Least-Squares Parabola Proof and Lagrange Interpolation: 04-BS-5, December 2016, Q4

Question 4: Least-Squares Parabola Proof and Lagrange Interpolation (20 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) $n$ data pairs $(X_i,Y_i)$ and the origin-anchored parabola model $Y=\alpha X+\beta X^2$. (B) four points, tabulated below.

Given data (Q4B)
$x$	$-3$	$-2$	$1$	$2$
$F(x)$	$0$	$4$	$-8$	$0$

Find. (A) derive the printed Cramer's-rule formulas for $\alpha,\beta$. (B) the interpolating cubic $P(x)$.

Approach. (A) minimize the sum of squared residuals by calculus (normal equations), then solve the resulting $2\times2$ linear system by Cramer's rule. (B) build the four cubic Lagrange basis polynomials and sum them weighted by the table values.

(A) Set up the normal equations. Minimize $S=\sum_i(Y_i-\alpha X_i-\beta X_i^2)^2$ over $\alpha,\beta$: $$\frac{\partial S}{\partial\alpha}=-2\sum X_i(Y_i-\alpha X_i-\beta X_i^2)=0\ \Rightarrow\ \alpha\sum X_i^2+\beta\sum X_i^3=\sum X_iY_i$$ $$\frac{\partial S}{\partial\beta}=-2\sum X_i^2(Y_i-\alpha X_i-\beta X_i^2)=0\ \Rightarrow\ \alpha\sum X_i^3+\beta\sum X_i^4=\sum X_i^2Y_i$$
(A) Solve the $2\times2$ system by Cramer's rule. Writing the system as $\begin{pmatrix}\sum X_i^2&\sum X_i^3\\\sum X_i^3&\sum X_i^4\end{pmatrix}\binom{\alpha}{\beta}=\binom{\sum X_iY_i}{\sum X_i^2Y_i}$ and replacing each column by the right-hand side in turn, $$\boxed{\alpha=\dfrac{\bigl(\sum X_iY_i\bigr)\bigl(\sum X_i^4\bigr)-\bigl(\sum X_i^2Y_i\bigr)\bigl(\sum X_i^3\bigr)}{\bigl(\sum X_i^2\bigr)\bigl(\sum X_i^4\bigr)-\bigl(\sum X_i^3\bigr)^2}}\ ,\qquad \boxed{\beta=\dfrac{\bigl(\sum X_i^2\bigr)\bigl(\sum X_i^2Y_i\bigr)-\bigl(\sum X_i^3\bigr)\bigl(\sum X_iY_i\bigr)}{\bigl(\sum X_i^2\bigr)\bigl(\sum X_i^4\bigr)-\bigl(\sum X_i^3\bigr)^2}}$$ — matching exactly the formulas printed on the exam.
(B) Build the Lagrange basis and sum. $L_i(x)=\prod_{j\ne i}\dfrac{x-x_j}{x_i-x_j}$, $P(x)=\sum_iF_i\,L_i(x)$. Expanding symbolically and simplifying, $$\boxed{P(x)=x^3+2x^2-5x-6}$$
(B) Check against every node. $P(-3)=-27+18+15-6=0$; $P(-2)=-8+8+10-6=4$; $P(1)=1+2-5-6=-8$; $P(2)=8+8-10-6=0$ — all four match the table exactly, confirming the construction.

Part	Result
(A) $\alpha,\beta$	Cramer's-rule formulas above (proved; matches source)
(B) $P(x)$	$x^3+2x^2-5x-6$

Topic: least-squares curve fitting (normal equations + Cramer's rule) and Lagrange polynomial interpolation.

Key relations: normal equations from $\partial S/\partial(\text{coeff})=0$; Cramer's rule $x_k=\det(A_k)/\det(A)$; $L_i(x)=\prod(x-x_j)/(x_i-x_j)$.

Why this works: minimizing the convex quadratic $S(\alpha,\beta)$ makes its unique stationary point the global minimum (its Hessian, the Gram matrix above, is positive semi-definite); each Lagrange basis $L_i$ equals $1$ at its own node and $0$ at every other node by construction, so the weighted sum reproduces all four tabulated values exactly — and with 4 points there is exactly one degree-$\le3$ polynomial that does so (Vandermonde uniqueness).

Common pitfall: confusing this origin-anchored parabola $Y=\alpha X+\beta X^2$ with the more familiar intercept form $Y=\alpha+\beta X$ — the two summation matrices differ, so the formulas are not interchangeable; for (B), using the node spacing symbolically instead of the actual (unequal, $1,3,1$) gaps.

Source: Kreyszig, Advanced Engineering Mathematics, Ch. 19 (Curve Fitting; Interpolation).

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

Question 4: Least-Squares Parabola Proof and Lagrange Interpolation (20 marks)