Question 4 of 7: Polynomial Interpolation — Newton Divided Differences and Lagrange
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Examination — 04-BS-5 Advanced Mathematics, December 2015. Closed-book, 3-hour exam;
candidates were permitted one 8.5"x11" aid sheet (both sides) and an approved calculator. The exam
instructs that any five questions constitute a complete paper (first five answers in the answer book are marked); every question is solved below as a full study resource.
Find. (A) The Newton divided-difference table and the resulting interpolating polynomial for
six data points; (B) the Lagrange interpolating polynomial through the four points above.
Approach. Build the divided-difference triangle level by level for (A); use the Lagrange
cardinal-basis formula directly for (B); both are cross-checked to reconstruct every tabulated value.
(A) Divided-difference table. With $x_0,\dots,x_5=-3,-2,0,3,5,6$ and
$F(x_0),\dots,F(x_5)=-28,0,8,-10,28,80$:
$$f[x_i,x_{i+1}]=\frac{F(x_{i+1})-F(x_i)}{x_{i+1}-x_i}:\quad 28,\ 4,\ -6,\ 19,\ 52$$
$$f[x_i,x_{i+1},x_{i+2}]:\quad -8,\ -2,\ 5,\ 11 \qquad f[\dots,x_{i+3}]:\quad 1,\ 1,\ 1$$
$$f[\dots,x_{i+4}]:\quad 0,\ 0 \qquad f[x_0,\dots,x_5]:\quad 0$$
The 4th and 5th divided differences vanish identically, so the true interpolating polynomial has degree
3, not 5 — "highest possible degree" here means the degree the data actually supports once the
higher differences collapse.
(A) Assemble Newton's form. Using the top diagonal $f[x_0]=-28$, $f[x_0,x_1]=28$,
$f[x_0,x_1,x_2]=-8$, $f[x_0,x_1,x_2,x_3]=1$:
$$P(x) = -28 + 28(x+3) - 8(x+3)(x+2) + 1\cdot(x+3)(x+2)(x-0).$$
Expanding,
$$\boxed{P(x) = x^{3} - 3x^{2} - 6x + 8}$$
which reproduces all six tabulated values exactly.
(B) Lagrange cardinal form. With two of the four ordinates zero ($x=-2,4$), only the $x=-5$
and $x=0$ terms contribute:
$$L_0(x)=\frac{(x+2)(x-0)(x-4)}{(-3)(-5)(-9)} = \frac{x(x+2)(x-4)}{-135}, \qquad
L_2(x)=\frac{(x+5)(x+2)(x-4)}{(5)(2)(-4)} = \frac{(x+5)(x+2)(x-4)}{-40}.$$
$$P(x) = -162\,L_0(x) + 8\,L_2(x) = \frac{6}{5}\,x(x+2)(x-4) - \frac{1}{5}(x+5)(x+2)(x-4).$$
(B) Factor and simplify. Both terms share the factor $(x+2)(x-4)$:
$$P(x) = (x+2)(x-4)\left[\frac{6x - (x+5)}{5}\right] = (x+2)(x-4)\cdot\frac{5x-5}{5} = (x+2)(x-4)(x-1).$$
$$\boxed{P(x) = (x+2)(x-4)(x-1) = x^{3} - 3x^{2} - 6x + 8}$$
Both interpolants collapse to the same cubic $x^{3}-3x^{2}-6x+8$ — the exam samples one
underlying cubic twice, at two different point sets, once via Newton's divided differences and once via Lagrange,
which is a strong internal cross-check that both methods were applied correctly.
