Computing $x_3$ using the secant method

Question

So given the function $f=x^2-1=0$, we must use secant method to find a root, I also have $x_1=-1+\epsilon, x_2=\frac{\epsilon}{2}$. I am to prove $x_3=-1-\epsilon$

$f(x_1)=-2\epsilon+\epsilon^2, f(x_2)=\frac{\epsilon^2}{4}-1$ By the secant method $x_3=x_2-f(x_2)\frac{x_2-x_1}{f(x_2)-f(x_1)}$

So $x_3=\epsilon/2-(\frac{\epsilon^2}{4}-1)\bigg(\frac{\frac{\epsilon}{2}-(-1-\epsilon)}{(\frac{\epsilon^2}{4}-1)-(-2\epsilon+\epsilon^2)}\bigg)$

But this now seems unsimplifiable to me and out of the ball park of the expected solution

Answer 1

As G.Gare mentioned, you have a few mistakes.

Correcting them and simplifying, you should get $$x_3=\frac{(\epsilon -1) \epsilon +2}{3 \epsilon -2}$$ Since $\epsilon$ is supposed to be small, use the long division or Taylor series to get $$x_3=-1-\epsilon -2 \epsilon ^2+O\left(\epsilon ^3\right)=-1-\epsilon +O\left(\epsilon ^2\right)$$

Answer 2

With a concrete equation, you can simplify the quotient by canceling $x_2-x_1$, as \begin{align} x_3=x_2-f(x_2)\frac{x_2-x_1}{f(x_2)-f(x_1)} &=x_2-(x_2^2-1)\frac{x_2-x_1}{x_2^2-x_1^2}\tag1\\ &=x_2-\frac{x_2^2-1}{x_2+x_1}=\frac{x_2x_1+1}{x_2+x_1}\tag2 \end{align} Only then insert the values into the simplified expression \begin{align} x_3=\frac{\fracϵ2(-1+ϵ)+1}{-1+\frac{3ϵ}2} =-\frac{1-\fracϵ2+\frac{ϵ^2}2}{1-\frac{3ϵ}2} &= -1-ϵ\,\frac{1+\fracϵ2}{1-\frac{3ϵ}2}\tag3\\ &=-1-ϵ\,\left(1+\frac{2ϵ}{1-\frac{3ϵ}2}\right)\tag4 \end{align} etc.