In the class we discussed Newton\'s Method for the solution os f(x) = 0. It\'s d
ID: 3288373 • Letter: I
Question
In the class we discussed Newton's Method for the solution os f(x) = 0. It's described in section 3.7. The reason it shows up here is that it leads to a (convergent or divergent) sequence. Newton's Method is given by x0 given, xn+1 = g(xn) = xn - f(xn) / f'(xn). Suppose f(x) = x2 - 2. Then g(x) = 0. If x0 = 1 the limit of the sequence x0, x1, x2, ... defined by Newton's method in this case is z = 0. If x0 = -1 the limit of the sequence x0, x1, x2, ... defined by Newton's method in this case is z = 0.Explanation / Answer
(1)
g(x) = x - f(x)/f'(x)
g(x) = x - (x^2 - 2)/(2x)
g(x) = 2x^2/(2x) - (x^2 - 2)/(2x)
g(x) = (x^2 + 2)/(2x)
(2)
x0 = 1
x1 = g(1) = 1.5
x2 = g(1.5) = 1.41667
x3 = g(1.41667) = 1.41422
x4 = g(1.41422) = 1.41421
The limit is approximately 1.414.
(3)
x0 = -1
x1 = g(-1) = -1.5
x2 = g(-1.5) = -1.41667
x3 = g(-1.41667) = -1.41422
x4 = g(-1.41422) = -1.41421
The limit is approximately -1.414.
(4)
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