Question: 18 (a) Sketch on the same axis the graphs of $y = e^{-x}$ and $y = \sin x$ where $0 \leq x \leq \pi$. [2]
(b) Hence state the number of roots of the equation $e^{-x} - \sin x = 0$ [1]
(c) Show that the smallest root of $e^{-x} - \sin x = 0$ lies between $x = 0$ and $x = \frac{\pi}{2}$. [3]
(c) Starting with $x_0 = 0.8$, use the Newton-Raphson method twice to estimate the smallest root of $e^{-x} - \sin x = 0$, giving the answer correct to three decimal places. [4]
1. **Problem Statement:**
We are asked to analyze the functions $y = e^{-x}$ and $y = \sin x$ on the interval $0 \leq x \leq \pi$.
2. **Part (a): Sketch the graphs**
- The graph of $y = e^{-x}$ starts at $y=1$ when $x=0$ and decreases exponentially towards 0 as $x$ approaches $\pi$.
- The graph of $y = \sin x$ starts at 0 when $x=0$, rises to 1 at $x=\frac{\pi}{2}$, and returns to 0 at $x=\pi$.
3. **Part (b): Number of roots of $e^{-x} - \sin x = 0$**
- The roots correspond to the points where the graphs intersect.
- Since $e^{-x}$ decreases from 1 to about $e^{-\pi} \approx 0.043$ and $\sin x$ rises from 0 to 1 and back to 0, the graphs intersect twice.
- Therefore, the equation has **2 roots**.
4. **Part (c): Show smallest root lies between $0$ and $\frac{\pi}{2}$**
- Evaluate the function $f(x) = e^{-x} - \sin x$ at the endpoints:
$$f(0) = e^{0} - \sin 0 = 1 - 0 = 1 > 0$$
$$f\left(\frac{\pi}{2}\right) = e^{-\frac{\pi}{2}} - \sin \frac{\pi}{2} = e^{-\frac{\pi}{2}} - 1 < 0$$
- Since $f(0) > 0$ and $f(\frac{\pi}{2}) < 0$, by the Intermediate Value Theorem, there is at least one root between $0$ and $\frac{\pi}{2}$.
5. **Part (d): Newton-Raphson method to estimate smallest root**
- Define $f(x) = e^{-x} - \sin x$.
- Its derivative is:
$$f'(x) = -e^{-x} - \cos x$$
- Newton-Raphson iteration formula:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
- Starting with $x_0 = 0.8$:
1. Calculate $f(0.8)$:
$$f(0.8) = e^{-0.8} - \sin 0.8 \approx 0.4493 - 0.7174 = -0.2681$$
2. Calculate $f'(0.8)$:
$$f'(0.8) = -e^{-0.8} - \cos 0.8 \approx -0.4493 - 0.6967 = -1.1460$$
3. Compute $x_1$:
$$x_1 = 0.8 - \frac{-0.2681}{-1.1460} = 0.8 - 0.2340 = 0.5660$$
- Second iteration:
1. Calculate $f(0.5660)$:
$$f(0.5660) = e^{-0.5660} - \sin 0.5660 \approx 0.5679 - 0.5363 = 0.0316$$
2. Calculate $f'(0.5660)$:
$$f'(0.5660) = -e^{-0.5660} - \cos 0.5660 \approx -0.5679 - 0.8443 = -1.4122$$
3. Compute $x_2$:
$$x_2 = 0.5660 - \frac{0.0316}{-1.4122} = 0.5660 + 0.0224 = 0.5884$$
- Rounded to three decimal places, the estimate for the smallest root is:
$$\boxed{0.588}$$