Use Newton’s method to find all roots (4 iterations each): (a) cos(x) =√x , (b) sin(x) =x^2−2

Answer 1

(a) For cos x - x^0.5 = 0, obtain d(cos x - x^0.5)/dx equal to (-sin x - 0.5x^-0.5).

Next, write x - [(cos x - x^0.5) ÷ (-sin x - 0.5x^-0.5)]. For the range of x in radians, 0 ≤ x ≤ 2π,

choose x=π/4 and compute a result for x - [(cos x - x^0.5) ÷ (-sin x - 0.5x^-0.5)]. Then use that

result as the next value of x for another calculation of x - [(cos x - x^0.5) ÷ (-sin x - 0.5x^-0.5)].

Feeding each succeeding result of x - [(cos x - x^0.5) ÷ (-sin x - 0.5x^-0.5)] back into

x - [(cos x - x^0.5) ÷ (-sin x - 0.5x^-0.5)] will finally give a result that equals the

last input value of x.

Repeated calculations of x - [(cos x - x^0.5) ÷ (-sin x - 0.5x^-0.5)] in a programmable calculator, starting at x=π/4 and using each new result as the succeeding value of x, will give x = 0.6417143709 (in radians) as a highly accurate solution for cos x - x^0.5 = 0.

(b) For sin x - (x² - 2) = 0, write d[sin x - (x² - 2)]/dx equal to cos x - 2x. Again, following the formula

x - f(x)/f'(x), obtain x - [sin x - (x² - 2)]/[cos x - 2x].

Repeated calculations of x - [sin x - (x² - 2)]/[cos x - 2x] in a programmable calculator, starting at x=π/4 and using each new result as the succeeding value of x , will give x = 1.728466319 (in radians) as one highly accurate solution for sin x - (x² - 2) = 0.

Inspection of a graph of sin x - (x² - 2) will show another x-intercept or negative "zero" root

for sin x - (x² - 2). Start with x = -π/4 and run succeeding results as the value of x in

x - [sin x - (x² - 2)]/[cos x - 2x] to obtain x = -1.061549775 (in radians) as a second highly

accurate solution for sin x - (x² - 2) = 0.

Answer 2

Newton's method approximates roots of functions using an iterative procedure. Typically, we want to approximate the root to some function, f(x) = 0. We make an initial guess at the root (x₀), and use the following formula to calculate the next term (x₁).

x₁ = x₀ - f(x₀)/f'(x₀), we can repeat this in general for as many iterations as we want until we converge on a solution.

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

In this problem, we do not have the format f(x) = 0, but instead we have g(x) = h(x). To get around this, we subtract one side to get 0 = g(x) - h(x). I will go through only part (a), as this process can be time consuming.

Our equation we are trying to solve is 0 = cos(x) - √x. In this case, we'll call f(x) = g(x) - h(x).

f(x) = cos(x) - √x

f'(x) = -sin(x) - 1/(2√x)

Next, we need to make an initial guess. It is helpful to graph cos(x) and √x on the same graph to see how many times they intersect and also a good initial guess. Lets try x₀ = 1 (if you graphed you will see it's clearly not correct but it's somewhat close).

f(x₀) = f(1) = cos(1) - √1 = -0.460

f'(x₀) = f'(1) = -sin(1) - 1/(2√1) = -1.341

x₁ = 1 - (-0.460/-1.341) = 0.657

Next, let's do the second iteration.

f(x₁) = f(0.657) = cos(0.657) - √0.657 = -0.019

f'(x₁) = f'(0.657) = -sin(0.657) - 1/(2√0.657) = -1.228

x₂ = 0.657 - (-0.019/-1.228) = 0.642

You would continue this process for two more iterations to find x₃ and x₄.

For part (b), it would look very similar to the above. One thing to watch out for is if you graph the functions in part (b) on the same graph, you will see there are two intersection points. Therefore, there will be two roots in this part and you will need to make approximate initial guesses for each root.