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.
43, 3484, 24 259 093
x_0&=6 f(x)& = 2x^2-5x+1 To do so, we will use a table.
| n | x_n | f(x)=2x^2-5x+1 | x_(n+1) |
|---|---|---|---|
| 0 | x_0= 6 | 2 x_0^2-5 x_0+1=2( 6)^2-5( 6)+1 | 43 |
| 1 | x_1= 43 | 2 x_1^2-5 x_1+1=2( 43)^2-5( 43)+1 | 3484 |
| 2 | x_2= 3484 | 2 x_2^2-5 x_2+1=2( 3484)^2-5( 3484)+1 | 24 259 093 |
| 3 | x_3= 24 259 093 | - | - |
Therefore, the first three iterates of the function are 43, 3484 and 24 259 093.