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

`x_2=1.433`

`x_3=1.383`

Step-by-step 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-(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-((1.6)^5-5)/(5(1.6)^4)\\x_2=1.433`

Then the second iteration:

`x_(2+1)=1.433-((1.433)^5-5)/(5(1.433)^4)\\x_3=1.383`

Best regards.

Answer 2

`x_(1) = 1.6`,
`x_(2) = 1.433`,
`x_(3) = 1.384`

Step-by-step explanation:

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

`x_(n+1) = x_(n) - (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)- (x_(n)^(5)-5)/(5\cdot x_(n)^(4))`

The first two iterations are presented herein:

`x_(2) = 1.6 - (1.6^(5)-5)/(5\cdot (1.6)^(4))`

`x_(2) = 1.433`

`x_(3) = 1.433 - (1.433^(5)-5)/(5\cdot (1.433)^(4))`

`x_(3) = 1.384`