Question 6 of 7: Newton’s Method and Fixed-Point Iteration
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 6: Newton’s Method and Fixed-Point Iteration (10+10 marks)
Check: as literally printed, $5^x+6x-3=0$ has $f(1.5)=5^{1.5}+6(1.5)-3=17.18$, nowhere near a root — because $f'(x)=5^x\ln5+6\gt0$ everywhere, $f$ is strictly increasing (and convex, $f''(x)=5^x(\ln5)^2\gt0$), so it has exactly one real root, at $x\approx0.2506$, not near $1.5$. This looks like a sign typo in the source (the mirror equation $5^x-6x-3=0$ does have a root at $x\approx1.57$, genuinely close to $x_0=1.5$). Newton’s method is applied exactly as instructed — equation as printed, $x_0=1.5$, three iterations — below; because $f$ is monotone and convex, the iteration is still well-posed and converges to the equation’s unique real root, just not one "close to" the stated $x_0$.
Find. (A) $x_1,x_2,x_3$ by Newton’s method. (B) $x_1,\dots,x_5$ by fixed-point iteration, and a justification of convergence.
Approach. (A) Apply $x_{k+1}=x_k-f(x_k)/f'(x_k)$ with $f'(x)=5^x\ln5+6$, rounding each iterate to 7 significant digits before the next step (as instructed). (B) Iterate $x_{k+1}=g(x_k)$ directly; convergence of a fixed-point scheme is governed by $|g'(x)|$ near the root — it converges iff this is below 1.
(A) Newton's method on $f(x)=5^x+6x-3$, $f'(x)=5^x\ln5+6$, from $x_0=1.5$.
$$x_1=1.5-\dfrac{f(1.5)}{f'(1.5)}=1.5-\dfrac{17.18034}{24.98055}=0.7839754$$
$$x_2=0.7839754-\dfrac{f(0.7839754)}{f'(0.7839754)}=0.3358832$$
$$x_3=0.3358832-\dfrac{f(0.3358832)}{f'(0.3358832)}=\boxed{0.2523191}$$
The sequence is decreasing monotonically toward the true root ($x^\ast\approx0.2505532$, found independently by bisection) — the expected behaviour for Newton’s method on a convex, increasing $f$ started at a point with $f(x_0)\gt0$.
(B) Fixed-point iteration $x_{k+1}=g(x_k)=(x_k^3+11x_k-3)/(7x_k)$ from $x_0=1.7$.
$$x_1=\dfrac{1.7^3+11(1.7)-3}{7(1.7)}=1.732185,\qquad x_2=1.752650,\qquad x_3=1.765727$$
$$x_4=1.774111,\qquad x_5=\boxed{1.779497}$$
The iterates increase monotonically, closing in on the true root of the cubic near this starting point, $x^\ast\approx1.789244$ (found independently by bisection on $x^3-7x^2+11x-3=0$).
Explain the convergence. Differentiating, $g'(x)=\dfrac{d}{dx}\!\left[\dfrac{x^3+11x-3}{7x}\right]=\dfrac{2x^3+3}{7x^2}$ (after simplification). Evaluated at the root, $g'(x^\ast)\approx0.645$, so $|g'(x^\ast)|\lt1$: by the fixed-point (contraction-mapping) theorem, the iteration $x_{k+1}=g(x_k)$ converges to $x^\ast$ for any $x_0$ close enough to it, at a linear rate governed by $|g'(x^\ast)|\approx0.645$ per step. Not every algebraic rearrangement of $x=g(x)$ has this property — e.g. solving for the $x^3$-derived branch $x=(7x^2-11x+3)^{1/3}$ instead would give $|g'|\gt1$ near the same root and diverge.
