- 2069 Bhadra · 4+4 marks
Bisection and N-R: e^x = 3x
Short answer
For f(x)=e^x-3x, bisection on (0,1) and (1,2) gives roots 0.619 and 1.512; Newton-Raphson from 0.5 and 1.5 gives the same values in 3-4 iterations.
Find a root of e^x = 3x using bisection method and Newton-Raphson method correct up to 3 decimal places.
Write the equation as f(x) = eˣ − 3x = 0.
f(0) = 1 > 0, f(1) = 2.7183 − 3 = −0.2817 < 0, f(2) = 7.3891 − 6 = 1.3891 > 0. There is a sign change in (0, 1) and another in (1, 2), so the equation has two real roots. Both are found below.
(a) Bisection method (c = (a + b)/2)
Root in (0, 1):
| n | a | b | c=(a+b)/2 | f(c) | root lies in |
|---|---|---|---|---|---|
| 1 | 0.0000 | 1.0000 | 0.5000 | 0.1487 | [c, b] |
| 2 | 0.5000 | 1.0000 | 0.7500 | -0.1330 | [a, c] |
| 3 | 0.5000 | 0.7500 | 0.6250 | -0.0068 | [a, c] |
| 4 | 0.5000 | 0.6250 | 0.5625 | 0.0676 | [c, b] |
| 5 | 0.5625 | 0.6250 | 0.5938 | 0.0295 | [c, b] |
| 6 | 0.5938 | 0.6250 | 0.6094 | 0.0112 | [c, b] |
| 7 | 0.6094 | 0.6250 | 0.6172 | 0.0021 | [c, b] |
| 8 | 0.6172 | 0.6250 | 0.6211 | -0.0023 | [a, c] |
| 9 | 0.6172 | 0.6211 | 0.6191 | -0.0001 | [a, c] |
| 10 | 0.6172 | 0.6191 | 0.6182 | 0.0010 | [c, b] |
| 11 | 0.6182 | 0.6191 | 0.6187 | 0.0005 | [c, b] |
Root in (1, 2):
| n | a | b | c=(a+b)/2 | f(c) | root lies in |
|---|---|---|---|---|---|
| 1 | 1.0000 | 2.0000 | 1.5000 | -0.0183 | [c, b] |
| 2 | 1.5000 | 2.0000 | 1.7500 | 0.5046 | [a, c] |
| 3 | 1.5000 | 1.7500 | 1.6250 | 0.2034 | [a, c] |
| 4 | 1.5000 | 1.6250 | 1.5625 | 0.0832 | [a, c] |
| 5 | 1.5000 | 1.5625 | 1.5312 | 0.0302 | [a, c] |
| 6 | 1.5000 | 1.5312 | 1.5156 | 0.0054 | [a, c] |
| 7 | 1.5000 | 1.5156 | 1.5078 | -0.0066 | [c, b] |
| 8 | 1.5078 | 1.5156 | 1.5117 | -0.0006 | [c, b] |
| 9 | 1.5117 | 1.5156 | 1.5137 | 0.0024 | [a, c] |
| 10 | 1.5117 | 1.5137 | 1.5127 | 0.0009 | [a, c] |
| 11 | 1.5117 | 1.5127 | 1.5122 | 0.0001 | [a, c] |
(b) Newton-Raphson method
f′(x) = eˣ − 3; x(n+1) = x(n) − f(x(n))/f′(x(n)).
Root near 0.5 (x0 = 0.5):
| n | x(n) | f(x(n)) | f'(x(n)) | x(n+1) |
|---|---|---|---|---|
| 0 | 0.5000 | 0.1487 | -1.3513 | 0.6101 |
| 1 | 0.6101 | 0.0104 | -1.1595 | 0.6190 |
| 2 | 0.6190 | 0.0001 | -1.1429 | 0.6191 |
| 3 | 0.6191 | 0.0000 | -1.1428 | 0.6191 |
Root near 1.5 (x0 = 1.5):
| n | x(n) | f(x(n)) | f'(x(n)) | x(n+1) |
|---|---|---|---|---|
| 0 | 1.5000 | -0.0183 | 1.4817 | 1.5124 |
| 1 | 1.5124 | 0.0003 | 1.5374 | 1.5121 |
| 2 | 1.5121 | 0.0000 | 1.5364 | 1.5121 |
Answer: the roots are ≈ 0.619 and ≈ 1.512. Bisection needed 11 halvings for each root to fix three decimals, while Newton-Raphson reached the same values in only 3 to 4 steps.
✓ Answer checked by Claude Opus 5.5 (high effort)
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