---
title: "N-R: 4 sin x - e^x"
subject: "Numerical Method"
chapter: 2
key: "semester_4/numerical_method:newton-raphson-4sinx-ex"
source: https://www.bctengineer.com/q/semester_4/numerical_method/newton-raphson-4sinx-ex
site: BCT Engineering
---

# N-R: 4 sin x - e^x

Numerical Method · Chapter 2: Solutions of Nonlinear Equations

- Asked in: 2068 Magh, 8 marks
- Video: [Newton's Method](https://www.youtube.com/watch?v=-5e2cULI3H8) — The Organic Chemistry Tutor
- 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-4sinx-ex
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-2

## Question

Evaluate the real root of f(x) = 4 sin x - e^x using Newton Raphson method. The absolute error of root in consecutive iteration should be less than 0.01%.

## Short answer

Apply Newton-Raphson to f(x)=4 sin x - e^x with x0=0.5. Iterations give x=0.370558, with relative change below 0.01%, so the smallest positive real root is approximately 0.3706.

## Answer

**Newton-Raphson formula:** x(n+1) = x(n) − f(x(n)) / f′(x(n)). Stop when the relative change |x(n+1) − x(n)| / |x(n+1)| × 100 % is below 0.01 %.

#### Step 1: Function, derivative and starting value
```text
f(x)  = 4 sin x − eˣ
f'(x) = 4 cos x − eˣ        (x in radians)
```
f(0) = −1 < 0 and f(0.5) = 4(0.4794) − 1.6487 = 0.2690 > 0, so a root lies in (0, 0.5). Take x0 = 0.5.
(A second root lies in (1, 2) because f(1) = 0.648 > 0 and f(2) = −3.75 < 0.)

#### Step 2: Iterations (six decimals shown)
| n | x(n) | f(x(n)) | f'(x(n)) | x(n+1) | error % = \|Δx / x(n+1)\| × 100 |
|---|---|---|---|---|---|
| 0 | 0.500000 | 0.268981 | 1.861609 | 0.355512 | 40.6424 |
| 1 | 0.355512 | -0.034630 | 2.322964 | 0.370419 | 4.0246 |
| 2 | 0.370419 | -0.000316 | 2.280360 | 0.370558 | 0.0374 |
| 3 | 0.370558 | -0.000000 | 2.279959 | 0.370558 | 0.0000 |

The error is below 0.01 % at the last row.

**Answer: smallest positive real root ≈ 0.3706**

Starting from x0 = 1.5 the same method converges to the other real root ≈ 1.3650 in 4 iterations.

## Check yourself

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

1. f'(x) for 4 sin x - e^x is:
   A. 4 cos x - e^x
   B. 4 cos x + e^x
   C. -4 sin x - e^x
   D. 4 sin x - e^x
   Answer: A. Differentiate each term. (Key point: Derivative)

2. Stop when:
   A. f = 1
   B. |Δx/x(n+1)| x 100 < 0.01 %
   C. x = 0.5
   D. Δx > 1
   Answer: B. Percent relative change criterion. (Key point: Stopping)

3. A second root lies in (1, 2) because:
   A. f(1) < 0
   B. f(2) > 0
   C. f(1) > 0 and f(2) < 0
   D. f'(1) = 0
   Answer: C. There is another sign change. (Key point: Two roots)

4. Starting at 0.5 gives root:
   A. 0.5
   B. 1.3650
   C. 0.3555
   D. 0.3706
   Answer: D. Converges to 0.3706. (Key point: Result)

## Related in Numerical Method

- [Limitations of Newton-Raphson](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alimitations-newton-raphson) — Numerical Method, chapter 2
- [Newton-Raphson algorithm](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Anewton-raphson-algorithm) — Numerical Method, chapter 2
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- [Absolute versus relative error](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aabsolute-relative-error-difference) — Numerical Method, chapter 1, asked 2×
- [Fixed point: sin x + 3x - 2](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Afixed-point-sinx-3x-2) — Numerical Method, chapter 2, asked 2×
- [Derive derivatives from Newton forward](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aderive-derivatives-newton-forward) — Numerical Method, chapter 5

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