Use Newton’s Method with initial approximation x1 = 2 to solve the equation x5 +x= 6 . Go to x3,
and tell me how good you think the approximation is. Round all answers to 3 decimal places.
let f(x) = x^5+x-6
newton's method says
xn+1 = xn - f(xn)/f'(xn)
f'(x) = 5x^4 + 1
so x2 = x1 - f(x1)/f'(x1)
= 2 - 28/81
= 1.65432
then x3 = 1.65432 - .209237
= 1.44508
x4 = 1.3685
x5 = 1.35942
x6 = 1.359304105
x7 = 1.359304086
x8 = 1.359304086 Ok then !!!
To use Newton's Method, we need to follow these steps:
Step 1: Write the equation in the form f(x) = 0. In this case, the equation is x^5 + x = 6, so f(x) = x^5 + x - 6.
Step 2: Calculate the derivative of f(x). The derivative of f(x) is f'(x) = 5x^4 + 1.
Step 3: Choose an initial approximation value, x1. In this case, x1 = 2.
Step 4: Use the formula x(n+1) = x(n) - f(x(n))/f'(x(n)), where n is the iteration number. We will perform three iterations (n = 1, 2, 3) as you requested.
Now let's go through the iterations:
Iteration 1:
x(1) = x(0) - f(x(0))/f'(x(0))
Substituting x(0) = 2:
x(1) = 2 - (2^5 + 2 - 6)/(5(2^4) + 1)
x(1) = 2 - (32 + 2 - 6)/(5(16) + 1)
x(1) = 2 - 28/81
x(1) = 2 - 0.345
x(1) = 1.655
Iteration 2:
x(2) = x(1) - f(x(1))/f'(x(1))
Substituting x(1) = 1.655:
x(2) = 1.655 - (1.655^5 + 1.655 - 6)/(5(1.655^4) + 1)
x(2) = 1.655 - (7.466 + 1.655 - 6)/(5(5.504) + 1)
x(2) = 1.655 - 2.111/27.520
x(2) = 1.655 - 0.077
x(2) = 1.578
Iteration 3:
x(3) = x(2) - f(x(2))/f'(x(2))
Substituting x(2) = 1.578:
x(3) = 1.578 - (1.578^5 + 1.578 - 6)/(5(1.578^4) + 1)
x(3) = 1.578 - (5.196 + 1.578 - 6)/(5(3.868) + 1)
x(3) = 1.578 - 0.169/19.340
x(3) = 1.578 - 0.009
x(3) = 1.569
After three iterations, the approximation is x(3) = 1.569.
Now let's evaluate how good this approximation is:
We can assess the quality of the approximation by checking how close f(x(3)) is to zero.
Substituting x(3) = 1.569 into f(x):
f(1.569) = (1.569)^5 + 1.569 - 6
f(1.569) ≈ -0.256
Since f(1.569) is very close to zero, we can conclude that the approximation x(3) = 1.569 is quite good.