Least-Squares Parabola and Lagrange Interpolation: 04-BS-5, May 2013, Q4

Question 4: Least-Squares Parabola and Lagrange Interpolation (A) 10, (B) 10 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (A) A least-squares model $y=\alpha x+\beta x^2$ (no constant term) and $n$ data points $(x_i,y_i)$. (B) Four tabulated values of a quadratic function $f$ (table below).

Given data (Part B)
$x$	-2	1	2	4
$F(x)$	11	-4	3	35

Find. (A) Derive $\alpha,\beta$ that minimize the sum of squared residuals. (B) The 2nd-degree Lagrange polynomial through the first three points, then check it against the fourth.

Approach. (A) Differentiate the sum of squares w.r.t. $\alpha,\beta$, set both to zero (normal equations), solve the resulting $2\times2$ linear system by Cramer's rule. (B) Build the Lagrange basis polynomials for the first three $x$-values and combine.

(A) Form the sum of squared residuals and differentiate. $$S=\sum_{i=1}^n(y_i-\alpha x_i-\beta x_i^2)^2$$ $$\frac{\partial S}{\partial\alpha}=-2\sum x_i(y_i-\alpha x_i-\beta x_i^2)=0,\qquad\frac{\partial S}{\partial\beta}=-2\sum x_i^2(y_i-\alpha x_i-\beta x_i^2)=0$$
(A) Rearrange into the normal equations. $$\alpha\sum x_i^2+\beta\sum x_i^3=\sum x_iy_i,\qquad\alpha\sum x_i^3+\beta\sum x_i^4=\sum x_i^2y_i$$
(A) Solve the $2\times2$ system by Cramer's rule. With $D=\big(\sum x_i^2\big)\big(\sum x_i^4\big)-\big(\sum x_i^3\big)^2$: $$\boxed{\alpha=\dfrac{\big(\sum x_i^4\big)\big(\sum x_iy_i\big)-\big(\sum x_i^3\big)\big(\sum x_i^2y_i\big)}{D}},\qquad\boxed{\beta=\dfrac{\big(\sum x_i^2\big)\big(\sum x_i^2y_i\big)-\big(\sum x_i^3\big)\big(\sum x_iy_i\big)}{D}}$$ matching the printed formulas exactly (all sums run $i=1$ to $n$). $\blacksquare$
(B) Write the Lagrange quadratic through $(-2,11),(1,-4),(2,3)$. $$L(x)=11\cdot\frac{(x-1)(x-2)}{(-2-1)(-2-2)}+(-4)\cdot\frac{(x+2)(x-2)}{(1+2)(1-2)}+3\cdot\frac{(x+2)(x-1)}{(2+2)(2-1)}$$
(B) Expand and simplify. $$L(x)=\frac{11(x-1)(x-2)}{12}+\frac{4(x^2-4)}{3}+\frac{3(x+2)(x-1)}{4}=\boxed{3x^2-2x-5}$$ (Check: $L(-2)=11,\ L(1)=-4,\ L(2)=3$ ✓ — reproduces all three data points exactly.)
(B) Check against the fourth point. $$L(4)=3(16)-2(4)-5=48-8-5=\boxed{35}$$ which equals $F(4)=35$ exactly — since all four tabulated values happen to lie on a single quadratic, the check point confirms $L(x)$ recovers the underlying function exactly, not merely a nearby approximation.

Quantity	Result
Normal equations	$\alpha\sum x^2+\beta\sum x^3=\sum xy$; $\alpha\sum x^3+\beta\sum x^4=\sum x^2y$
$L(x)$	$3x^2-2x-5$
Check: $L(4)$	$35=F(4)$ ✓

Concept & Theory

Topic. Linear least-squares curve fitting through the origin, and Lagrange polynomial interpolation.

Key relations. Normal equations follow from $\nabla_{(\alpha,\beta)}S=0$; the Lagrange basis is $\ell_i(x)=\prod_{j\ne i}\dfrac{x-x_j}{x_i-x_j}$, and $L(x)=\sum_i y_i\ell_i(x)$ is the unique polynomial of degree $\le n-1$ through $n$ points.

Why this works. Because $y=\alpha x+\beta x^2$ is linear in the unknowns $\alpha,\beta$, minimizing $S$ reduces to a linear system solvable in closed form; a 3-point Lagrange interpolant reproduces any TRUE quadratic exactly, so a 4th check point that also lies on the curve confirms both the fit and the data's quadratic nature.

Common pitfall. Forgetting the model has no constant term ($y=\alpha x+\beta x^2$, not $\alpha+\beta x+\gamma x^2$), which shifts every power of $x$ in the normal-equation sums by one; in Lagrange interpolation, mismatching a numerator factor $(x-x_j)$ with the point it should skip ($j=i$).

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 19 (§19.3–19.4 Interpolation, Least Squares).

Question 4 of 7: Least-Squares Parabola and Lagrange Interpolation

Question 4: Least-Squares Parabola and Lagrange Interpolation (A) 10, (B) 10 marks