# Homework Help: Newton's method of estimation - using derivatives

1. Dec 6, 2012

### Bruce3

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. Dec 7, 2012

### Simon Bridge

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