Are there some sufficient conditions for secant method $$x_{k+1}=x_k-\dfrac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}f(x_k)$$
to converge?
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Does this answer your question? Convergence of Secant method – Simply Beautiful Art Nov 26 '20 at 22:02
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tl;dr the secant method converges where Newton's method converges, but may also converge elsewhere. A nice example of their difference can be seen from $f(x)=\sqrt[3]x$, where Newton's method diverges everywhere and the secant method converges wherever $|x_1|\le|x_0|$. – Simply Beautiful Art Nov 26 '20 at 22:06
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Does this answer your question? Question about newton's interpolation and rootfinding – Carl Christian Nov 27 '20 at 08:41
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You can easily deduce a local result. The error satisfies $$ e_{n+1} = -\dfrac{f''(\xi_n)}{2f'(\eta_n)} e_{n} e_{n-1}. $$
If $f\in C^2$ and $f'(z)\ne 0$, the method will converge if $x_0, x_1$ are sufficiently close to $z$. the rate of convergence will be $\alpha = (1+\sqrt{5})/2 \approx 1.618$.
PierreCarre
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1There is a non-trivial step at the start where you have to ensure that $x_{n+1}$ is not too far away from the previous iterates, or that the whole sequence stays inside some "nice" set, so that bounds on the first and second derivative are applicable at the intermediate points. – Lutz Lehmann Nov 27 '20 at 11:32