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$$f(x)=\sqrt{x^2+1}-1$$

because the function does not change signs both bisection and REGULA-FALSI can not be used, so I have used Newton Rapshon.

  1. If will look at the minimum point we can see that $x=0$, can it be translated to a numerical method?

  2. With Newton Rapshon expression where of the kind $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ where f'(x_n) tend to zero, is it a numerical problem?

newhere
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    Is it intentional that you have both catastrophic cancellation and a double root that under perturbation might not be a root in the same task? Are you allowed to rewrite as $f(x)=\frac{x^2}{1+\sqrt{1+x^2}}$? – Lutz Lehmann Dec 15 '19 at 20:46
  • @Dr.LutzLehmann correct, I must look solve for $\frac{x^2}{1+\sqrt{1+x^2}}$ – newhere Dec 15 '19 at 20:51
  • @Dr.LutzLehmann it is weird the Newton Rapshon still worked – newhere Dec 15 '19 at 20:54
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    Then this should be easy, Newton should converge linearly to zero, halving the distance in every step. Close to zero this looks more-or-less like $x^2/2$ so that the Newton step is $x_{n+1}=x_n-\frac{x_n^2/2}{x_n}=\frac{x_n}2$. – Lutz Lehmann Dec 15 '19 at 20:55
  • Your question is still lacking context. The equation is trivially solvable symbolically. It would make sense to use this function to test some numerical method, but you do not say which method. Or if that is a wider research task, what are other classes of test functions that have to be tried out? But then that question might be better located at scicomp.SE. – Lutz Lehmann Dec 15 '19 at 21:43
  • @Dr.LutzLehmann it was given in numerical method course, it just asked to solve it, and explain which method we used, there is a hint to look at the optimum point – newhere Dec 15 '19 at 21:57
  • Yes, that might be one scenario to try, test some random point $x$, find a positive value, look for a or the minimum in the hope to find a negative value and then apply some bracketing method to the smallest to date interval with a sign change in the value of $f$. Thus apply root finding to $f'$ or $g(x)=f(x+h)-f(x-h)$. – Lutz Lehmann Dec 15 '19 at 22:07

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