Newton's Method to find xn

Question

I am not very familiar with Newton's method, could you please show the steps so that I can learn how to solve this problem? thank you

Answer 1

(a)

The function is g(x) = x² - cosx

Given X₁ = 1

According to newton Raphson Method 

g(1) = 1² - cos 1

g(1) = 1 - 0.9998

g(1) = 0.0002

g'(x) = 2x + sin x

g'(1) = 2(1) + sin 1

g'(1) = 2 + 0.01745

g'(1) = 2.01745

 

According to newton Raphson Method X₂ = 0.9999

 

Answer 2

(b)

The function is g(x) = X⁷ - 4

Given X₁ = 1

According to newton Raphson Method 

g'(x) = 7X⁶

^{Now solve for}  X₂

g(1) = (1)⁷ - 4 = 1 - 4 = -3

g'(1) = 7(1)⁶  = 7

X_{2 = 1 - (-3/7) =1.428}

 

^{Now solve for}  X₃

g(1.428) = (1.428)⁷ - 4 = 12.108 - 4 = 8.108

g'(1.428) = 7(1.428)⁶  = 59.35

X_{3 = 1.428 - (8.108/59.35) =1.291}

 

^{Now solve for}  X₄

g(1.291) = (1.291)⁷ - 4 = 5.977 - 4 =1.977

g'(1.291) = 7(1.291)⁶  = 32.408

X_{4  = 1.291 - (1.977/32.408) =1.229}

 

According to newton Raphson Method the value of X₄ = 1.229