Question

find a positive root 17^1/3 correct to 4 decimal places by Newton Raphson Method take x0= 2.5​

Answer 1

The Required root is 2.5712 correct to four Decimal Places

Given,

positive root= 17^1/3

$x_{0}$ = 2.5

To Find,

a positive root 17^1/3 correct to 4 decimal places by using Newton Raphson method.

Assuming f(x)= x^3 - 17

The iterative formula is  $x_{n+1}$ = 1 / 3($2x_{n}$+ a / x$x^{2} _{n}$)

Here a= 17

$x_{0}$= 2.5   because ∛8 =2 and ∛27=3

$x_{1}$= 1 / 3(2*2.5+ 17/ (2.5)^2) = 1 / 3(5+17/6.25)= 2.5733

$x_{2}$= 1 / 3(2*2.5733+ 17 /(2.5733)^2= 1 /3 (5.1466+17 / 6.6220)=2.5713

$x_{3}$= 1 / 3(2*2.5713+17 / (2.5713)^2 = 1 / 3(5.1426+17/6.61158)=2.57128

$x_{4}$=1/3(2*2.57128+17 / (2.57128)^2=1 / 3(5.14256+17/6.61148)=2.57128

Thus ,the Required root is 2.5712 correct to four Decimal Places

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