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The equationhas a solution somewhere between 1 and 2, since if

and if

If we call this solutionthensatisfies

We can rearrange this equation in various ways. Two such are

and

We can use these rearrangements as iteration formulae to attempt to findto 2 decimal places say. The iteration formulae are

and

Withthese formulae give respectively

Iterate

1.5

1.5

1.5753

1.447

1.683

1.409

1.839

1.383

2.071

1.364

2.436

1.351

3.070

1.342

4.421

1.335

9.011

1.330

90.485

1.327

The progress of the iterates is shown below in the two cases. In the graph below left, the gradient of the graph is greater than 1 and the graph crosses the linefrom below. The iterates diverge. In the graph below right the gradient of the graph is less than 1 and the graph crosses the line from above. The iterates converge to