Solving equations using iteration – Higher tier

Approximate solutions to more complex equations can be found using a process called iteration. Iteration means repeatedly carrying out a process. To solve an equation using iteration, start with an intial value and substitute this into the equation to obtain a new value, then use the new value for the next substitution, and so on.

Example

Find the solution to the equation x^3 + 5x = 20 using the initial value x_0=2, giving the answer to 3 decimal places.

First, rearrange the equation to leave x on its own on one side of the equation.

One way to do this is:

x^3 + 5x = 20

x^3 = 20 - 5x

x = \sqrt[3]{20 - 5x}

To solve the equation, use the iterative formula x_{n+1} \sqrt[3]{20-5x_n}

We are given the initial value x_0=2

Substituting this into the iterative formula gives x_1 = \sqrt[3]{20-5 \times 2} = \sqrt[3]{10} = 2.154...

Substituting iteratively gives:

x_2 = \sqrt[3]{20-5 \times 2.154} = \sqrt[3]{9.227...} = 2.097 (3 dp)

x_3 = \sqrt[3]{20-5 \times 2.097} = \sqrt[3]{9.512...} = 2.118 (3 dp)

x_4 = \sqrt[3]{20-5 \times 2.118} = \sqrt[3]{9.405...} = 2.111 (3 dp)

x_5 = \sqrt[3]{20-5 \times 2.111} = \sqrt[3]{9.445...} = 2.114 (3 dp)

x_6 = \sqrt[3]{20-5 \times 2.114} = \sqrt[3]{9.430...} = 2.113 (3 dp)

x_7 = \sqrt[3]{20-5 \times 2.113} = \sqrt[3]{9.436...} = 2.113 (3 dp)

Since x_6 and x_7 give the same value to 3 decimal places, the iteration stops. The solution to the equation x^3 + 5x = 20 is 2.113 to 3 decimal places.

curriculum-key-fact
If your calculator has an ANS button, use it to keep the value from one iteration to substitute into the next iteration. To solve the equation on a calculator with an ANS, type 2 =, then type \sqrt[3]{20-5\times \text{ANS}} = to find the first iteration, then just press = for the next iteration.

Example

Show that the equation x^3+2x=7 has a solution that lies between 1 and 2.

Use the iterative equation x_{n+1} = \sqrt[3]{7-2x_n} to find the solution to the equation to 3 decimal places.

Substituting x = 1 into the left hand side of the equation gives 1^3+2×1=3

Substituting x = 2 into the left hand side of the equation gives 2^3+2×2=12

The value on the right hand side of the equation, 7, lies between these two values, 3 and 12, so the solution to the equation must be between 1 and 2.

Using the initial value x_0=1 and substituting this iteratively gives:

x_1 = \sqrt[3]{7-2\times1} = \sqrt[3]{5} = 1.710 (3dp)

x_2 = \sqrt[3]{7-2\times1.710}... = \sqrt[3]{3.580}... = 1.530 (3 dp)

x_3 = \sqrt[3]{7-2\times1.530}... = \sqrt[3]{3.940}... = 1.579 (3 dp)

x_4 = \sqrt[3]{7-2\times1.579}... = \sqrt[3]{3.841}... = 1.566 (3 dp)

x_5 = \sqrt[3]{7-2\times1.566}... = \sqrt[3]{3.867}... = 1.570 (3 dp)

x_6 = \sqrt[3]{7-2\times1.570}... = \sqrt[3]{3.860}... = 1.569 (3 dp)

x_7 = \sqrt[3]{7-2\times1.569}... = \sqrt[3]{3.862}... = 1.569 (3 dp)

So the solution to the equation x^3+2x=7 is 1.569 to 3 decimal places.

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