Newton's method of estimation - using derivatives

[Bruce3]

Homework Statement

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 x_n+1 = x_n – f(x_n)/f’(x_n). Under “good” conditions the sequence x_n 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.

Homework 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 x_n+1 = x_n – f(x_n)/f’(x_n). Under “good” conditions the sequence x_n 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
x_n+1 = x_n – (n^2-k)/(2n)
x_n+1 = x_n – n/2 – k/(2x)
x_n+1 – x_n + n/2 = – k/(2x)
(x_n+1 – x_n + n/2)*(-2n) = k

Any help on the work is greatly appreciated.

 

[Simon Bridge]

Any help on the work is greatly appreciated.

Cool! Um... what seems to be the problem?

...
f(x) = x^2 –k
f’(x) = 2x
x_n+1 = x_n – (n^2-k)/(2n)

... this is where you lose me: what is your reasoning here?