# Secant Method for $f(x) = \ln(x-1) +\cos(x-1)$, interval $1.3 \leq x \leq 2$

I am trying to see if I got the right answer so far for:

Using the Secant Method for $f(x) = \ln(x-1) +\cos(x-1)$, interval $1.3 \leq x \leq 2$

$p_0 = 1.3$
$p_1 = 2$
$p_2 = 1.520607$
$p_3 = 1.2043557751$

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For the Secant Method for approximating a root of $f(x)$, the iteration turns out to be $$x_{n+1}=x_n - f(x_n)\frac{x_n-x_{n-1}}{f(x_n)-f(x_{n-1}}.$$ The derivation is not difficult. The number $x_{n+1}$ is where the line through $(x_{n-1},f(x_{n-1}))$ and $(x_{n},f(x_{n}))$ meets the $x$-axis.
I calculated $p_2$. It agrees with what you wrote, exactly. (My cheap calculator only displayed $6$ digits after the decimal point.)
For $p_3$, I got $1.204357$. There is obvious agreement. Of course you have a bunch more digits, but they don't mean much, since we are clearly not yet close to the root.