Numerical solution of equations — practice questions

(i) Let $f(x) = x^3 + 2x - 1$. $f(0.4) = (0.4)^3 + 2(0.4) - 1 = 0.064 + 0.8 - 1 = -0.136$ $f(0.5) = (0.5)^3 + 2(0.5) - 1 = 0.125 + 1 - 1 = 0.125$ Since $f(x)$ is a continuous function and there is a change of sign between $x=0.4$ and $x=0.5$, a root must lie in the interval $(0.4, 0.5)$. [M1 for both evaluations, A1 for conclusion]

(ii) We are given the iterative formula $x_{n+1} = \frac{1}{x_n^2 + 2}$ and $x_0 = 0.45$. $x_1 = \frac{1}{(0.45)^2 + 2} = \frac{1}{2.2025} = 0.45403...$ $x_2 = \frac{1}{(0.45403...)^2 + 2} = 0.45340...$ $x_3 = \frac{1}{(0.45340...)^2 + 2} = 0.45352...$ $x_4 = \frac{1}{(0.45352...)^2 + 2} = 0.45350...$ $x_5 = \frac{1}{(0.45350...)^2 + 2} = 0.45350...$ [M1 for using the formula correctly at least once]

The values are converging. To show the root is $0.454$ to 3 d.p., we check the interval $[0.4535, 0.4545]$. $f(0.4535) = (0.4535)^3 + 2(0.4535) - 1 = -0.00018... < 0$ $f(0.4545) = (0.4545)^3 + 2(0.4545) - 1 = +0.0020... > 0$ There is a sign change in the interval $[0.4535, 0.4545]$, so the root is $0.454$ correct to 3 decimal places. [A1 for showing iterations converge, A1 for correct check and final answer]

Let $f(x) = e^x - 5x$. Then $f'(x) = e^x - 5$. The Newton-Raphson formula is $x_{n+1} = x_n - \frac{e^{x_n} - 5x_n}{e^{x_n} - 5}$. [M1 for correct $f'(x)$ and formula]

We start with $x_0 = 2$. $x_1 = 2 - \frac{e^2 - 5(2)}{e^2 - 5} = 2 - \frac{7.389056 - 10}{7.389056 - 5} = 2 - \frac{-2.610944}{2.389056} = 2 - (-1.09288) = 3.09288$ [M1 for first correct application of formula]

$x_2 = 3.09288 - \frac{e^{3.09288} - 5(3.09288)}{e^{3.09288} - 5} = 3.09288 - \frac{22.0295 - 15.4644}{22.0295 - 5} = 3.09288 - \frac{6.5651}{17.0295} = 3.09288 - 0.38551 = 2.70737$

$x_3 = 2.70737 - \frac{e^{2.70737} - 5(2.70737)}{e^{2.70737} - 5} = 2.70737 - \frac{14.9888 - 13.53685}{14.9888 - 5} = 2.70737 - \frac{1.45195}{9.9888} = 2.70737 - 0.14535 = 2.56202$

Continuing this process (often done on a calculator): $x_4 = 2.54303$ $x_5 = 2.54264$ $x_6 = 2.54264$

The values have converged. To check for 4 d.p., we test the interval $[2.542635, 2.542645]$. $f(2.542635) = e^{2.542635} - 5(2.542635) = -0.000003... < 0$ $f(2.542645) = e^{2.542645} - 5(2.542645) = +0.000009... > 0$ Since there is a sign change, the root is $2.5426$ correct to 4 decimal places. [A1 for showing convergence, A1 for correct answer to 4 d.p. after a valid check]