5. Recursion and Iteration
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Use the initial value as the input of a function rule. This will give us the first output value.
4.25, 29.5625, 936.0664
x_0&= 12 f(x)& = x^2+2x+3 To do so, we will use a table.
| n | x_n | f(x)=x^2+2x+3 | x_(n+1) |
|---|---|---|---|
| 0 | x_0= 1/2 | x_0^2+2 x_0+3=( 1/2)^2+2( 1/2)+3 | 4.25 |
| 1 | x_1= 4.25 | x_1^2+2 x_1+3=( 4.25)^2+2( 4.25)+3 | 29.5625 |
| 2 | x_2= 29.5625 | x_2^2+2 x_2+3=( 29.5625)^2+2( 29.5625)+3 | ≈ 936.0664 |
| 3 | x_3≈ 936.0664 | - | - |
Therefore, the first three iterates of the function are 4.25, 29.5625 and 936.0664.