use newtons method to approximate the indicated root

Question

Use Newton's method to approximate the indicated root of the following equation correct to six decimal places.

The root of 2x^3−6x^2+3x+1=0in the interval [2,3].

Answer 1

Let's neaten things up:

 

f(x) = 2x³ - 6x² + 3x + 1

 

f'(x) = 6x² - 12x + 3

 

We are iterating:

 

x_n+1 = x_n - (2x³ - 6x² + 3x + 1)/(6x² - 12x + 3)

 

Note that convergence is quadratic. That is, there is a positive constant C such that in the long run, the errors obey |e_n+1| ≤ Ce_n². Thus the number of digits correctly calculated roughly doubles each iteration:

 

x₀ = 2.5000000000000000000

x₁ = 2.2857142857142857143

x₂ = 2.2288242730720606827

x₃ = 2.2247650958354280934

x₄ = 2.2247448718925248504

x₅ = 2.2247448713915890494

x₆ = 2.2247448713915890491

 

Note that the cubic factors as (x - 1)(2x² - 4x - 1) and it's three roots are:

 

x = 1, x = (2 ± √6)/2

 

I think you can guess which one we computed.

 

Answer 2

Newton's method means take a starting point (since the problem says in the interval [2, 3], let's start at 2.5).

 

Then x n+1 = x n - f(x n)/f'(x n)

Of course, with this problem f'(x) = 6x^2 - 12x + 3.

 

Thus, we may write x n+1 = x n - (2x^3-6x^2+3x+1)/(6x^2 - 12x + 3)

 

x f(x) f'(x) step=-f(x)/f'(x)
2.5 2.25 10.5 -0.214285714
2.285714286 0.393586006 6.918367347 -0.056890013
2.228824273 0.024598836 6.060054565 -0.004059177
2.224765096 0.00012135 6.00029724 -2.02239E-05
2.224744872 3.00561E-09 6.000000007 -5.00936E-10