Question 4 of 7: Newton Divided-Difference Interpolation and Numerical Differentiation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2014 — 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 of equal value (20 marks each, with internal splits as marked); 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), Newton divided-difference interpolation, numerical differentiation, Romberg integration, Newton’s method and fixed-point iteration (Ch. 19), Doolittle/Crout LU factorization and forward/back substitution (Ch. 7).
Question 4: Newton Divided-Difference Interpolation and Numerical Differentiation (14+6 marks)
Given. (A) Seven data pairs $(x_i,f_i)$, unequally spaced. (B) A second, separate table of ten equally-spaced pairs ($h=1$) and four forward-difference derivative formulas.
(A) Data for the divided-difference polynomial
x
−5
−4
−3
−1
0
2
3
f(x)
−175
−156
−65
93
100
0
−23
(B) Data for the derivative estimates at x₀=−2, h=1
x
−4
−3
−2
−1
0
1
2
3
4
5
f(x)
−55
−7
11
11
5
5
23
71
161
305
Find. (A) The Newton divided-difference interpolating polynomial through all seven points, of the highest degree the data supports. (B) $f'(-2)$, $f''(-2)$, $f'''(-2)$, $f^{(4)}(-2)$ from the second table.
Approach. (A) Build the divided-difference triangle from the seven $(x_i,f_i)$ pairs, then assemble the Newton forward form $P(x)=f[x_0]+f[x_0,x_1](x-x_0)+\cdots$; the degree is set by where the differences stop being zero, not by the point count. (B) Read the five needed function values straight off the second table ($x_0=-2$ through $x_0+4h=2$) and substitute into the four supplied formulas.
(A) Build the divided-difference table. With $f[x_i]=f_i$ and $f[x_i,\dots,x_{i+k}]=\dfrac{f[x_{i+1},\dots,x_{i+k}]-f[x_i,\dots,x_{i+k-1}]}{x_{i+k}-x_i}$, the leading diagonal (the coefficients actually used in the Newton form) is:
Leading diagonal of the divided-difference triangle
Order
0
1
2
3
4
5
6
f[x₀,…]
−175
19
36
−10
1
0
0
The 5th- and 6th-order divided differences are exactly zero — a telltale sign that the seven points lie exactly on a polynomial of degree 4, not 6.
Assemble the Newton form. With nodes taken in table order ($x_0=-5,x_1=-4,x_2=-3,x_3=-1,x_4=0$):
$$P(x)=-175+19(x+5)+36(x+5)(x+4)-10(x+5)(x+4)(x+3)+1\cdot(x+5)(x+4)(x+3)(x+1)$$
(the last two terms drop out since their coefficients are 0).
Expand to standard form. Multiplying out:
$$\boxed{P(x)=x^4+3x^3-25x^2-20x+100}$$
“Highest possible degree”: with 7 points the Newton form could in principle carry terms up to degree 6, but because the two highest divided differences vanish, the interpolating polynomial that actually reproduces the data has degree exactly 4 — that quartic is the highest-degree polynomial the data supports (any degree-6 form would just re-derive the same quartic with two zero leading coefficients).
(B) Read off the five function values needed from the second table at $x_0=-2,\,x_0+h=-1,\,x_0+2h=0,\,x_0+3h=1,\,x_0+4h=2$: $f(x_0)=11,\ f(x_0+h)=11,\ f(x_0+2h)=5,\ f(x_0+3h)=5,\ f(x_0+4h)=23$, with $h=1$.
Substitute into each forward-difference formula.
$$f'(-2)\approx\tfrac{1}{12}\big[-25(11)+48(11)-36(5)+16(5)-3(23)\big]=\tfrac{1}{12}(84)=\boxed{7}$$
$$f''(-2)\approx\tfrac{1}{12}\big[35(11)-104(11)+114(5)-56(5)+11(23)\big]=\tfrac{1}{12}(-216)=\boxed{-18}$$
$$f'''(-2)\approx\tfrac{1}{2}\big[-5(11)+18(11)-24(5)+14(5)-3(23)\big]=\tfrac{1}{2}(24)=\boxed{12}$$
$$f^{(4)}(-2)\approx\big[11-4(11)+6(5)-4(5)+23\big]=\boxed{0}$$
A vanishing 4th-derivative estimate is consistent with the underlying data being close to a cubic near $x=-2$ (the same "differences settle to a low-degree polynomial" behaviour seen in part A).
