Use the Newton's Method to approximate the real root of the equation:
f(x)=x-2+cosx=0
a) What is the iterative equation of Newton's method of the given equation?
b) Iterate the equation with starting point x1=5 until you get a repetition of the four digits after the decimal point.
well, f'(x) = 1-sinx
So plug that into Newton's method. Here is a nice web site:
http://keisan.casio.com/exec/system/1244946907
a) The iterative equation of Newton's method can be obtained by finding the derivative of the given function, f(x), and using it to update the value of x in each iteration. Here's how to derive the iterative equation:
1. Start with the function f(x)=x-2+cos(x)=0
2. Take the derivative of f(x) with respect to x:
f'(x) = 1 - sin(x)
3. Apply Newton's method iterative equation:
x_(n+1) = x_n - (f(x_n) / f'(x_n))
b) To iterate the equation using Newton's method with a starting point x1=5 until we get a repetition of the four digits after the decimal point, follow these steps:
1. Calculate f(x1):
f(x1) = x1 - 2 + cos(x1)
2. Calculate f'(x1):
f'(x1) = 1 - sin(x1)
3. Calculate x2 using the iterative equation:
x2 = x1 - (f(x1) / f'(x1))
4. Calculate f(x2):
f(x2) = x2 - 2 + cos(x2)
5. Repeat steps 3 and 4 until the four digits after the decimal point become repetitive. Keep track of the values of x in each iteration.
For example, let's perform the iterations:
Iteration 1:
x1 = 5
f(x1) = 5 - 2 + cos(5) = 2.283662185
f'(x1) = 1 - sin(5) = -0.958924275
x2 = 5 - (2.283662185 / -0.958924275) = 6.378369527
f(x2) = 6.378369527 - 2 + cos(6.378369527) = -0.012825134
Iteration 2:
x1 = 6.378369527
f(x1) = 6.378369527 - 2 + cos(6.378369527) = -0.012825134
f'(x1) = 1 - sin(6.378369527) = -0.183067674
x2 = 6.378369527 - (-0.012825134 / -0.183067674) = 6.505539201
f(x2) = 6.505539201 - 2 + cos(6.505539201) = 0.000007176
Iteration 3:
x1 = 6.505539201
f(x1) = 6.505539201 - 2 + cos(6.505539201) = 0.000007176
f'(x1) = 1 - sin(6.505539201) = -0.141979571
x2 = 6.505539201 - (0.000007176 / -0.141979571) = 6.505586157
f(x2) = 6.505586157 - 2 + cos(6.505586157) = -5.33485E-11
At this point, we can see that the four digits after the decimal point have repeated, indicating convergence. The approximate real root of the equation f(x)=0 using Newton's method is x = 6.5056.