calculus, math

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Use Newton's method to approximate the value of

(543)^(1/5) as follows:
Let x1=2 be the initial approximation.
find x2 and x3 =? approximation

• typo? -

The question is not completely defined.
Newton's method is for solving an equation.
(543)^(1/5) is an expression that does not contain unknowns.

• calculus, math -

543^(1/5) is the solution to
x^5 = 543 or
x^5 - 543 = 0
let y = x^5 - 543

y' = 5x^4

newx = x - y/y' = x - (x^5 - 543)/(5x^4)
= (4x^5 + 543)/(5x^4)

x2 = 8.3875
x3 = 6.7319
x4 = 5.4384
x5 = 4.3507
x6 = 3.480576
x7 = 3.524449
x8 = 3.5233845
x9 = 3.523384

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