Question 4 of 7: Newton's Divided-Difference and Lagrange Interpolation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2019 — 04-BS-5 Advanced Mathematics, 3 hours, closed book (approved Casio/Sharp calculator and one double-sided aid sheet permitted). Any five of the seven questions constitute a complete paper and all questions are of equal value; all seven are answered below as a full study resource.
Reference texts: Kreyszig, Advanced Engineering Mathematics, 10th ed. (Wiley) — Ch. 11 (Sturm–Liouville Problems, Fourier Series, Fourier Integrals and Transforms), Ch. 19 (Numerics in General: interpolation, numerical differentiation, Romberg integration, iterative equation solving), Ch. 20 (Numeric Linear Algebra: LU factorization). Supporting: Chapra & Canale, Numerical Methods for Engineers, 7th ed. — Ch. 5–6 (bracketing and open root-finding methods), Ch. 18 (interpolation), Ch. 22 (Romberg integration).
Given. (A) Six data pairs $(x_i,F(x_i))$ tabulated below. (B) Four data pairs (stated on the following source page): $(-5,-162),\ (-2,0),\ (0,8),\ (4,0)$.
$x$
$-3$
$-2$
$0$
$3$
$5$
$6$
$F(x)$
$-28$
$0$
$8$
$-10$
$28$
$80$
Find. (A) Newton's divided-difference table and the interpolating polynomial of the highest degree the data actually supports. (B) The Lagrange polynomial through the four points.
Approach. (A) Build the divided-difference table level by level; if a level's entries all vanish, the true polynomial degree is lower than the naive $n-1=5$. (B) Apply the Lagrange formula $L(x)=\sum_i F(x_i)\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}$ directly.
(A) Build the divided-difference table. With $f[x_i]=F(x_i)$ and $f[x_i,\dots,x_{i+k}]=\dfrac{f[x_{i+1},\dots,x_{i+k}]-f[x_i,\dots,x_{i+k-1}]}{x_{i+k}-x_i}$:
$x$
$F(x)$
1st DD
2nd DD
3rd DD
4th DD
5th DD
$-3$
$-28$
$28$
$-8$
$1$
$0$
$0$
$-2$
$0$
$4$
$-2$
$1$
$0$
$0$
$8$
$-6$
$5$
$1$
$3$
$-10$
$19$
$11$
$5$
$28$
$52$
$6$
$80$
Read off the leading diagonal. $f[x_0]=-28,\ f[x_0,x_1]=28,\ f[x_0,x_1,x_2]=-8,\ f[x_0,\dots,x_3]=1,\ f[x_0,\dots,x_4]=0,\ f[x_0,\dots,x_5]=0$. The 4th and 5th divided differences vanish identically, so the data is exactly cubic — the "polynomial of highest possible degree" that these six points actually support has degree $3$, not $5$.
Assemble Newton's form.
$$P(x)=f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)+f[x_0,\dots,x_3](x-x_0)(x-x_1)(x-x_2)$$
$$=-28+28(x+3)-8(x+3)(x+2)+1\cdot(x+3)(x+2)(x)$$
Expanding,
$$\boxed{P(x)=x^3-3x^2-6x+8}$$
which reproduces all six tabulated values exactly (verified by direct substitution).
(B) Apply the Lagrange formula. With $(x_0,x_1,x_2,x_3)=(-5,-2,0,4)$ and $(y_0,y_1,y_2,y_3)=(-162,0,8,0)$,
$$L(x)=\sum_{i=0}^{3}y_i\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}$$
Only $y_0=-162$ and $y_2=8$ are non-zero, so
$$L(x)=-162\cdot\frac{(x+2)(x)(x-4)}{(-3)(-5)(-9)}+8\cdot\frac{(x+5)(x+2)(x-4)}{5\cdot2\cdot(-4)}$$
Expanding each term and summing:
$$\boxed{L(x)=x^3-3x^2-6x+8}$$
Check Parts (A) and (B) produce the identical cubic $x^3-3x^2-6x+8$, even though 4(B)'s four points are not a subset of 4(A)'s six — both data sets were generated from the same underlying cubic. This is confirmed: $L(-5)=-162$, $L(4)=0$, and $P(x)$ reproduces all six of 4(A)'s tabulated values exactly. It is a strong consistency check, not a coincidence to be second-guessed.
