1. Choose a suitable start value for $x$, then find the corresponding $y$-value on the curve.
2. Move over to the $y = x$ curve. Reuse this $x$-value.
3. Repeat this process until suitably close to a root (assuming the process converges).

📑Example:

📑Example The curve $y = x^3 + 4x - 3$ intersects the $x$-axis at the point $A$ where $x = \alpha$. (a) Show that $\alpha$ lies between 0.5 and 1.0.

(b) Show that the equation $x^3 + 4x - 3 = 0$ can be rearranged into the form $x = \frac{3 - x^3}{4}$.

(c)

(i) Use the iteration $x_{n+1} = \frac{3 - x_n^3}{4}$ with $x_1 = 0.5$ to find $x_3$, giving your answer to two decimal places.

(ii) The sketch on Figure 1 shows parts of the graphs of $y = \frac{3 - x^3}{4}$ and $y = x$, and the position of $x_1$.

On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of $x_2$ and $x_3$ on the $x$-axis.

Solution:

(a)

Change of sign. Since the curve is continuous in the interval $(0.5, 1)$, then a root lies in this interval.

(b)

Here we have rearranged the equation we wanted to solve into $2$ separate equations, one of which is $y = x$.

• We are solving simultaneously:
• LHS: $y = x$
• RHS: $y = \frac{3 - x^3}{4}$ (c) (i)

(c) (ii)

📑Example : (b) The curve $y = 3^x$ intersects the line $y = x + 3$ at the point where $x = \alpha$.

(i) Show that $\alpha$ lies between 0.5 and 1.5.

(This would involve substituting $x = 0.5$ and $x = 1.5$ into the equation and showing the signs of the function change.)

(ii) Show that the equation $3^x = x + 3$ can be rearranged into the form:

(iii) Use the iteration $x_{n+1} = \frac{\ln(x_n + 3)}{\ln 3}$ with $x_1 = 0.5$ to find $x_3$, giving your answer to two significant figures.

(iv) The sketch on Figure 1 shows part of the graphs of $y = \frac{\ln(x + 3)}{\ln 3}$ and $y = x$, and the position of $x_1$.

On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of $x_2$ and $x_3$ on the $x$-axis.

Solution

(b) (i)