# Secant Method Minimum Estimation

## Homework Statement

I am having some trouble understanding the following solved problem:

Use two iterations of the Secant method to estimate where $x^2 -x$ has a local minimum. Start with $x^{(0)} =1$, $x^{(1)} =1/2.$ ## Homework Equations

Secant iteration:

$$x_k = x_{k-1} - \frac{x_{k-1} - x_{k-2}}{f'_{k-1} - f'_{k-2}}$$

## The Attempt at a Solution

So I understand that for the minimum we need to solve at the derivative of the function, which is

$$3x^2 -1$$

Using the given initial values $x^{(0)} =1$, $x^{(1)} =1/2,$ the first iteration of the Secant method becomes

$$x^{(3)} = \frac{1}{2} - \frac{(1/2)- 1}{(3(1/2)^2 -1) - (3(1)^2 -1)} = 0.2778$$

So, how did they get $0.556$ for the first iteration?

Also, why did they use $1/4$ instead of $1/2$ as $x_{k-1}$?

I can't really follow the given answer, so I am wondering if there is an error in the solution. Any help is greatly appreciated.

Related Calculus and Beyond Homework Help News on Phys.org
BvU
Homework Helper
2019 Award
I have learned a different secant method. Please check it out.

In your calculation, however, in fact you correctly use the function, not the derivative .

 my mistake. Not the function nor the derivative. The derivative of $x^2-x$ is $2x-1$ !

But you do forget to multiply with f(1/2).
And you confuse yourself by calling it x(3)

• roam
Thank you so much for the explanation, it makes perfect sense now. I got the right answer, but I believe there is a typo in the solution in that $x_{k-1}$ in the first iteration should be $1/2$ instead of $1/4.$

BvU
I think you've got it right ! Well observed and no wonder you were a bit disoriented • 