Secant Method Exercise

This exercise is also on numericalmethods.xlsx

Given:  $ f(x) = x^2-10x+9$, find the roots using the secant method.

Computation:

1. We choose two points for $x_{-1}$ and $x_{0}$.
2. Solve for the next points using the formula: \[  x_n = x_{n-1} - f(x_{n-1}) \frac{x_{n-1}-x_{n-2}}{f(x_{n-1})-f(x_{n-1})}   \]

This formula is derived from getting the slope of the secant lines through the points $(x_n,f(x_n))$, $(x_{n-1},f(x_{n-1}))$ and  $(x_{n-2},f(x_{n-2}))$.

\[  \frac{f(x_n) - f(x_{n-1})}{x_n - x_{n-1}} = \frac{f(x_{n-1}) - f(x_{n-2})}{x_{n-1} - x_{n-2}} \]

Since $f(x_n)=0$, then
\[ \frac{0 - f(x_{n-1})}{x_n - x_{n-1}} = \frac{f(x_{n-1}) - f(x_{n-2})}{x_{n-1} - x_{n-2}} \]

Solve for $x_n.$

\[ \begin{split}
- f(x_{n-1})(x_{n-1} - x_{n-2})      &= (x_n - x_{n-1})( f(x_{n-1}) - f(x_{n-2})) \\
\frac{- f(x_{n-1})(x_{n-1} - x_{n-2})}{ f(x_{n-1}) - f(x_{n-2})}    &=  x_n - x_{n-1} \\
 x_{n-1} - \frac{ f(x_{n-1})(x_{n-1} - x_{n-2})}{ f(x_{n-1}) - f(x_{n-2})}    &=  x_n \\
 x_n &=  x_{n-1} - \frac{ f(x_{n-1})(x_{n-1} - x_{n-2})}{ f(x_{n-1}) - f(x_{n-2})} \end{split}\]

3. Continue to iterate until $f(x) = 0$ or we've applied the stopping rules.

This is my answer.

I also applied secant method on $ f(x) = cos(x) - x$ and arrived at this result.

So far in excel it works. How about applying these methods in Visual Basic? Will it work? What can be my possible problems?