Use Newton's method to approximate a root of the equation 4sin(x)=x as follows

Question

Use Newton's method to approximate a root of the equation 4sin(x)=x as follows. 

Let x1=1 be the initial approximation.

Answer 1

Newton's Method is a way of approximating a zero. Notice if you take 4sin(x) = x and subtract "x" from both sides, you get 4sin(x) - x = 0 and this very much looks like what you would get if you had the function f(x) = 4sin(x) - x and you set it equal to zero to find the x-intercepts. So, for us to use Newton's method, we consider f(x) = 4sin(x) - x.

Newton's Method uses a straight line to estimate a function value close by a given value of "x" and iterates. First you select a starting point and calculate a "new" value of "x" that is then used in your calculation that produces a newer value of "x" and so forth until the value of "x" doesn't change anymore.

Since Newton's Method uses a straight line, we need a slope. The slope that is used is the derivative. The equation that is used is a variant of the point-slope form and looks like this:

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

Meaning that we use our initial value of "x" (which in this case is x = 1) to find a new value of "x" and we keep repeating.

Since this problem uses f(x) = 4sin(x) - x, we also need the derivative which is:

f '(x) = 4cos(x) - 1

So, to start with, we do:

x₂ = x₁ - f(x₁)/f '(x₁)

where x₁ = 1 meaning f(x₁) = f(1) = 4sin(1) - 1 = 2.365883939 and f '(x₁) - f '(1) = 4cos(1) - 1 = 1.161209223

So x₂ = -1.037431233

Then we calculate x3 using:

x₃ = x₂ - f(x₂)/f '(x₂)

or x₃ = 1.29....

Then we calculate x4 using:

x₄ = x₃ - f(x₃)/f '(x₃)

or x₄ = -23.10...

And you keep going:

x₅ = -13.47. . .

x₆ = -20.52. . .

x₇ = -8.69. . .

x₈ = -7.17. . .

x₉ = -9.84. . .

x₁₀ = -7.38. . .

x₁₁ = -11.99. . .

x₁₂ = -17.98. . .

x₁₃ = -31.16. . .

x₁₄ = -42.36. . .

x₁₅ = -4.22. . .

x₁₆ = -1.55. . .

x₁₇ = -4.23. . .

x₁₈ = -1.52. . .

x₁₉ = -4.65. . .

x₂₀ = 2.33. . .

x₂₁ = 2.48. . .

x₂₂ = 2.474593

x₂₃ = 2.474577

x₂₄ = 2.474577

It takes a LONG time to get to values that don't change anymore using x₁ = 1

Answer 2

The equation for Newton's method is x₂=x₁-(4sin x₁-x₁)/cos x₁

Use x₁=1.

You make the substitution.

Then your answer becomes x₁ repeatedly until you get the result to the desired accuracy.

Of course x=0 is a root and DESMOS approximates the root you want as 2.47458.