Question 4 of 7: Newton's Divided Differences and Finite-Difference Derivatives
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 04-BS-5 Advanced Mathematics. Three-hour, closed-book exam (one double-sided 8.5"×11" aid sheet permitted; approved Casio/Sharp calculator allowed). Format: seven questions (Q1 20, Q2 20, Q3 20, Q4 20, Q5 20, Q6 20, Q7 20 marks, per the printed marking scheme); any five constitute a complete paper and only the first five appearing in the answer book are marked. All seven are solved below for completeness.
Reference texts: Kreyszig, Advanced Engineering Mathematics (10th ed., Wiley) — power-series solutions of ODEs about an ordinary point (Ch. 5), Fourier series and the Fourier transform (Ch. 11), numerics in general: interpolation, root-finding, Romberg integration (Ch. 19), numeric linear algebra: LU factorization (Ch. 20); Burden & Faires, Numerical Analysis (9th ed., Cengage) — Newton divided-difference interpolation, finite-difference derivative stencils, Newton–Raphson/bisection/fixed-point convergence theory, Romberg extrapolation, Doolittle LU factorization.
Question 4: Newton's Divided Differences and Finite-Difference Derivatives (A) 12, (B) 8 marks
Find (A). Newton's divided-difference interpolating polynomial through all 7 points, of the highest degree the data support.
Approach (A). Build the divided-difference table $f[x_i]=F(x_i)$, $f[x_i,\dots,x_{i+j}]=\dfrac{f[x_{i+1},\dots,x_{i+j}]-f[x_i,\dots,x_{i+j-1}]}{x_{i+j}-x_i}$, then assemble Newton's form from the top diagonal.
Build the divided-difference table. Working column by column from the raw data:
Divided-difference table
$x_i$
$f[x_i]$
$f[x_i,x_{i+1}]$
2nd
3rd
4th
5th
6th
-5
0
3
-5
1
0
0
0
-3
6
-12
0
1
0
0
-2
-6
-12
5
1
0
0
-30
8
10
1
2
-14
38
14
3
24
66
4
90
Every divided difference of order 4 and higher is identically zero.
Assemble Newton's polynomial from the top-diagonal entries. $f[x_0]=0,\ f[x_0,x_1]=3,\ f[x_0,x_1,x_2]=-5,\ f[x_0,\dots,x_3]=1$, all higher terms zero:
$$P(x)=0+3(x+5)-5(x+5)(x+3)+1\cdot(x+5)(x+3)(x+2)+0+0+0$$
$$\boxed{P(x)=x^3+5x^2-6x-30}$$
Identify the highest possible degree. Although 7 data points could in principle require a degree-6 polynomial, the 4th, 5th and 6th divided differences vanish identically, so the data are exactly consistent with (and no more complex than) a cubic. $P(x)$ reproduces all 7 tabulated values exactly, so degree 3 is both necessary and sufficient — adding the higher-order Newton terms would only add zero.
Given data (B): $f(x)$ at $x_0=-2$, $h=2$
$x$
-6
-4
-2
0
2
4
6
$f(x)$
-30
10
-6
-30
-14
90
330
Find (B). $f'(-2),\,f''(-2),\,f'''(-2),\,f''''(-2)$ using the supplied 5-point forward-difference stencils with nodes $x_0,x_0+h,\dots,x_0+4h=-2,0,2,4,6$.
Approach (B). Read off $f_0=f(-2)=-6,\ f_1=f(0)=-30,\ f_2=f(2)=-14,\ f_3=f(4)=90,\ f_4=f(6)=330$ and substitute directly into each supplied formula with $h=2$.
First derivative.
$$f'(-2)\approx\dfrac{1}{12(2)}\big[-25(-6)+48(-30)-36(-14)+16(90)-3(330)\big]=\dfrac{-336}{24}=\boxed{-14}$$
Second derivative.
$$f''(-2)\approx\dfrac{1}{12(2)^2}\big[35(-6)-104(-30)+114(-14)-56(90)+11(330)\big]=\dfrac{-96}{48}=\boxed{-2}$$
Third derivative.
$$f'''(-2)\approx\dfrac{1}{2(2)^3}\big[-5(-6)+18(-30)-24(-14)+14(90)-3(330)\big]=\dfrac{96}{16}=\boxed{6}$$
Fourth derivative.
$$f''''(-2)\approx\dfrac{1}{2^4}\big[(-6)-4(-30)+6(-14)-4(90)+330\big]=\dfrac{0}{16}=\boxed{0}$$
A vanishing 4th derivative is consistent with the underlying data being (at most) a cubic in the neighbourhood of $x_0=-2$ — the same conclusion the 4th divided difference reached in part (A) for the companion table.
