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

Nov 8, 2016

The value of ${x}_{3} = \frac{3}{4}$

#### Explanation:

The Newton's method is ${x}_{n + 1} = {x}_{n} - f \frac{{x}_{n}}{f ' \left({x}_{n}\right)}$
Here $f \left(x\right) = 4 x - 3$ and $f ' \left(x\right) = 4$
And ${x}_{0} = 2$
$\therefore {x}_{1} = {x}_{0} - f \frac{{x}_{0}}{f ' \left({x}_{0}\right)} = 2 - \frac{5}{4} = \frac{3}{4}$

${x}_{2} = {x}_{1} - f \frac{{x}_{1}}{f ' \left({x}_{1}\right)} = \frac{3}{4} - 0 = \frac{3}{4}$

${x}_{3} = {x}_{2} - f \frac{{x}_{2}}{f ' \left({x}_{2}\right)} = \frac{3}{4} - 0$
We obtain the same valuebecause $f \left(x\right)$ is a linear function and the intercept with the x-axis is $\left(\frac{3}{4} , 0\right)$

Nov 11, 2016

${x}_{1} = 5$, ${x}_{2} = 17$, ${x}_{3} = 65$

#### Explanation:

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

$\left\{\begin{matrix}{x}_{0} = 2 \\ {x}_{n} = f \left({x}_{n - 1}\right) = 4 {x}_{n - 1} - 3 \text{ for } n \ge 1\end{matrix}\right.$

We find:

${x}_{0} = 2$

${x}_{1} = 4 {x}_{0} - 3 = 4 \left(2\right) - 3 = 8 - 3 = 5$

${x}_{2} = 4 {x}_{1} - 3 = 4 \left(5\right) - 3 = 20 - 3 = 17$

${x}_{3} = 4 {x}_{2} - 3 = 4 \left(17\right) - 3 = 68 - 3 = 65$

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:

${x}_{n + 1} = 4 {x}_{n} - 3 = 4 \left({4}^{n} + 1\right) - 3 = {4}^{n + 1} + 4 - 3 = {4}^{n + 1} + 1$