A suffix notation is a way of identifying different versions of the same variable. This is often used when repeating the same steps many times, a process called iteration.
In step 1 of an iteration, a variable might be shown as `x_1`, with the small 1 (the suffix) indicating that that is the value of `x` in step 1. If the value of `x` changes in step 2, then the new value is shown as `x_2`.
An initial value for `x` would be shown as `x_0`.
Some processes are recursive: that is, the answer from one step can be used as the input to the next step, with each step in the process taking you closer to the answer. Typically, the formula would be shown as `x_(n+1)=root(3)(20 - x_n^2 - x_n)`.
Carry out the iterative formula `x_(n+1)=root(3)(x_n^3 - x_n)` with a starting value of `x_0` for 5 iterations to obtain an approximate solution to the formula `x^3 + 2x = 30`.
Start with a value of `x_0 = 2`.
Answer:
| step 1 | 2 | `root(3)(30-2(2))` | 2.962496 |
| step 2 | 2.963496 | `root(3)(30-2(2.962496))` | 2.887501 |
| step 3 | 2.887501 | `root(3)(30-2(2.887501))` | 2.893485 |
| step 4 | 2.893485 | `root(3)(30-2(2.893485))` | 2.893009 |
| step 5 | 2.893009 | `root(3)(30-2(2.893009))` | 2.893046 |
By using an iterative method, or otherwise, solve the equation `x^3 + 2x = 40` to three decimal places.
Rearrange the formula
`x^3 + 2x = 40`
`x^3 = 40 - 2x`
`x = root(3)(40 - 2x)`
The iterative formula is `x_(n+1) = root(3)(40 - 2_n)`
| step 1 | 2 | `root(3)(40 - 2(2))` | 3.301927 | 3.302 (3dp) |
| step 2 | 3.301927 | `root(3)(40 - 2(3.301927))` | 3.220318 | 3.220 (3dp) |
| step 3 | 2.220318 | `root(3)(40 - 2(3.220318))` | 3.225556 | 3.226 (3dp) |
| step 4 | 3.225556 | `root(3)(40 - 2(3.225556))` | 3.225220 | 3.225 (3dp) |
| step 5 | 3.225220 | `root(3)(40 - 2(3.225220))` | 3.225242 | 3.225 (3dp) |
Answer: 3.225