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
    f(x)=(-2-x)^(1/3)
    f(x)=-2-x^3
    and all kinds of other ones.
    The one that seems to converge best is
    first make
    x^3=-2-x
    divide by x and take the square root to get
    f(x)=sqrt(-1-2/x)
    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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