Question

Consider the following Polynomial

Compute the root x=1 of the above function with actual error less than 0.00001 by:



1- Bisection Method .

2- Secant Method .

3- False Position Method .

4- Newton's Method .

5- Fixed point iterative Method ( use convergent re-arrangment)

Use the synthetic division to compute f(x) and it’s derivative​

Answer 1

Given:

f(x) = x⁴ - 3x³ - 10x² + x + 2

To find:

Find a root of an equation x⁴ - 3x³ - 10x² + x + 2 by:

1) Bisection Method

2) Secant Method

3) False Position Method

4) Newton's Method

5) Fixed point iterative Method

Solution:

1) Bisection Method

f(x) = x⁴ - 3x³ - 10x² + x + 2

x = 0, f(x) = 2

x = 1, f(x) = -9

1st Iteration:

Here f(0) = 2 > 0 and f(1) = -9 < 0

∴ Now, Root lies between 0 and 1

$x_0 = \frac{0+1}{2} = 0.5$

f(x₀) = f(0.5) = 0.5⁴ - 3(0.5)³ - 10(0.5)² + 0.5 + 2 = -0.3125 < 0

2nd Iteration:

Here f(0) = 2 > 0 and f(0.5) = -0.3125 < 0

∴ Now, Root lies between 0 and 0.5

$x_1 = \frac{0 + 0.5}{2} = 0.25$

f(x₁) = f(0.25) = 0.25⁴ - 3(0.25)³ - 10(0.25)² + 0.25 + 2 = 1.58203 > 0

3rd Iteration:

Here f(0.25) = 1.58203 > 0 and f(0.5) = -0.3125 < 0

∴ Now, Root lies between 0.25 and 0.5

$x_2 = \frac{0.25 + 0.5}{2} = 0.375$

f(x₂) = f(0.375) = 0.375⁴ - 3(0.375)³ - 10(0.375)² + 0.375 + 2 = 0.83032 > 0

Similarly we will do for more number of iterations.

For 19th Iteration:

Here f(0.46983) = 0.00003 > 0 and f(0.46983) = -0.00001 < 0

∴ Now, Root lies between 0.46983 and 0.46983

$x_1_8 = \frac{0.46983+0.46983}{2} = 0.46983$

f(x₁₈) = f(0.46983) = 0.00001 > 0

20th Iteration:

Here f(0.46983) = 0.00001 > 0 and f(0.46983) = -0.00001 < 0

∴ Now, Root lies between 0.46983 and 0.46983

$x_1_9 = \frac{0.46983+0.46983}{2} = 0.46983$

f(x19) = f(0.46983) = 0 < 0

Approximate root of the equation x⁴ - 3x³ - 10x² + x + 2 using Bisection method is 0.46983

2) Secant Method

x = 0, f(x) = 2

x = 1, f(x) = -9

1st iteration:

x₀ = 0 and x₁ = 1

f(x₀) = f(0) = 2 and f(x₁) = f(1) = -9

$x_2 = x_0 - f(x_0)\frac{x_1 - x_0}{f(x_1) -- f(x_0)}$

x₂ = 0.18182

∴f(x₂) = f(0.18182) = 1.8343

2nd iteration:

x₁ = 1 and x₂ = 0.18182

f(x₁) = f(1) = -9 and f(x₂) = f(0.18182) = 1.8343

$x_3 = x_1 - f(x_1)\frac{x_2 - x_1}{f(x_2) - f(x_1)}$

x₃ = 0.32034

∴f(x3) = f(0.32034) = 1.20607

Similarly we will do for more number of iterations.

At 7th Iteration:

x₆ = 0.46667 and x₇ = 0.46993

(x₆) = f(0.46667) = 0.03143 and f(x₇) = f(0.46993) = -0.00099

$x_8 = x_6 - f(x_6)\frac{x_7 - x_6}{f(x_7) - f(x_6)}$

x₈ = 0.46983

∴f(x₈) = f(0.46983) = 0

Approximate root of the equation x⁴ - 3x³ - 10x² + x + 2 using Secant method is 0.46983

Please ask one method at a time.