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This is an old revision of this page, as edited by SineBot (talk | contribs) at 04:59, 21 May 2011 (Signing comment by Intellec7 - "secant method iteration requires single function evaluation?: new section"). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

Convergence Rate for Repeated Roots?

Is there a fixed order of convergence for repeated roots with the secant method? For instance, with the Newton-Raphson method, R=2 (quadratic) for simple roots and R=1 for repeated roots. For the Secant Method, R=1.618.... for simple roots, but what about repeated/complex roots? Computer Guru (talk) 21:40, 26 May 2008 (UTC)[reply]

Incorrect image?

Either I'm going crazy, or the image on the page isn't correct. Shouldn't the second secant go from f(x0) to f(x1), rather than between f(x2) and f(x1) as it appears to be doing now? --VPeric (talk) 17:34, 18 March 2009 (UTC)[reply]

You are probably confusing the Secant method with the False position method. Tovrstra (talk) 12:27, 22 October 2009 (UTC)[reply]

secant method iteration requires single function evaluation?

Assuming that evaluation of a function and evaluation of its derivative takes the same amount of time, the article writes that an iteration of the secant method is twice as quick as an iteration of Newton's method. Doesn't the secant method require evaluating the function at two points, though? —Preceding unsigned comment added by Intellec7 (talkcontribs) 04:58, 21 May 2011 (UTC)[reply]