Question 6 of 7: Root-Finding by Bisection, Newton's Method, and Fixed-Point Iteration

Question 6: Root-Finding by Bisection, Newton's Method, and Fixed-Point Iteration (A(a): 6; A(b): 6; B: 8 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. (A) $h(x)=2e^{-x}-3\cos x$, root bracketed in $[-1,0]$. (B) $p(x)=x^4-3x^3+5$, root near $x_0=2$.

Find. (A)(a) The bracket after 3 bisections (6-digit arithmetic). (b) Two further Newton iterations from that point (7-digit arithmetic). (B) Five fixed-point iterations of a valid rearrangement $x=g(x)$ (7-digit arithmetic).

Approach. (A) Confirm the sign change, bisect three times tracking the sub-interval that keeps the sign change, then switch to Newton's method for quadratic convergence. (B) Rearrange $p(x)=0$ into a form $x=g(x)$ with $|g'(\text{root})|\lt1$ near $x_0=2$, then iterate.

1. (A)(a) Confirm the bracket and bisect three times. $h(-1)=2e^{1}-3\cos(1)=5.436564-1.620907=3.815657\gt0$; $h(0)=2-3=-1\lt0$, confirming a sign change on $[-1,0]$.

   Bisection (6-digit arithmetic)

   Iter.	Interval	Midpoint	$h(\text{mid})$	New interval
   1	$[-1,\ 0]$	$-0.5$	$+0.664695$	$[-0.5,\ 0]$
   2	$[-0.5,\ 0]$	$-0.25$	$-0.338686$	$[-0.5,\ -0.25]$
   3	$[-0.5,\ -0.25]$	$-0.375$	$+0.118460$	$[-0.375,\ -0.25]$

   $$\boxed{x\approx\dfrac{-0.375+(-0.25)}{2}=-0.312500\ \text{after 3 bisections}}$$
2. (A)(b) Switch to Newton's method. With $h'(x)=-2e^{-x}+3\sin x$, starting from $x_0=-0.3125$ and using $x_{k+1}=x_k-h(x_k)/h'(x_k)$ (7-digit arithmetic): $$x_1=-0.312500-\dfrac{h(-0.312500)}{h'(-0.312500)}=\boxed{-0.345604}$$ $$x_2=-0.345604-\dfrac{h(-0.345604)}{h'(-0.345604)}=\boxed{-0.344804}$$ The true root (by high-precision root-finding) is $-0.344804$ to six decimals — the second Newton iterate already matches to 6 significant figures, illustrating Newton's quadratic convergence once bisection has supplied a good starting bracket.
3. (B) Rearrange into a convergent fixed-point form. Writing $x^4-3x^3+5=0$ as $x^3(x-3)=-5$, i.e. $x-3=-5/x^3$, gives $$\boxed{g(x)=3-\dfrac{5}{x^3}}$$ Its derivative $g'(x)=15/x^4$ is small near the root ($g'\approx0.257$ at the root $\approx2.76$, found below), so $|g'(\text{root})|\lt1$ and the iteration converges there (a rearrangement like $x=(3x^3-5)^{1/4}$ would instead diverge from $x_0=2$, since its derivative exceeds 1 in magnitude near the root).
4. Iterate five times from $x_0=2$ (7-digit arithmetic).

   Fixed-point iterates $x_{k+1}=g(x_k)$

   $k$	0	1	2	3	4	5
   $x_k$	2.0000000	2.3750000	2.6267678	2.7241297	2.7526644	2.7602767

   $$\boxed{x_5=2.7602767\ \text{after 5 iterations}}$$ (the true root, found by bisection/Newton to high precision, is $2.7629419$; the iteration is converging steadily towards it, with the error shrinking by roughly the factor $g'(\text{root})\approx0.257$ each step, consistent with linear convergence.)