Is there periodic sequence generated by secant method? Is there $f(x)$, $x_0, x_1$ such that $\{x_n\}$ is periodic sequence?
$$x_n=x_{n-1}-f(x_{n-1})\frac{x_{n-1}-x_{n-2}}{f(x_{n-1})-f(x_{n-2})}$$
 A: Yes there are such sequences. It can be constructed as follows. Let $(x_1,y_1)$ and $(x_2,y_2)$ be the two starting points. We look for the next two points, $(x_3,y_3)$ and $(x_4,y_4)$, to be positioned symmetrically across origin. With the cycle consisting of these four points. (There could be other solutions and of course there would be solutions with different or longer cycles. So this is far from a complete classification.)
Let the iteration formula be noted by $\displaystyle a_3=Int(a_1,b_1,a_2,b_2)={{b_2 a_1 - b_1 a_2}\over {b_2-b_1}}$, i.e. the $x$-intercept of line through $(a_1,b_1)$ and $(a_2,b_2)$ is $a_3$.
Assume $x_1,x_2,y_1,y_2$  are the four variables to be found from the following equations:
Symmetry consitions:
$x_3=-x_1$ and  $y_3=-y_1$ $\hskip 1in$ (eq 1)
$x_4=-x_2$ and  $y_4=-y_2$ $\hskip 1in$ (eq 2)
Intercept conditions
$x_3=Int(x_1,y_1,x_2,y_2)$ $\hskip 1in$ (eq 3)
$x_4=Int(x_2,y_2,x_3,y_3)$ $\hskip 1in$ (eq 4)
Cyclic conditions $(x_5,y_5)=(x_1,y_1)$ and $(x_6,y_6)=(x_2,y_2)$
$x_5=Int(x_3,y_3,x_4,y_4)$  $\hskip 1in$ (eq 3)
$x_6=Int(x_4,y_4,x_5,y_5)$ $\hskip 1in$ (eq 5)
As it turns out these equations have  a two-parameter family of solutions. (The two degrees of freedom are just the scaling of the graph in $x$ and $y$ directions.)
The intercept formula together with symmetry conditions give the following
For $(x_3,y_3)$ point: $\displaystyle x_3= {{y_2x_1 - y_1 x_2}\over {y_2-y_1}}$ simplifies to $\displaystyle -x_1= {{y_2x_1 - y_1 x_2}\over {y_2-y_1}}$$\hskip 1in$ (eq 6) 
For $(x_4,y_4)$ point: $\displaystyle x_4= {{y_3x_2 - y_2 x_3}\over {y_3-y_2}}$ simplifies to $\displaystyle -x_2= {{-y_1x_2 + y_2 x_1}\over {-y_1-y_2}}$  $\hskip 1in$ (eq 7) 
For $(x_5,y_5)$ point: $\displaystyle x_5= {{y_4 x_3 - y_3 x_4}\over {y_4-y_3}}$ simplifies to  $\displaystyle x_1= {{y_2 x_1 - y_1 x_2}\over {-y_2+y_1}}$ $\hskip 1in$ (eq 8) 
For $(x_6,y_6)$ point: $\displaystyle x_6= {{y_5 x_4 - y_4 x_5}\over {y_5-y_4}}$ simplifies to  $\displaystyle x_2= {{-y_1 x_2 + y_2 x_1}\over {y_1+y_2}}$ $\hskip 1in$ (eq 9)
(eq 6) and (eq 8) are same and  give  $2x_1y_2=y_1(x_1+x_2)$ $\hskip 1in$ (eq 10)
(eq 7) and  (eq 9) are same and give $2x_2y_1=y_2(x_1-x_2)$ $\hskip 1in$  (eq 11)
multiplying the (eq 10) and (eq 11) and dividing by $y_1y_2$ gives $4x_1x_2=x_1^2-x_2^2$, upon solving this we end up with
$x_2=x_1(-2\pm \sqrt 5)$ and $\displaystyle y_2=y_1 ({{-1\pm \sqrt{5}}\over 2})$.
