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 using the initial value , giving the answer to 3 decimal places.

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

One way to do this is:

To solve the equation, use the iterative formula

We are given the initial value

Substituting this into the iterative formula gives

Substituting iteratively gives:

(3 dp)

(3 dp)

(3 dp)

(3 dp)

(3 dp)

(3 dp)

Since and give the same value to 3 decimal places, the iteration stops. The solution to the equation is 2.113 to 3 decimal places.

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 to find the first iteration, then just press = for the next iteration.

Example

Show that the equation has a solution that lies between 1 and 2.

Use the iterative equation to find the solution to the equation to 3 decimal places.

Substituting into the left hand side of the equation gives

Substituting into the left hand side of the equation gives

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 and substituting this iteratively gives:

(3dp)

(3 dp)

(3 dp)

(3 dp)

(3 dp)

(3 dp)

(3 dp)

So the solution to the equation is 1.569 to 3 decimal places.

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