This time consider hx2x22x13 Find the exact root by using th

This time, consider h(x)=2x^2-2x+13 Find the exact root by using the quadratic formula. Use x0=1 and perform 2 Newton iterations. Sketch and label your iterates on the complex plane. Does it look like the process will converge? Use z=i, and perform 2 Newton iterations. Simplify each iteration to a+bi form.) Does it look like the process will converge? Use z0=1+i, and perform 2 Newton iterations. Simplify each iteration to a+bi form.) Does it look like the process will converge?

Solution

The given function is quadratic and we can easily find its zeros using the quadratic formulas. However we start with this example in order to be able to compare the zero found using Newton\'s method with the one using the quadratic formulas.

How to use Newton\'s method to find the largest zero of f?

First we need to find a close approximation to the zero. This can be done graphically. The graph of f below clearly shows that f has two zeros, both of them negative and the largest one is closer to zero. We can use zero as a starting value in the Newton\'s method procedure

We now calculate the first derivative f \'

f \'(x) = 2x - 2

Now we start the procedure as follows:

Let x₀ = 0 . This is the starting approximate value to the largest zero. You might decide to take another value as long as it is close to the zero you are approximating.

We now calculate x₁ using the above procedure for n = 0 as follows:

x₁ = x₀ - f(x₀) / f \'(x₀)

Substitute x₀ by its value 0 and calculate x₁

x₁ = 0 - f(0) / f \'(0)

= - ((0) ² + (0)x + 13) / (2(0) - 2)

= -13/2

We now calculate x₂ as follows

x₂ = x₁ - f(x₁) / f \'(x₁)

x₂ = -1/3 - f(-1/3) / f \'(-1/3) -0.38095238

We now calculate x₃ as follows

x₃ = x₂ - f(x₂) / f \'(x₂)

x₃ -0.38196555

Continue with the procedure to find

x₄ -0.38196601

x₅ -0.38196601

Since x₅ x₄ there is no need to continue as we will not be able to make any more progress in approximating the zero.

We now compare x₅ to the exact value of the largest zero which is ( -3 + sqrt (5)) / 2 -0.38196601 and we can say that they are equal up to 8 decimal places. Another way to check the accuracy of our approximation is to compute

f(x₅) 2.8 10^-9

Since f(x₅) is very close to zero, x₅ -0.38196601 is a good approximate value to the zero of f.