What would be the Newton's method in the form $x_{k+1}=g(x_k)$ to solve the equation $$f(x)=x^2-2bx+b^2-d^2=0$$ in which both $b>0,d>0$ are parameters? Additionally, I need to show that $|g'(x)|\le 1/2$ whenever $|x-b|\ge d/\sqrt{2}$ and also that $|g(x)-b|\ge d/\sqrt{2}$ whenever $|x-b|\ge d/\sqrt{2}$.
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If you already have the equation in that form, what do you need Newton's method for? The quadratic formula immediately gives the exact roots $b\pm d$. – hmakholm left over Monica Feb 10 '16 at 21:40
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@HenningMakholm So, does that mean that there isn't a way to put it into the form $x_{k+1}=g(x_k)$? – user310546 Feb 10 '16 at 21:42
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There is. Convergence is another problem. – Bernard Feb 10 '16 at 21:43
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@user310546: Why would you want to do that? If you already have $b$ and $d$, simply adding and subtracting those known numbers will give you the exact roots immediately, with less effort than any conceivable stepwise approximation method would need. – hmakholm left over Monica Feb 10 '16 at 21:44
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in this example you don't need the Newton method – Dr. Sonnhard Graubner Feb 10 '16 at 21:44
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@HenningMakholm I understand. But I need it to show the rest of my question. – user310546 Feb 10 '16 at 21:45
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@HenningMakholm I know that $$g(x)=x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}$$ and we can use that to find $g$. But I can't figure out the second question still. – user310546 Feb 10 '16 at 21:52