = (t₂ - t₁) - a

c = t₁

In our example above, this would give us:

a = ½ (25 - 24 + 3) = 2

b = (12 - 9) - 2 = 7

c = 3

Therefore, the quadratic equation to generate the sequence 3, 12, 25 is:

In an early post, I introduced the Lazy Caterer Sequence, which starts: 1, 2, 4. Solving for a, b, c yields:

a = ½ (t₃ - 2t₂ + t₁) = ½ (4 - 4 + 1) = ½

b = (t₂ - t₁) - a = (2 - 1) - ½ = ½

c = t₁ = 1