Question:

For Newton's method....?

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How do you make a good initial guess? For example, for

x^4 - x - 1 = 0

what would be a good guess for x be and how do you know?

How about for x^3-3x^2+3x-1=0

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3 ANSWERS


  1. You dont have to have a good first guess.

    As long as the function isnt cyclic, discontinuous or piecewise, you really dont need to worry about accuracy.

    The worse your guess, the more iterations you will have to do. But so what? Newtons Method isnt 100% accurate unless the graph is linear or you do an infinite number of guesses anyway.

    Pick any guess... you will find at least one root eventually. And when you do, factor it out until you reduce the polynomial.

    ===

    There are modified versions of Newtons Method that converge faster. I forget their names and equations, but wikipedia article should link to them in the Newtons Method article. So if accuracy, iterations, time, efficiency... if those are your concerns, look up the other forms.

    ===

    There are easier ways of finding the correct root to 100% accuracy... that is, exact roots... for cubics and quartics without having to use Newtons Method.


  2. thats a tough one

  3. The easiest method is to graph the equation on a graphing

    calculator and estimate the zeros.

    If you haven't got one, then plug in some simple values

    for x, such as -2, -1, 0, 1, 2, and note where the function is

    positive and where it is negative. A zero occurs near where

    the function changes sign, from pos to neg, or neg to pos.

    So for x^4 - x - 1 = 0, let f(x) = x^4 - x -1

    For x = -2, f(x) = 17

    For x = -1, f(x) = 1

    For x = 0, f(x) = -1

    For x = 1, f(x) = -1

    For x = 2, f(x) = 13

    There is a change in the sign of f(x), for x between -1 and 0,

    and between x = 1 and 2.

    So there are zeros at -1 < x < 0 and 1 < x < 2.

    Good enough guesses would probably be -0.5 and 1.5

    Same goes for f(x) = x^3 - 3x^2 + 3x -1.

    For x = 0, f(x) = -1

    For x = 1, f(x) = 0 (well, we've already found the zero here)

    For x = 2, f(x) = 1

    So there's a change of sign in f(x) at 0 < x < 2,

    which we already know is at x = 1.

    Do the simple numbers first. If you don't get any joy, widen

    your choice of numbers, not only to larger numbers, but to

    smaller differences between your choices as well, in case

    there are two or more zeros quite close to each other.

    For example, you could choose two values, wide enough,

    where the graph is above the X-axis for both, but in reality,

    the graph may have dipped below the X-axis between these

    values, so you won't know unless you take more values

    in between these two. This can be very time-consuming,

    so a graphing calculator is the quicker way to go.

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