---
title: "N-R: sin x = e^-x with limitations"
subject: "Numerical Method"
chapter: 2
key: "semester_4/numerical_method:newton-raphson-sinx-exp"
source: https://www.bctengineer.com/q/semester_4/numerical_method/newton-raphson-sinx-exp
site: BCT Engineering
---

# N-R: sin x = e^-x with limitations

Numerical Method · Chapter 2: Solutions of Nonlinear Equations

- Asked in: 2073 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/newton-raphson-sinx-exp
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-2

## Question

Find a real root of the equation sin x = e^(-x) correct up to four decimal places using N-R method. What are the limitations of this method?

## Short answer

With f(x)=sin x - e^{-x}, f'(x)=cos x + e^{-x}, from x0=0.6, the root is 0.5885. Limitations: requires derivative; may fail if f'≈0; needs close initial guess; no guaranteed convergence without bracketing; linear convergence at multiple roots; no error bound.

## Answer

#### Part 1: Root by Newton-Raphson
```text
f(x)  = sin x − e^(−x)
f'(x) = cos x + e^(−x)
x(n+1) = x(n) − f(x(n)) / f'(x(n))     (x in radians)
```
f(0.5) = 0.4794 − 0.6065 = −0.1271 < 0 and f(1) = 0.8415 − 0.3679 = 0.4736 > 0, so a root lies in (0.5, 1). Take x0 = 0.6.

| n | x(n) | f(x(n)) | f'(x(n)) | x(n+1) |
|---|---|---|---|---|
| 0 | 0.6000 | 0.0158 | 1.3741 | 0.5885 |
| 1 | 0.5885 | -0.0001 | 1.3870 | 0.5885 |
| 2 | 0.5885 | -0.0000 | 1.3869 | 0.5885 |

Two successive iterates agree to four decimals.

**Answer: root ≈ 0.5885**

#### Part 2: Limitations of the method
1. It needs f′(x), which must be derived and evaluated each step.
2. If f′(x(n)) = 0 or is very small, the step is undefined or very large (here f′ = cos x + e^(−x) can become small, so a bad guess can fail).
3. The initial guess must be close to the root; otherwise the iteration may diverge, oscillate or jump to another root (this function has many roots, such as 3.0964, 6.2850, ...).
4. Convergence is guaranteed only when |f·f″/f′²| < 1 near the root.
5. At a multiple root the convergence is only linear.
6. There is no bracketing, so no error bound is available in advance.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For f = sin x - e^(-x), f'(x) is:
   A. cos x + e^(-x)
   B. cos x - e^(-x)
   C. -cos x + e^(-x)
   D. sin x + e^(-x)
   Answer: A. Derivative of -e^(-x) is +e^(-x). (Key point: Derivative)

2. The root is:
   A. 0.6000
   B. 0.5885
   C. 0.4794
   D. 1.0000
   Answer: B. Iteration from 0.6 converges to 0.5885. (Key point: Result)

3. If f'(x(n)) is zero or very small:
   A. Convergence is faster
   B. The root is found
   C. The step is undefined or very large
   D. The method becomes bisection
   Answer: C. The tangent is nearly horizontal. (Key point: Limitation: f' zero)

4. With a poor starting guess the method may:
   A. Always find the nearest root
   B. Always converge
   C. Never use f'
   D. Diverge, oscillate or jump to another root
   Answer: D. This function has many roots, e.g. 3.0964. (Key point: Limitation: guess)

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