Question

Alculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. (Round your answers to three decimal places.)
f(x) = x5 − 5, x1 = 1.6

Answer 1

Answer:

$x_2=1.433$

$x_3=1.383$

Explanation:

Hello,

In this case, for the Newton-Raphson method, we use the following equation in order to show the iterations i:

$x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$

Hence the derivative of the given function is:

$f'(x)=5x^4$

In such a way, the first iteration:

$x_{1+1}=1.6-\frac{(1.6)^5-5}{5(1.6)^4}\\x_2=1.433$

Then the second iteration:

$x_{2+1}=1.433-\frac{(1.433)^5-5}{5(1.433)^4}\\x_3=1.383$

Best regards.

Answer 2

Answer:

$x_{1} = 1.6$, $x_{2} = 1.433$, $x_{3} = 1.384$

Explanation:

The expression for the approximation via Newton's Method has the following form:

$x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$

The function and its derivative are, respectively:

$f(x_{n}) = x^{5}-5$

$f'(x_{n})= 5\cdot x ^{4}$

After substituting the known variable, the Newton's expression is left as follows:

$x_{n+1} =x_{n}- \frac{x_{n}^{5}-5}{5\cdot x_{n}^{4}}$

The first two iterations are presented herein:

$x_{2} = 1.6 - \frac{1.6^{5}-5}{5\cdot (1.6)^{4}}$

$x_{2} = 1.433$

$x_{3} = 1.433 - \frac{1.433^{5}-5}{5\cdot (1.433)^{4}}$

$x_{3} = 1.384$