Use Newton's method with initial approximation x1 = 0 to find x2, x3, x4 to approximate the solution of x^3 - 3x = 1.

Question

Give your answers to 6 decimal places. Be sure to clearly state what function, derivative, and formula you are using.

Answer 1

Newton's Method is a linear estimation of a function, i.e., instead of thinking of the function as a curved line, you consider it as a straight line at points very close to your point in question. It starts with the point-slope form of a line:

y - y₁ = m(x - x₁) but uses a few special replacements.

1) We use the point (x₁, f(x₁) as the point (x₁, y₁), 2) We use the derivative at x = x₁ as the slope, and 3) We will be looking for the value of "x" that yields a y-value of zero.

That makes the point-slope form turn into: 0 - f(x₁) = f '(x₁)•(x - x₁) or, solving for "x", we get:

x = x₁ - f(x₁)/f '(x₁) where "x" is the next iteration. So, you could make it generic by making it:

x_n+1 = x_n - f(x_n)/f '(x_n)

At any rate, we first need a function instead of an equation. So, we turn x³ - 3x = 1 into 0 = x³ - 3x - 1 and then into:

f(x) = x³ - 3x - 1 in which we are looking for the "zero"

This makes f '(x) = 3x² - 3

We start with x₁ = x_n = 0, which means f(x_n) = f(x₁) = f(0) = -1 and f '(x_n) = f '(x₁) = f '(0) = -3

So x_n+1 = x_n - f(x_n)/f '(x_n) becomes x₂ = 0 - (-1)/(-3) = -1/3

So x₂ = -0.333333

Then we iterate.

For n = 2, the equation becomes x₃ = x₂ - f(x₂)/f '(x₂) with x₂ = -1/3, f(x₂) = -1/27, and f '(x₂) = -8/3

So x_n+1 = x_n - f(x_n)/f '(x_n) becomes x₃ = -1/3 - (-1/27)/(-8/3) = -25/72

So x₃ = -0.347222

For n = 3, the equation becomes x₄ = x₃ - f(x₃)/f '(x₃) with x₃ = -0.347222, f(x₃) = -0.000196, and f '(x₃) = -2.638310

So x_n+1 = x_n - f(x_n)/f '(x_n) becomes x₄ = -0.347222 - (-0.000196)/(-2.638310) = -0.347296

So x₄ = -0.347296