How do you find the first three iterate of the function $f(x)=4x-3$ for the given initial value $x_0=2$ ?

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Answer 1 · 2016-11-08T08:58:03

The value of $x_3=3/4$

The Newton's method is $x_(n+1)=x_n-f(x_n)/(f'(x_n))$
Here $f(x)=4x-3$ and $f'(x)=4$
And $x_0=2$
$:. x_1=x_0-f(x_0)/(f'(x_0))=2-5/4=3/4$

$x_2=x_1-f(x_1)/(f'(x_1))=3/4-0=3/4$

$x_3=x_2-f(x_2)/(f'(x_2))=3/4-0$
We obtain the same valuebecause $f(x)$ is a linear function and the intercept with the x-axis is $(3/4,0)$

Answer 2 · 2016-11-11T00:35:57

$x_1 = 5$ , $x_2 = 17$ , $x_3 = 65$

If I understand the question correctly, it is talking about a sequence defined by:

${ (x_0 = 2), (x_n = f(x_(n-1)) = 4x_(n-1)-3 " for " n >= 1) :}$

We find:

$x_0 = 2$

$x_1 = 4x_0-3 = 4(2)-3 = 8-3 = 5$

$x_2 = 4x_1-3 = 4(5)-3 = 20-3 = 17$

$x_3 = 4x_2-3 = 4(17)-3 = 68-3 = 65$

The formula for a general term of the sequence is:

$x_n = 4^n+1$

as can be proved by induction:

Base case:

$x_0 = 2 = 1 + 1 = 4^0 + 1$

Induction step:

$x_(n+1) = 4x_n - 3 = 4(4^n+1) - 3 = 4^(n+1) + 4 - 3 = 4^(n+1) + 1$

How do you find the first three iterate of the function f(x)=4x-3f(x)=4x-3 for the given initial value x_0=2x_0=2?