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Newton's method states that, if there is a root $r$ of $f(x)$ that we want to calculate and $r_0$ is an approximation, then a better approximation is $$r_1=r_0-\frac{f(r_0)}{f'(r_0)}$$We could substitute this into the formula again to get $$r_2=r_1-\frac{f(r_1)}{f'(r_1)}=r_0-\frac{f(r_0)}{f'(r_0)}-\frac{f\left(r_0-\frac{f(r_0)}{f'(r_0)}\right)}{f'\left(r_0-\frac{f(r_0)}{f'(r_0)}\right)}$$We could keep substituting like so. How can we turn this into an infinite series? I can't find an explicit formula for $a_k$ such that $$r=\sum_{k=0}^\infty a_k$$Because the $a_n$ depends on all other terms before it.

Edit: The recursive relation for $a_n$ is $a_n=\frac{f\left(\sum_{k=1}^na_{k-1}\right)}{f'\left(\sum_{k=1}^na_{k-1}\right)}$ where $a_0=r_0$, but can we find a closed form for $a_n$?

Kamal Saleh
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  • Not sure what you are hoping for. It isn't strictly true that Newton always returns a superior estimate. Sometimes the method fails to converge. Are you assuming convergence here? And what would be the value of writing the limiting process as a sum? – lulu Dec 20 '23 at 21:13
  • Yes, I thought that if we keep applying newton's method, we approach the root. – Kamal Saleh Dec 20 '23 at 21:15
  • Sadly, no. Not always. here are some examples of the failure of Newton, you can find many more online. – lulu Dec 20 '23 at 21:17
  • But if I specify that the closest root to $r_0$ is $r$, then it will work? – Kamal Saleh Dec 20 '23 at 21:19
  • No, again, not always. here is another counterexample. And here – lulu Dec 20 '23 at 21:20
  • The point of Newton is that it is easily shown that any fixed point of the process must be a zero of the function. So, if your process seems convergent numerically, then you can be pretty sure you found a root. But a priori one knows very little about the convergence. – lulu Dec 20 '23 at 21:22
  • And, possibly more to the point, in practice one really doesn't know much about values of the function. If it were easy to compute lots of values, you could just search for zeroes with a mesh. The usual problem is that computing the function, and derivatives, in question is "costly", so that numerical approximations are all you have. – lulu Dec 20 '23 at 21:25
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    It is not obvious how one would turn the nested functions into a series unless we use another method like series inversion – Тyma Gaidash Dec 21 '23 at 00:35

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