Use Newton's method to approximate a root of the equation

Question

2x^7+7x^4+2 =0 as follows. Let x_1 = 1 be the initial approximation. x_2 x_3?

Answer 1

a)

The function is 2x⁷ + 7x⁴+ 2 = 0

Let  f(x) = 2x⁷ + 7x⁴+ 2

Given x₁ = 1

Find x₂ = ?  and  x₃ = ?

f '(x) = 2*7x^7-1 + 7*4x^4-1

f '(x) = 14x⁶ + 28x³

According to Newton Method 

x₂ = x₁ – [ f(x₁) / f’(x₁) ]

f(x₁) = 2(x₁)⁷ + 7(x₁)⁴+ 2

f(x₁) = 2(1)⁷ + 7(1)⁴+ 2

=  2 + 7 + 2

= 11

f '(x₁) = 14(x₁)⁶ + 28(x₁)³

= 14(1)⁶ + 28(1)³

= 14 + 28

= 42

x₂ = x₁ – [ f(x₁) / f’(x₁) ]

x₂ = 1 – [ 11 / 42 ]

x₂ = 1 – 0.2619

x₂ = 0.738

Solution :

x₂ = 0.738

Answer 2

b)

The function is 2x⁷ + 7x⁴+ 2 = 0

Let  f(x) = 2x⁷ + 7x⁴+ 2

Given x₁ = 1

Find x₂ = ?  and  x₃ = ?

f '(x) = 2*7x^7-1 + 7*4x^4-1

f '(x) = 14x⁶ + 28x³

According to Newton Method 

x₂ = x₁ – [ f(x₁) / f’(x₁) ]

f(x₁) = 2(x₁)⁷ + 7(x₁)⁴+ 2

f(x₁) = 2(1)⁷ + 7(1)⁴+ 2 =  2 + 7 + 2 = 11

f '(x₁) = 14(x₁)⁶ + 28(x₁)³ = 14(1)⁶ + 28(1)³ = 14 + 28 = 42

x₂ = x₁ – [ f(x₁) / f’(x₁) ]

x₂ = 1 – [ 11 / 42 ]

x₂ = 1 – 0.2619

x₂ = 0.738

x₃ = x₂ – [ f(x₂) / f’(x₂) ]

f(x₂) = 2(x₂)⁷ + 7(x₂)⁴+ 2

f(x₂) = 2(0.738)⁷ + 7(0.738)⁴+ 2

=  0.238 + 2.076 + 2

= 4.314

f '(x₂) = 14(x₂)⁶ + 28(x₂₁)³

= 14(0.738)⁶ + 28(0.738)³

= 2.26 + 11.25

= 13.51

x₃ = x₃ – [ f(x₃) / f’(x₂) ]

x₃ = 1 – [ 4.314 / 13.51 ]

x₃ = 1 – 0.319

x₃ = 0.681

Solution :

x₃ = 0.681