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

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)[reply]

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[reply]

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 $x_{0}=1$ and $x_{1}=2$ . 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)[reply]

I changed some math formatting to be consistent, but I'm not sure it's better; think about it.

You probably need one more iteration so that people see the pattern.

Mjmohio (talk) 19:21, 18 September 2012 (UTC)[reply]

Advantages of secant method

It converges at faster than a linear rate, so that it is more rapidly convergent than the bisection method.
It does not require use of the derivative of the function, something that is not available in a number of applications.
It requires only one function evaluation per iteration, as compared with Newton’s method which requires two.

Disadvantages of secant method

It may not converge.
There is no guaranteed error bound for the computed iterates.
It is likely to have difficulty if f′(α) = 0. This means the x-axis is tangent to the graph of y = f (x) at x = α.
Newton’s method generalizes more easily to new methods for solving simultaneous systems of nonlinear equations.

la740411ohio (talk) 11:55, 18 September 2012 (UTC)[reply]

Changed to a list.

Add in cross-links to mentioned things like Newton's Method and references as available.

Some of this duplicates Secant Method#Comparison with other root-finding methods; it would be better to improve that section than to add new sections.

Mjmohio (talk) 00:29, 19 September 2012 (UTC)[reply]