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Calculus/Newton-Raphson method

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Question
Dear Scotto,

I have this problem in Newton Raphson method and would like your help on how to solve it so as to understand some things better.

The problem aks us to demonstrate that the Newton-Raphson method iteration function for a(x) = x^(-1) - A, where A > 0, is
φ(x) = x(2 - Ax).
In addition, we have to check that X = A−1 is a fixed point of φ(x).( Χ is the solution to the function)

Answer
The Newton-Raphson method is x(n+1) = xn - a(xn)/a'(xn) where xn is on term in the series
and x(n+1) is the next term.

In this problem, we are give a(x) = 1/x - A.  This says a'(x) = -1/x².

Taking xn as just x, the method says that the next term will be x - a(x)/a'(x),
and that is x - (1/x - A)/(-1/x²).  Inverting and multiplying gives x + x - Ax².

Combining terms gives us 2x - Ax².

Factoring out x gives us .... the answer.


If X = A-1, the a(x) = 1/(A-1) - A = (1 + A(A-1))/(A-1) = (A² - A + 1)/(A-1).
Looking at φ(x) for x=A-1, we get
φ(x) = (A-1)(2-A(A-1))
    = (A-1)(2-A²+A)
    = -(A-1)(A²-A-2)
V   = -(A-1)(A-2)(A+1), which is different from A-1, so A-1 is not a solution.

Calculus

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