Question 5 of 7: Newton's Method, Bisection, and Fixed-Point Iteration

Question 5: Newton's Method, Bisection, and Fixed-Point Iteration (A) 8, (B)(i) 6, (B)(ii) 6 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). $g(x)=x^2-2-\sin^2x$, $x_0=1.8$.

Find (A). $x_1,x_2,x_3$ by Newton's method, $x_{n+1}=x_n-g(x_n)/g'(x_n)$.

Approach (A). $g'(x)=2x-2\sin x\cos x=2x-\sin(2x)$; iterate three times from $x_0=1.8$, carrying eight digits.

1. Set up the derivative. $g'(x)=2x-\sin(2x)$.
2. Iterate three times. $$x_1=1.8-\dfrac{g(1.8)}{g'(1.8)}=1.7278616,\qquad x_2=1.7278616-\dfrac{g(x_1)}{g'(x_1)}=1.7252125$$ $$\boxed{x_3=1.7252125-\dfrac{g(x_2)}{g'(x_2)}=1.7252089}$$ Substituting back, $g(x_3)\approx2.6\times10^{-11}$ — the iteration has converged to the full eight-digit precision requested.

Given (B)(i). $g(x)=\ln(1+x^2)-x^2+3x+4$ (the difference of the two given curves), bracket $[a,b]=[4.5,4.6]$ with $g(4.5)\gt0,\ g(4.6)\lt0$.

Find (B)(i). The bisection midpoint after 5 iterations.

Approach (B)(i). Halve the bracket, keeping the sub-interval whose endpoints still bracket the sign change, five times.

1. Confirm the bracket. $g(4.5)=0.3063569\gt0$, $g(4.6)=-0.2617111\lt0$: a root lies in $(4.5,4.6)$.
2. Bisect five times.

   Bisection iterations

   Iter.	$[a,b]$	midpoint $m$	$g(m)$
   1	$[4.5,4.6]$	4.55	$+0.024927$
   2	$[4.5,4.55]$	4.575	$-0.117741$
   3	$[4.55,4.575]$	4.5625	$-0.046244$
   4	$[4.55,4.5625]$	4.55625	$-0.010618$
   5	$[4.55,4.55625]$	4.553125	$+0.007165$

   After 5 bisections the bracket has narrowed to $[4.553125,4.55625]$, width $0.03125=(0.1)/2^5$, with best estimate $$\boxed{x\approx4.5531250}$$

Given (B)(ii). $\varphi(x)=\dfrac{\ln(1+x^2)+3x+4}{x}$, start value $x_0=4.5531250$ (the last bisection midpoint from (i)).

Find (B)(ii). $x_1,x_2,x_3$ by fixed-point iteration $x_{n+1}=\varphi(x_n)$, carrying seven digits.

Approach (B)(ii). Iterate $\varphi$ three times from the bisection result; the sequence should settle near the same root found in (i).

1. Iterate three times. $$x_1=\varphi(4.5531250)=4.5546987,\qquad x_2=\varphi(x_1)=4.5543063$$ $$\boxed{x_3=\varphi(x_2)=4.5544041}$$ The fixed-point sequence is converging toward the same root the bisection method bracketed ($\approx4.5544$), consistent between both methods to four decimal places after only a few iterations.