Numerical solution of equations — practice questions

Let $f(x) = e^{2x} - 5x - 3$.

To show a root exists between 1 and 2, we evaluate $f(1)$ and $f(2)$.

For $x=1$: $f(1) = e^{2(1)} - 5(1) - 3 = e^2 - 8 \approx 7.389 - 8 = -0.611$

For $x=2$: $f(2) = e^{2(2)} - 5(2) - 3 = e^4 - 10 - 3 = e^4 - 13 \approx 54.598 - 13 = 41.598$

We have $f(1) < 0$ and $f(2) > 0$. Since the function $f(x)$ is continuous and there is a change of sign in the interval $[1, 2]$, there must be at least one root between $x=1$ and $x=2$.

(i) Let $f(x) = x^3 + x - 5$. $f(1) = 1^3 + 1 - 5 = -3$ $f(2) = 2^3 + 2 - 5 = 8 + 2 - 5 = 5$ Since $f(x)$ is continuous and there is a change of sign between $x=1$ and $x=2$, a root lies in the interval $(1, 2)$.

(ii) Start with the original equation: $x^3 + x - 5 = 0$ $x^3 = 5 - x$ $x = \sqrt[3]{5-x}$

(iii) Using the iterative formula $x_{n+1} = \sqrt[3]{5-x_n}$ with $x_0 = 1.5$. $x_0 = 1.5$ $x_1 = \sqrt[3]{5 - 1.5} = \sqrt[3]{3.5} \approx 1.518294$ $x_2 = \sqrt[3]{5 - 1.518294} \approx 1.515993$ $x_3 = \sqrt[3]{5 - 1.515993} \approx 1.516328$ $x_4 = \sqrt[3]{5 - 1.516328} \approx 1.516281$ $x_5 = \sqrt[3]{5 - 1.516281} \approx 1.516288$

The values are converging. Both $x_4$ and $x_5$ round to 1.516. To be certain, we can check the sign change around this value. Let's check the interval $[1.5155, 1.5165]$. $f(1.5155) = (1.5155)^3 + 1.5155 - 5 \approx -0.0051$ $f(1.5165) = (1.5165)^3 + 1.5165 - 5 \approx +0.0054$ There is a sign change, so the root lies in this interval. Therefore, the root $\alpha = 1.516$ correct to 3 decimal places.