Post a New Question

calculus, math

posted by on .

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.
    Please check for typo.

  • 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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question