Consider we are approxinating a root by the secant method.
Then, the interation is given by $x_{n+1}=x_n - f(x_n)\frac{x_n - x_{n-1}}{f(x_n)-f(x_{n-1})}$.
In my text (Atkinson), it's written that:
After some manipulation, it's possible to show that $\alpha -x_{n+1}=(\alpha - x_n)(\alpha - x_{n+1})\frac{ -f''(\xi_n)}{2f'(\zeta_n)}$
I completely have no idea how the above equation is derived from the iteration formula.. How?
There’s more than one way to obtain this expression. Below, I’ll fill out the details outlined by Atkinson:
Suppose
$x_{n + 1} = x_n - \frac{{f\left( {x_n } \right)\left( {x_n - x_{n - 1} } \right)}}{{f\left( {x_n } \right) - f\left( {x_{n - 1} } \right)}}$
and let $\alpha$ be a zero of $f$ (you neglected to mention this property of $\alpha$ in your question).
Then,
$\alpha - x_{n + 1} = \alpha - x_n + \frac{{f\left( {x_n } \right)\left( {x_n - x_{n - 1} } \right)}}{{f\left( {x_n } \right) - f\left( {x_{n - 1} } \right)}} = \alpha - x_n + \frac{{f\left( {x_n } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}}$
where in the denominator of the last term, I’m using standard divided difference notation. This can further be rewritten as
$\alpha - x_n + \frac{{f\left( {x_n } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}} = \frac{{\left( {\alpha - x_n } \right)\left( {\alpha - x_{n - 1} } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}}\left( {\frac{{f\left[ {x_{n - 1} ,x_n } \right]}}{{\left( {\alpha - x_{n - 1} } \right)}} + \frac{{f\left( {x_n } \right)}}{{\left( {\alpha - x_n } \right)\left( {\alpha - x_{n - 1} } \right)}}} \right)$
Note that we can rewrite the parenthesized term in the right hand expression as
$\left( {\frac{{f\left[ {x_{n - 1} ,x_n } \right]}}{{\left( {\alpha - x_{n - 1} } \right)}} + \frac{{f\left( {x_n } \right)}}{{\left( {\alpha - x_n } \right)\left( {\alpha - x_{n - 1} } \right)}}} \right) = - \left( {\frac{{f\left[ {x_{n - 1} ,x_n } \right] - \frac{{f\left( \alpha \right) - f\left( {x_n } \right)}}{{\left( {\alpha - x_n } \right)}}}}{{\left( {x_{n - 1} - \alpha } \right)}}} \right) = - \left( {\frac{{f\left[ {x_{n - 1} ,x_n } \right] - f\left[ {x_n ,\alpha } \right]}}{{\left( {x_{n - 1} - \alpha } \right)}}} \right)$
This is the negative of a second-order divided difference that we can write as $- f\left[ {x_{n - 1} ,x_n ,\alpha } \right]$. Therefore,
$\alpha - x_n + \frac{{f\left( {x_n } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}} = - \frac{{\left( {\alpha - x_n } \right)\left( {\alpha - x_{n - 1} } \right)}}{{f\left[ {x_{n - 1} ,x_n } \right]}}f\left[ {x_{n - 1} ,x_n ,\alpha } \right]$
Atkinson’s expression then follows from the mean value theorem for divided differences. Note that in Atkinson’s expression, $\xi _n$ is some (not any) point which lies between the min and max of the elements of $x_{n - 1}$, $x_n$, and $\alpha$ and $\zeta_n$ is some (not any) point which lies between the min and max of the elements of $x_{n - 1}$ and $x_n$.
