Talk:Secant method

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)

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)

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

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 (talk • contribs) 04:58, 21 May 2011 (UTC)

Citation for order of convergence

The article states that the order of convergence is equal to the golden ratio. However, I seem to miss a direct citation of a reference where this is demonstrated. Mjpnijmeijer (talk) 16:53, 16 December 2011 (UTC)

Really Cool History Missing

3000 years of history and the basis of other algorithms? It seems like there must be a history section missing. Anyone know it? I checked Wikipedia and couldn't find anything....173.242.89.38 (talk) 23:25, 8 August 2012 (UTC)EAZen

Numerical Example?

It may be useful to some readers to see the secant method applied in a numerical example. An example (maybe similar to the one below) could help clarify the method and the iterative process....thoughts?

A numerical example

Consider $f(x)=x^{3}-2$. We know the exact solution to be $x={\sqrt[{3}]{2}}$. To approximate this solution using the secant method, let's let x₀ = 1 and x₁ = 2. Then f(x₀) = f(1) = -1 and f(x₁) = f(2) = 6. Now use the formula to calculate x₂:

$x_{2}=x_{1}-f(x_{1}){x_{1}-x_{0} \over f(x_{1})-f(x_{0})}=2-6({2-1 \over 6+1})=2-{6 \over 7}={8 \over 7}\approx 1.1429$

In the next step use x₁ and x₂ together with f(x₁) = 6 and f(x₂) = -174/343 to calculate x₃. We can continue on in this manner until we have a solution correct to our desired level of precision.

Brmcvet (talk) 00:54, 11 September 2012 (UTC)