Use Newton’s method to find the root of the equation x5 − x4 + 3x2 − 3x − 2 = 0 in the interval [1,2]

Question

. (a) Use Newton’s method to find the root of the equation x⁵ − x⁴ + 3x² − 3x − 2 = 0 in the interval [1,2] correct to six decimal places.

(b) Use Newton’s method to find the root of the equation 3x⁴ −8x³ + 2 = 0 in the interval [2,3] correct to six decimal places

Answer 1

(a)We create function: N(x) = x - f(x)/f'(x), where f(x) = x⁵ - x⁴ + 3x² - 3x - 2 and f'(x) = 5x⁴ - 4x³ + 6x - 3.

Thyen we put in TI-84 Y= this function and then 2nd/1-value to get table (x₁ = 1.5 from interval {1, 2}

x_n N(x_n)

1.5 1.3438596

1.3438596 1.3003202

1.3003202 1.2923958

1.2923958 1.2974195

1.2974195 1.2973834

1.2973834 1.2973834

So,x = 1.297383 with accuracy to 6 digits

(b) f(x) = 3x⁴ - 8x³ + 2; f'(x) = 12x³ - 24x²; N(x) = (3x⁴ - 8x³ + 2)/(12x³ - 24x²); x₁ = 2.5

x_n N(x_n)

1.5 2.655

2.655 2.6307249

2.6307249 2.6300208

2.6300208 2.6300203

2.6300203 2.6300203

So, x = 2.630020 with accuracy to 6 decimals

Answer 2

1) Start with 1.5

f(x) = x⁵ - x⁴ + 3x² - 3x - 2

f'(x) = 5x⁴ - 4x³ + 6x - 3

The Newton method is

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

x_n+1 = x_n - (x_n⁵ - x_n⁴ + 3x_n² - 3x_n - 2)/ (5x_n⁴ - 4x_n³ + 6x_n -3)

and our progression is 1.500000;1.343860;1.300320;1.297396;1.297383, with a final Newton step of 2.21E-10

2) Start with 2.5

f(x) = 3x⁴ - 8x³ + 2

f'(x) = 12x³ - 24x² + 2

x_n+1 = x_n - (3x_n⁴ - 8x_n³ + 2)/ (12x_n³ - 24x_n² + 2)

and out progression is

2.500000;2.655000;2.630725;2.630021;2.630020

The final Newton step is 4.0E-13