2071 Bhadra · 8 marks

Bisection: sin x = 1/x

Short answer

With f(x)=sin x -1/x, bracket [1,2]. Biseection yields root 1.114 after 11 iterations (interval width <0.001).

Use bisection method to find a real positive root of sin x = 1/x correct up to three decimal places.

Bisection method: if f(a)·f(b) < 0, the root lies in [a, b]. Take c = (a + b)/2; if f(a)·f(c) < 0 the root is in [a, c], otherwise in [c, b]. Repeat until the interval is shorter than the required accuracy.

Step 1: Bracket the root

f(x) = sin x − 1/x (radians) f(1) = 0.8415 − 1 = −0.1585 < 0 and f(2) = 0.9093 − 0.5 = 0.4093 > 0, so a root lies in (1, 2). Take a = 1, b = 2.

Step 2: Iterations

n	a	b	c=(a+b)/2	f(c)	root lies in
1	1.0000	2.0000	1.5000	0.3308	[a, c]
2	1.0000	1.5000	1.2500	0.1490	[a, c]
3	1.0000	1.2500	1.1250	0.0134	[a, c]
4	1.0000	1.1250	1.0625	-0.0676	[c, b]
5	1.0625	1.1250	1.0938	-0.0259	[c, b]
6	1.0938	1.1250	1.1094	-0.0060	[c, b]
7	1.1094	1.1250	1.1172	0.0038	[a, c]
8	1.1094	1.1172	1.1133	-0.0011	[c, b]
9	1.1133	1.1172	1.1152	0.0013	[a, c]
10	1.1133	1.1152	1.1143	0.0001	[a, c]
11	1.1133	1.1143	1.1138	-0.0005	[c, b]

After 11 iterations the interval width is 0.00049 (< 0.001), so the root is fixed to 3 decimals.

Answer: root ≈ 1.114

✓ Answer checked by Claude Opus 5.5 (high effort)

Bisection: 3x - √(1+sin x) · Numerical Method, chapter 2
Bisection and N-R: e^x = 3x · Numerical Method, chapter 2
N-R: sin x = e^-x with limitations · Numerical Method, chapter 2
Bisection method pseudo-code · Numerical Method, chapter 2 · asked 3×
Bisection method working principle · Numerical Method, chapter 2
Romberg: (e^x + sin x)/(1+x²) · Numerical Method, chapter 5 · asked 2×
Asynchronous mod-12 up counter, positive edge · Digital Logic, chapter 8 · asked 3×
ANOVA: three oil spill clean-up systems · Probability and Statistics, chapter 6

All Numerical Method chapter 2 questions →

Bisection: sin x = 1/x

Step 1: Bracket the root

Step 2: Iterations

Related questions