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Calculus 1 Newton's Method

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Using Newton's method, approximate the root of the equation x^3+x+2=0 with the initial guess x1=-1 gives: x2=? and x3=? answers are not 0 or -1

  • Calculus 1 Newton's Method -

    let f(x) = y = x^3 + x +2
    dy/dx = 3x^2 + 1

    newton's formula

    x new = x old - f(xold/f '(xold)

    x2 = -1 - (-1-1+2)/(3+1) = 1-0 = -1

    since x1 = x1 exactly, x = -1 is a root

  • Calculus 1 Newton's Method -

    Check for typo's in the question.
    I believe there is only one real root for the given equation at -1.

    So your initial guesses have to be complex, such as 1+i. Convergence will depend on the form of the iteration equation.
    You can try
    and all kinds of other ones.
    The one that seems to converge best is
    first make
    divide by x and take the square root to get
    With a starting value of 1+i, ou should converge quite well, as the iterations alternate between the targeted root of (1/2)±sqrt(7)/2.

    Once you have converged to one, you can take the conjugate for the other, without having to do the same things all over again.

  • Calculus 1 Newton's Method-corr -

    the targeted root of (1/2)±sqrt(7)/2i.

  • Calculus 1 Newton's Method -

    My last line should say:

    since x2 = x1 exactly, x = -1 is a root

  • Calculus 1 Newton's Method-corr -

    I'll get it right this time!
    the targeted root of (1/2)±isqrt(7)/2.

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