Numerical solution of equations

Numerical solution of equations

Why numerical methods

Many equations, such as $x^3 + x - 3 = 0$ or $\mathrm{e}^x = 3 - x$, have no neat algebraic solution. A numerical method finds a root to whatever accuracy is needed.

Locating a root from a graph

The number of roots of $\mathrm{f}(x) = \mathrm{g}(x)$ is the number of times $y = \mathrm{f}(x)$ and $y = \mathrm{g}(x)$ cross. Sketching the two graphs shows how many roots there are and roughly where.

$y = \mathrm{e}^x$ is always increasing and $y = 3 - x$ is always decreasing, so they cross exactly once.

Locating a root by a sign change

If $\mathrm{f}$ is continuous and $\mathrm{f}(a)$ and $\mathrm{f}(b)$ have opposite signs, there is a root between $a$ and $b$.

$\mathrm{f}(1) = -1 < 0$ and $\mathrm{f}(2) = 7 > 0$. The sign changes and $\mathrm{f}$ is continuous, so there is a root in the interval.

Iterative formulae

An iterative formula $x_{n+1} = \mathrm{F}(x_n)$ uses each value to produce the next. If the sequence converges to a limit $\alpha$, then $\alpha = \mathrm{F}(\alpha)$, so the limit is a root of $x = \mathrm{F}(x)$.

At the limit, $x = (3 - x)^{\frac{1}{3}}$. Cubing gives $x^3 = 3 - x$, so $x^3 + x - 3 = 0$.

Using an iteration

Store each answer and use the calculator's answer key, so that no accuracy is lost. Keep at least two more decimal places than the question asks for.

$x_2 = 1.2599 \quad x_3 = 1.2028 \quad x_4 = 1.2158 \quad x_5 = 1.2129 \quad x_6 = 1.2135 \quad x_7 = 1.2134$

The root is 1.21 to 2 decimal places.

Checking the accuracy

To prove that 1.21 is correct to 2 decimal places, test the two ends of the rounding interval:

$\mathrm{f}(1.205) = -0.045 < 0 \qquad \mathrm{f}(1.215) = 0.009 > 0$

The sign changes, so the root lies between 1.205 and 1.215, and it rounds to 1.21.

Common forms of question

Questions in this topic often follow the same path.

1. Set up an equation from a context, often from a sector, a triangle or a stationary point: "Show that $\alpha$ satisfies $x = \dots$".
2. Show by a sign change that the root lies in a given interval.
3. Use the iteration to find it, with each step shown.