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October 24, 2014

Posted by **kayla** on Tuesday, November 22, 2011 at 1:22am.

(543)^(1/5) as follows:

Let x1=2 be the initial approximation.

find x2 and x3 =? approximation

- typo? -
**MathMate**, Tuesday, November 22, 2011 at 7:21amThe question is not completely defined.

Newton's method is for solving an equation.

(543)^(1/5) is an expression that does not contain unknowns.

Please check for typo.

- calculus, math -
**Reiny**, Tuesday, November 22, 2011 at 9:09am543^(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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