Question 4 of 7: Newton's Divided-Difference Interpolating Polynomial

Question 4: Newton's Divided-Difference Interpolating Polynomial (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. Seven tabulated ordinates of an unknown polynomial $P$ at seven distinct, unequally spaced nodes.

Find. The full divided-difference table, the Newton form of the interpolating polynomial, its highest attainable degree and expanded form, and the values $P(-3)$ and $P(0)$ at the two abscissae that are not tabulated.

Approach. Build the divided-difference table column by column; the first column that is identically zero reveals the true degree. Assemble Newton's forward form from the top diagonal, expand it, and substitute the two required abscissae.

1. Definition of the divided differences. With $f[x_{i}] = P(x_{i})$, higher orders follow from $$f[x_{i},\dots,x_{i+k}] = \frac{f[x_{i+1},\dots,x_{i+k}] - f[x_{i},\dots,x_{i+k-1}]}{x_{i+k}-x_{i}} .$$ Unequal spacing is no obstacle: the denominator carries the actual node separation.
2. Build the table. Working left to right, for example $f[x_{0},x_{1}] = (66-256)/(-4+5) = -190$ and $f[x_{0},x_{1},x_{2}] = (-37-(-190))/(-2+5) = 51$:

   Divided-difference table

   $x_{i}$	$f[\,]$	1st	2nd	3rd	4th	5th	6th
   $-5$	$256$	$-190$	$51$	$-9$	$1$	$0$	$0$
   $-4$	$66$	$-37$	$15$	$-3$	$1$	$0$
   $-2$	$-8$	$8$	$0$	$3$	$1$
   $-1$	$0$	$8$	$12$	$8$
   $1$	$16$	$44$	$44$
   $2$	$60$	$132$
   $3$	$192$

   The fourth differences are constant at $1$ and the fifth and sixth differences are identically zero.
3. Interpret the vanishing columns. A constant $k$-th divided difference is the signature of a degree-$k$ polynomial, because for $P(x)=c_{k}x^{k}+\cdots$ the $k$-th divided difference equals the leading coefficient $c_{k}$ exactly. Here the fourth column is constant at $1$ and everything beyond it vanishes, so although seven nodes would in principle support a degree-six interpolant, the data are consistent with $$\boxed{\;\deg P = 4,\qquad c_{4} = 1\;}$$ and that quartic is the polynomial of highest possible degree the data genuinely determine.
4. Assemble Newton's form from the top diagonal. $$P(x) = 256 - 190(x+5) + 51(x+5)(x+4) - 9(x+5)(x+4)(x+2) + (x+5)(x+4)(x+2)(x+1) .$$ This is already the answer to the first part of the question; the zero fifth and sixth terms are simply omitted.
5. Expand into standard form. Multiplying out the products, $$(x+5)(x+4)(x+2)(x+1) = x^{4}+12x^{3}+49x^{2}+78x+40 ,$$ and collecting all four contributions term by term gives $$\boxed{\;P(x) = x^{4} + 3x^{3} + x^{2} + 5x + 6\;}$$ This reproduces all seven tabulated ordinates exactly, which is the natural check on the expansion.
6. Evaluate at the two requested abscissae. Substituting directly, $$P(-3) = 81 - 81 + 9 - 15 + 6 = \boxed{\,0\,},\qquad P(0) = 0 + 0 + 0 + 0 + 6 = \boxed{\,6\,} .$$ Both are interpolations rather than extrapolations, since $-3$ and $0$ lie inside the node range $[-5,3]$; the quartic accordingly gives them with no loss of reliability. That $x=-3$ turns out to be a root of $P$ is a genuine feature of the data, not a coincidence of the arithmetic: $P(x) = (x+3)(x^{3}+x-2) \cdot 1 + 0$ upon division.