1. Use the Newton-Raphson method to find an approximate root of the equation x^3 - x - 1 = 0 starting from an initial guess x0 = 1.5. Perform three iterations and express each step clearly. 2. Use the Secant method to estimate the root of the equation cos(x) - x = 0 using initial guesses x0 = 0.5 and x1 = 1.0. Perform three iterations and show all the calculations involved in each step.
Solution:
Question 1: Newton-Raphson Method
We need to solve x 3 − x − 1 = 0 x^3 - x - 1 = 0 x 3 − x − 1 = 0 x 0 = 1.5 x_0 = 1.5 x 0 = 1.5
Formula :
x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n ) f ( x ) = x 3 − x − 1 f(x) = x^3 - x - 1 f ( x ) = x 3 − x − 1 f ′ ( x ) = 3 x 2 − 1 f'(x) = 3x^2 - 1 f ′ ( x ) = 3 x 2 − 1
Steps :
Iteration 1 (x 0 = 1.5 x_0 = 1.5 x 0 = 1.5 f ( 1.5 ) = ( 1.5 ) 3 − 1.5 − 1 = 1.875 − 1.5 − 1 = − 0.625 f(1.5) = (1.5)^3 - 1.5 - 1 = 1.875 - 1.5 - 1 = -0.625 f ( 1.5 ) = ( 1.5 ) 3 − 1.5 − 1 = 1.875 − 1.5 − 1 = − 0.625 f ′ ( 1.5 ) = 3 ( 1.5 ) 2 − 1 = 3 ( 2.25 ) − 1 = 6.75 − 1 = 5.75 f'(1.5) = 3(1.5)^2 - 1 = 3(2.25) - 1 = 6.75 - 1 = 5.75 f ′ ( 1.5 ) = 3 ( 1.5 ) 2 − 1 = 3 ( 2.25 ) − 1 = 6.75 − 1 = 5.75 x 1 = 1.5 − − 0.625 5.75 = 1.5 + 0.1087 = 1.6087 x_1 = 1.5 - \frac{-0.625}{5.75} = 1.5 + 0.1087 = 1.6087 x 1 = 1.5 − 5.75 − 0.625 = 1.5 + 0.1087 = 1.6087
Iteration 2 (x 1 = 1.6087 x_1 = 1.6087 x 1 = 1.6087 f ( 1.6087 ) = ( 1.6087 ) 3 − 1.6087 − 1 = 4.163 − 1.6087 − 1 = 0.045 f(1.6087) = (1.6087)^3 - 1.6087 - 1 = 4.163 - 1.6087 - 1 = 0.045 f ( 1.6087 ) = ( 1.6087 ) 3 − 1.6087 − 1 = 4.163 − 1.6087 − 1 = 0.045 f ′ ( 1.6087 ) = 3 ( 1.6087 ) 2 − 1 = 3 ( 2.588 ) − 1 = 7.764 − 1 = 6.764 f'(1.6087) = 3(1.6087)^2 - 1 = 3(2.588) - 1 = 7.764 - 1 = 6.764 f ′ ( 1.6087 ) = 3 ( 1.6087 ) 2 − 1 = 3 ( 2.588 ) − 1 = 7.764 − 1 = 6.764 x 2 = 1.6087 − 0.045 6.764 = 1.6087 − 0.00665 = 1.6020 x_2 = 1.6087 - \frac{0.045}{6.764} = 1.6087 - 0.00665 = 1.6020 x 2 = 1.6087 − 6.764 0.045 = 1.6087 − 0.00665 = 1.6020
Iteration 3 (x 2 = 1.6020 x_2 = 1.6020 x 2 = 1.6020 f ( 1.6020 ) = ( 1.6020 ) 3 − 1.6020 − 1 = 4.114 − 1.602 − 1 = 0.0005 f(1.6020) = (1.6020)^3 - 1.6020 - 1 = 4.114 - 1.602 - 1 = 0.0005 f ( 1.6020 ) = ( 1.6020 ) 3 − 1.6020 − 1 = 4.114 − 1.602 − 1 = 0.0005 f ′ ( 1.6020 ) = 3 ( 1.6020 ) 2 − 1 = 3 ( 2.566 ) − 1 = 7.698 − 1 = 6.698 f'(1.6020) = 3(1.6020)^2 - 1 = 3(2.566) - 1 = 7.698 - 1 = 6.698 f ′ ( 1.6020 ) = 3 ( 1.6020 ) 2 − 1 = 3 ( 2.566 ) − 1 = 7.698 − 1 = 6.698 x 3 = 1.6020 − 0.0005 6.698 = 1.6020 − 0.000075 = 1.6019 x_3 = 1.6020 - \frac{0.0005}{6.698} = 1.6020 - 0.000075 = 1.6019 x 3 = 1.6020 − 6.698 0.0005 = 1.6020 − 0.000075 = 1.6019
Approximate Root after 3 Iterations: x ≈ 1.6019 x \approx 1.6019 x ≈ 1.6019 .
Question 2: Secant Method
We need to solve cos ( x ) − x = 0 \cos(x) - x = 0 cos ( x ) − x = 0 x 0 = 0.5 x_0 = 0.5 x 0 = 0.5 x 1 = 1.0 x_1 = 1.0 x 1 = 1.0
Formula :
x n + 1 = x n − f ( x n ) ( x n − x n − 1 ) f ( x n ) − f ( x n − 1 ) x_{n+1} = x_n - \frac{f(x_n)(x_n - x_{n-1})}{f(x_n) - f(x_{n-1})} x n + 1 = x n − f ( x n ) − f ( x n − 1 ) f ( x n ) ( x n − x n − 1 ) f ( x ) = cos ( x ) − x f(x) = \cos(x) - x f ( x ) = cos ( x ) − x
Steps :
Iteration 1 (x 0 = 0.5 , x 1 = 1.0 x_0 = 0.5, x_1 = 1.0 x 0 = 0.5 , x 1 = 1.0 f ( 0.5 ) = cos ( 0.5 ) − 0.5 = 0.8776 − 0.5 = 0.3776 f(0.5) = \cos(0.5) - 0.5 = 0.8776 - 0.5 = 0.3776 f ( 0.5 ) = cos ( 0.5 ) − 0.5 = 0.8776 − 0.5 = 0.3776 f ( 1.0 ) = cos ( 1.0 ) − 1.0 = 0.5403 − 1.0 = − 0.4597 f(1.0) = \cos(1.0) - 1.0 = 0.5403 - 1.0 = -0.4597 f ( 1.0 ) = cos ( 1.0 ) − 1.0 = 0.5403 − 1.0 = − 0.4597 x 2 = 1.0 − − 0.4597 ( 1.0 − 0.5 ) − 0.4597 − 0.3776 = 1.0 − − 0.22985 − 0.8373 = 1.0 − 0.2745 = 0.7255 x_2 = 1.0 - \frac{-0.4597(1.0 - 0.5)}{-0.4597 - 0.3776} = 1.0 - \frac{-0.22985}{-0.8373} = 1.0 - 0.2745 = 0.7255 x 2 = 1.0 − − 0.4597 − 0.3776 − 0.4597 ( 1.0 − 0.5 ) = 1.0 − − 0.8373 − 0.22985 = 1.0 − 0.2745 = 0.7255
Iteration 2 (x 1 = 1.0 , x 2 = 0.7255 x_1 = 1.0, x_2 = 0.7255 x 1 = 1.0 , x 2 = 0.7255 f ( 0.7255 ) = cos ( 0.7255 ) − 0.7255 = 0.7441 − 0.7255 = 0.0186 f(0.7255) = \cos(0.7255) - 0.7255 = 0.7441 - 0.7255 = 0.0186 f ( 0.7255 ) = cos ( 0.7255 ) − 0.7255 = 0.7441 − 0.7255 = 0.0186 f ( 1.0 ) = − 0.4597 f(1.0) = -0.4597 f ( 1.0 ) = − 0.4597 x 3 = 0.7255 − 0.0186 ( 0.7255 − 1.0 ) 0.0186 − ( − 0.4597 ) = 0.7255 − 0.0186 ( − 0.2745 ) 0.4783 = 0.7255 + 0.0107 = 0.7362 x_3 = 0.7255 - \frac{0.0186(0.7255 - 1.0)}{0.0186 - (-0.4597)} = 0.7255 - \frac{0.0186(-0.2745)}{0.4783} = 0.7255 + 0.0107 = 0.7362 x 3 = 0.7255 − 0.0186 − ( − 0.4597 ) 0.0186 ( 0.7255 − 1.0 ) = 0.7255 − 0.4783 0.0186 ( − 0.2745 ) = 0.7255 + 0.0107 = 0.7362
Iteration 3 (x 2 = 0.7255 , x 3 = 0.7362 x_2 = 0.7255, x_3 = 0.7362 x 2 = 0.7255 , x 3 = 0.7362 f ( 0.7362 ) = cos ( 0.7362 ) − 0.7362 = 0.7391 − 0.7362 = 0.0029 f(0.7362) = \cos(0.7362) - 0.7362 = 0.7391 - 0.7362 = 0.0029 f ( 0.7362 ) = cos ( 0.7362 ) − 0.7362 = 0.7391 − 0.7362 = 0.0029 f ( 0.7255 ) = 0.0186 f(0.7255) = 0.0186 f ( 0.7255 ) = 0.0186 x 4 = 0.7362 − 0.0029 ( 0.7362 − 0.7255 ) 0.0029 − 0.0186 = 0.7362 − 0.0029 ( 0.0107 ) − 0.0157 = 0.7362 + 0.0020 = 0.7382 x_4 = 0.7362 - \frac{0.0029(0.7362 - 0.7255)}{0.0029 - 0.0186} = 0.7362 - \frac{0.0029(0.0107)}{-0.0157} = 0.7362 + 0.0020 = 0.7382 x 4 = 0.7362 − 0.0029 − 0.0186 0.0029 ( 0.7362 − 0.7255 ) = 0.7362 − − 0.0157 0.0029 ( 0.0107 ) = 0.7362 + 0.0020 = 0.7382
Approximate Root after 3 Iterations: x ≈ 0.7382 x \approx 0.7382 x ≈ 0.7382 .
Next Steps
Would you like detailed explanations or further clarifications on any of these methods?
Expanded Questions:
What is the convergence criteria for the Newton-Raphson and Secant methods?
How can we interpret the errors in each iteration for these methods?
What are the limitations of the Newton-Raphson method?
How does the choice of initial guesses affect the Secant method?
Can these methods handle discontinuous functions effectively?
Tip:
The Newton-Raphson method converges faster but requires the derivative, while the Secant method is slower but doesn't need the derivative, making it more versatile for some applications.