The equation
has a solution somewhere between 1 and 2, since if![]()
![]()
and if![]()
![]()
If we call this solution
then
satisfies![]()
We can rearrange this equation in various ways. Two such are
and![]()
We can use these rearrangements as iteration formulae to attempt to find
to 2 decimal places say. The iteration formulae are
and![]()
With
these 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 line
from 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![]()
