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In Secant method - $$x_{n} = x_{n-1} - f(x_{n-1})\frac{x_{n-1}-x_{n-2}}{f(x_{n-1})-f(x_{n-2})}$$

I have to choose 2 initial - $x_0$ and $x_1$ .

How could I know whether with those 2 $x$'s the sequence is going to converge to the zero of $f(x)$ without calculating the above iterations ?

Jean Marie
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1 Answers1

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Wikipedia says:

If the initial values are not close enough to the root, then there is no guarantee that the secant method converges. There is no general definition of "close enough", but the criterion has to do with how "wiggly" the function is on the interval $[x_0, x_1]$. For example, if is differentiable on that interval and there is a point where $f'=0$ on the interval, then the algorithm may not converge.

The relevant page is here.

If you want a method that is guaranteed to converge to a root, then you should look into bisection or regula falsi. They require initial values that "bracket" the root you're seeking, though.

For some fascinating (and unusual) discussions of root-finding, I recommend papers by Bill Kahan, especially this one.

bubba
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  • If the $x_0$, $x_1$ straddle one root, and you modify the secant method so this is always true for $x_{n + 1}$ and $x_n$, it does converge. – vonbrand Apr 24 '13 at 15:35
  • Thanks. I thought that might be the case, but couldn't find a reference, and didn't have the energy to attempt a proof. – bubba Apr 24 '13 at 16:32
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    @bubba Dear Sir, I was interested by your recommended paper of Bill Kahan (last line of your answer), but the thread is to a document in japanese and I am unable to find my way through it. Could you tell me how I can do. Thank you very much by advance. – Jean Marie Oct 25 '17 at 10:14
  • @JeanMarie -- I don't know what happened to the original link. Anyway, I changed it to point to a better location (though same document). – bubba Oct 25 '17 at 12:19
  • @bubba Thank you very much. Very interesting article. – Jean Marie Oct 25 '17 at 13:18