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Newton's method of estimation - using derivatives

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Newton devised the following method for approximating a real root of the equation f(x) = 0. i.e. a real number for which f(r) = 0. We begin by guessing an approximation, say x1, to the real root r.

    (i) Find the equation of the line tangent to the graph of y = f(x) at the point (x1,f(x1)). Assume f is differentiable and f’(x1) ≠ 0.

    (ii) Newton felt that the x-intercept of the tangent line in part(i) would be a better approximation to r than x1. Label the point of intersection of the tangent line in (i) with the x-axis (x2,0). Find x2 in terms of x1, f(x1), and f’(x1).

    (iii) The above procedure may now be repeated using the tangent line at (x2,f(x2)). If f’(x2) ≠ 0 this leads to a third approximation, x3, where (x3,0) is the point of intersection of the x-axis with the tangent line at (x2,f(x2)). Find x3 in terms of x2, f(x2), and f’(x2).

    If done correctly, you have just derived Newton’s method, which can be written as:
    To approximate the real root of r of f(x)=0, begin by guessing an initial approximation to r, say x1, For n = 1, 2, 3, … let xn+1 = xn – f(xn)/f’(xn). Under “good” conditions the sequence xn will converge to r.

    (iv) Approximate k^.5 for k > 0 by applying Newton’s method to f(x) = x^2 – k with the initial guess x1, You have just derived the method Heron used around 100 A.D. to approximate k^.5.

    2. Relevant equations
    To approximate the real root of r of f(x)=0, begin by guessing an initial approximation to r, say x1, For n = 1, 2, 3, … let xn+1 = xn – f(xn)/f’(xn). Under “good” conditions the sequence xn will converge to r.




    My work
    (i) for (x1, f(x1))
    y – f(x1) = f’(x1)(x-x1)
    y = f’(x1)(x-x1)+f(x1)
    (ii) for (x2,0)
    0=f’(x1)(x2-x1) + f(x1)
    -f(x1) = f’(x1)(x2-x1)
    x2 – x1 = -f(x1)/f’(x1)
    x2 = x1 – f(x1)/f’(x1)
    (iii) for (x2, f(x2))
    y – f(x2) = f’(x2)(x-x2)
    y = f’(x2)(x-x2)+f(x2)
    (ii) for (x3,0)
    0=f’(x2)(x3-x2) + f(x2)
    -f(x2) = f’(x2)(x3-x2)
    x3 – x2 = -f(x2)/f’(x2)
    x3 = x2 – f(x2)/f’(x2)
    (iv)
    f(x) = x^2 –k
    f’(x) = 2x
    xn+1 = xn – (n^2-k)/(2n)
    xn+1 = xn – n/2 – k/(2x)
    xn+1 – xn + n/2 = – k/(2x)
    (xn+1 – xn + n/2)*(-2n) = k

    Any help on the work is greatly appreciated.
     
    Last edited by a moderator: Dec 6, 2012
  2. jcsd
  3. Dec 7, 2012 #2

    Simon Bridge

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    Cool! Um... what seems to be the problem?
    ... this is where you lose me: what is your reasoning here?
     
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