Use Newton's Method to approximate the positive root of the function
f(x) = (x^5) - 20
Show each iteration until successive iterations agree to six decimal places.
To use Newton's Method to approximate the positive root of the function f(x) = (x^5) - 20, you will need to follow these steps:
1. Start by selecting an initial guess for the root. Let's say we start with x0 = 2.
2. Calculate the derivative of the function, f'(x), which in this case is f'(x) = 5(x^4).
3. Use the formula for Newton's Method to find the next approximation:
x1 = x0 - (f(x0) / f'(x0))
4. Iterate further by using the same formula, substituting x1 as x0 in the formula:
x2 = x1 - (f(x1) / f'(x1))
5. Repeat this process until the successive iterations agree to six decimal places. In other words, keep calculating xn+1 using xn until the difference between xn+1 and xn is less than or equal to 0.000001 (which is 1e-6, representing six decimal places of precision).
Now, let's perform the calculation:
Iteration 1:
x1 = x0 - (f(x0) / f'(x0))
x1 = 2 - ((2^5) - 20) / (5(2^4))
x1 = 2 - (-4) / 80
x1 = 2 + 0.05
x1 = 2.05
Iteration 2:
x2 = x1 - (f(x1) / f'(x1))
x2 = 2.05 - ((2.05^5) - 20) / (5(2.05^4))
x2 = 2.05 - (-0.1606) / 82.01
x2 = 2.05 + 0.001955
x2 = 2.051955
Iteration 3:
x3 = x2 - (f(x2) / f'(x2))
x3 = 2.051955 - ((2.051955^5) - 20) / (5(2.051955^4))
x3 = 2.051955 - (-0.0000032) / 82.31436
x3 = 2.051955 + 0.0000000391
x3 = 2.0519550391
Iteration 4:
x4 = x3 - (f(x3) / f'(x3))
x4 = 2.0519550391 - ((2.0519550391^5) - 20) / (5(2.0519550391^4))
x4 = 2.0519550391 - (0.0000000000027) / 82.80189706
x4 = 2.0519550391 + 0.000000000000033
x4 = 2.051955039133
Iteration 5:
x5 = x4 - (f(x4) / f'(x4))
x5 = 2.051955039133 - ((2.051955039133^5) - 20) / (5(2.051955039133^4))
x5 = 2.051955039133 - (0.0000000000000000009) / 83.186602
x5 = 2.051955039133 + 0.00000000000000000001
x5 = 2.051955039133
As you can see, in iteration 5, the value of x5 is the same as x4 to six decimal places. Therefore, the value 2.051955 approximates the positive root of the function (x^5) - 20 using Newton's Method.