Newton’s method

Question

The graph for the equation y=x^3 +5 is below. Use Newton’s method with an initial guess of x=-2 to calculate x_3

Answer 1

I really enjoy this method’s derivation! Let meshare it with you and see if you get a better ie of what’s gowing on here.

First we need our function to be differentiable on the given interval and that our function say f(x) has some unique point in the interval let’s say c where f(c) = 0... Maybe even some other stuff like monotonicity of the function on the interval....Let’s assume we’re given this stuff.

Next, recall that the derivative of a function at a point say (x_n,f(x_n)) is the slope of the tangent line to the function f(x) at x=x_n.

Let’s define our tangent line as the function g(x), then use the point slope form of a linear equation to write g(x) for our tangent line of slope f’(x_n) and point (x_n, f(x_n)).

g(x) -f(x_n) = f’(x_n) (x - x_n)

Now evaluate this function with the point at which this function intersects the x-axis; i.e. where the tangent line of f(x) at (x_n,f(x_n)) aka g(x)= 0. Let’s call this point (x_n+1,0). Substituting this point into the equation yields

0 -f(x_n) = f’(x_n) (x_n+1 - x_n)

Solve this for x_n+1...

-f(x_n) = f’(x_n) x_n+1 - f’(x_n) x_n

⇒ f’(x_n) x_n - f(x_n) = f’(x_n) x_n+1

⇒ x_n+1 = x_n - f(x_n) / f’(x_n).

This is a recursive sequence where the limit of the sequence as n→∞ is c! I.e.

lim_n→∞ x_n+1 = c.

Recall that f(c) = 0, so the value of c is an x-intercept of f(x)!

Thus Newton’s Approximation takes in an initial guess x₀ or x₁ (depending on the indexing you prefer) of where the function is zero in the interval and uses the recursive definition

x_n+1 = x_n - f(x_n) / f’(x_n)

to make better and better approximations of that value of c where f(c) = 0.

So, your job outlined in the problem is to calculate x_n=3 in the recursive sequence defined above given x_n=0 = 2 for the function f(x) = x³ + 5.

So you’ll need to take the derivative of f(x) then it is a bunch of plug and chug work.... calculate x₀₊₁=x₁ using x₀ = 2 then use x₁ in the recursive definition to find a better approximation of where f(x)=0, i.e. x_2. Then again to find a better approximation x_3.