Accuracy of Quadratic Regression

[Albert Chan]

If we use data relative to its mean, and X is uniformly spaced, by symmetry, Σ(x^(2k+1)) = 0
This saved a lot of calculations!

X = [2003,2004,2005,2006,2007,2008,2009,2010,2011,2012]
Y = [813,941,962,1053,1132,1194,1205,1244,1254,1262]

x = X - X̅ = X - 2007.5
y = Y - Y̅ = Y - 1106.0

a*Σx³ + b*Σx² + c*Σx = Σxy         // uniform X spacing, a, c terms goes away

b = Σxy / Σx² = 4080 / 82.5 = 544/11

a*Σx² + b*Σx + c*Σ1 = Σy = 0      // uniform X spacing, b terms goes away
a*Σx⁴ + b*Σx³ + c*Σx² = Σx²y

a = (Σy Σx² - Σ1 Σx²y) / (Σx² Σx² - Σ1 Σx⁴) = (0 - 10*-2888) / (82.5^2 - 10*1208.625) = -361/66

c = (Σy - b*Σx - a*Σx²) / Σ1 = -a*Σx² / Σ1 = 361/66 * 82.5 / 10 = 361/8

(Y-1106.0) = (-361/66)*(X-2007.5)^2 + (544/11)*(X-2007.5) + 361/8

--> Y ≈ -5.4696969697*X^2 + 22010.2878788*X - 22141315.3333