Bisection, a Second-Order Iterative Method, and Fixed-Point Iteration: 04-BS-5, December 2016, Q6

Question 6: Bisection, a Second-Order Iterative Method, and Fixed-Point Iteration (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. (A) $f(x)=5^x+x^2-16$ bracketed on $[1,2]$, plus the stated second-order (Chebyshev-type) update using $f,f',f''$. (B) $x^3-6x^2+9x-3=0$ rearranged to $x=g(x)=(x^3+9x-3)/(6x)$, starting guess $x_0=1.6$.

Find. (A)(a) the bracket after 3 bisections; (A)(b) two further Newton-type iterates; (B) five fixed-point iterates.

Approach. Confirm the sign change, halve the bracket three times, switch to the faster second-order formula for two steps, then separately iterate the given fixed-point map five times from $x_0=1.6$.

(A)(a) Confirm the bracket, then bisect three times. $f(1.0)=5+1-16=-10.000$, $f(2.0)=25+4-16=13.000$ — sign change confirms a root in $(1,2)$.
Iteration 1: $c_1=1.5000$, $f(c_1)=5^{1.5}+2.25-16=-2.5697$ (negative) $\Rightarrow$ root in $(1.5,2)$.
Iteration 2: $c_2=1.7500$, $f(c_2)=5^{1.75}+3.0625-16=3.7810$ (positive) $\Rightarrow$ root in $(1.5,1.75)$.
Iteration 3: $c_3=1.6250$, $f(c_3)=5^{1.625}+2.6406-16=0.31244$ (positive, small) $\Rightarrow$ root in $(1.5,1.625)$. $$\boxed{x\approx c_3=1.6250}\quad\text{after three bisections (bracket width }0.125\text{)}$$
(A)(b) Two steps of the second-order formula from $x_0=1.625000$. With $f'(x)=5^x\ln5+2x$ and $f''(x)=5^x(\ln5)^2+2$:
Iteration 1: $x_1=1.612514$.
Iteration 2: $x_2=1.612513$. $$\boxed{x\approx1.612513}$$ matching the true root $1.6125132$ to six decimal places after only two steps — the cubic-order convergence of this method versus bisection's linear rate.
(B) Five fixed-point iterates of $g(x)=(x^3+9x-3)/(6x)$ from $x_0=1.6$. $$x_1=1.614167,\quad x_2=1.624498,\quad x_3=1.632045,\quad x_4=1.637564,\quad \boxed{x_5=1.641605}$$ The sequence climbs steadily toward the cubic's actual root near $1.6$ (confirmed by root-bracketing on the same cubic to be $1.652704$); the slow, monotone approach is expected because $|g'(\text{root})|$ is close to (but under) $1$ here — the exam asks only to demonstrate convergence toward that root in five steps, not to fully resolve it.

Part	Result
(A)(a) after 3 bisections	$c_3=1.6250$, bracket $(1.5,1.625)$
(A)(b) after 2 refined steps	$x\approx1.612513$ (true root $1.612513$)
(B) after 5 fixed-point steps	$x_5\approx1.641605$ (converging to true root $1.652704$)

Topic: root-finding — bisection, a second-order (Chebyshev-type) local method, and fixed-point iteration, applied to related equations.

Key relations: bisection halves the bracket every step (guaranteed, linear); $x_{n+1}=x_n-f/(f'-\tfrac{ff''}{2f'})$ converges near-cubically near a simple root; fixed point $x=g(x)$ converges linearly iff $|g'(\text{root})|<1$.

Why this works: pairing a slow-but-safe bracketing method with a fast local method is standard practice — bisection supplies a starting point already inside the fast method's convergence basin. For the fixed-point map, dividing $x^3-6x^2+9x-3=0$ by $6x$ (rather than, say, $6x^2$) is exactly the rearrangement that keeps $|g'|<1$ near this root.

Common pitfall: starting the fast second-order formula from a point still far from the root (it can diverge outside its basin); for fixed-point iteration, choosing an algebraically-equivalent but numerically unstable rearrangement of $x=g(x)$, which would make $|g'(\text{root})|>1$ and diverge instead of converge.

Source: Kreyszig, Advanced Engineering Mathematics, Ch. 19 (Solution of Equations by Iteration).

Question 6 of 7: Bisection, a Second-Order Iterative Method, and Fixed-Point Iteration

Question 6: Bisection, a Second-Order Iterative Method, and Fixed-Point Iteration (20 marks)