1)Using Newton's Method with A=5, approximate the square root of 27. Do 2 iterations.
2)Using Newton's Method with A=2, approximate the cube root of 9. Do 2 iterations.
visit
http://keisan.casio.com/exec/system/1244946907
and you can use the function
f(x) = x^2-27
f'(x) = 2x
to approximate √27
To approximate the square root of a number using Newton's Method, you need to follow these steps:
1) Choose a starting point, which is an initial guess for the square root.
2) Set up the iteration formula, which is x(n+1) = (x(n) + A/x(n))/2. Here, A represents the number you want to find the square root of, and x(n) represents the current approximation.
3) Apply the iteration formula twice to improve the approximation.
Let's apply these steps to the first question:
1) Choose a starting point: Let's choose 5 as the starting point.
2) Set up the iteration formula: x(n+1) = (x(n) + 27/x(n))/2
3) Apply the iteration formula twice:
1st iteration:
x(0) = 5
x(1) = (5 + 27/5)/2 = 4.7
2nd iteration:
x(1) = 4.7
x(2) = (4.7 + 27/4.7)/2 = 5.038
Therefore, after 2 iterations, the approximate square root of 27 using Newton's Method with A = 5 is 5.038.
The process to approximate the cube root using Newton's Method is similar. Here are the steps:
1) Choose a starting point, which is an initial guess for the cube root.
2) Set up the iteration formula, which is x(n+1) = (2x(n) + A/(x(n)^2))/3. Here, A represents the number you want to find the cube root of, and x(n) represents the current approximation.
3) Apply the iteration formula twice to improve the approximation.
Let's apply these steps to the second question:
1) Choose a starting point: Let's choose 2 as the starting point.
2) Set up the iteration formula: x(n+1) = (2x(n) + 9/(x(n)^2))/3
3) Apply the iteration formula twice:
1st iteration:
x(0) = 2
x(1) = (2*2 + 9/(2^2))/3 = 2.259
2nd iteration:
x(1) = 2.259
x(2) = (2*2.259 + 9/(2.259^2))/3 = 2.080
After 2 iterations, the approximate cube root of 9 using Newton's Method with A = 2 is 2.080.