secant method in maple 
secant method in maple. Find a root of the statement $x^3-3x^2+4x-1=0$ with the initial value $x_0=0$ and $x_1=1$ with 5 digits point approximation.

 A: If this is homework and you're supposed to program it yourself then you could show what you've accomplished on your own already.
If you just want to see answers from such an algorithm then you could use the Student:-NumericalAnalysis command. Eg,
[Student:-NumericalAnalysis:-Secant(x^3-3*x^2+4*x-1,x=[0,1],
                                    tolerance=1e-5,
                                    maxiterations=50)]:
evalf[5](%)[];

                                0.31767


[Student:-NumericalAnalysis:-Secant(x^3-3*x^2+4*x-1,x=[0,1],
                                    tolerance=1e-5,
                                    output=sequence,
                                    maxiterations=50)]:
evalf[5](%)[];

 0., 1., 0.50000, 0.20000, 0.33624, 0.31947, 0.31764, 0.31767, 0.31767

Programming questions about Maple are better asked on stackoverflow, unless it is the mathematics behind the algorithm that is your central concern.
A: another one: 

eps_step := 0.00001;
eps_abs := 0.00001;
f := x -> x^3-3*x^2+4*x-1;
x[0] := 1.0;
x[1] := 1.5;
for i from 2 to 100 do
x[i] := (x[i - 2]*f(x[i - 1]) - x[i - 1]*f(x[i - 2]))/(f(x[i - 1]) - f(x[i - 2]));
if abs( x[i] - x[i - 1] ) < eps_step and abs( f( x[i] ) )  < eps_abs then 
eps_abs then break;
elif i = 100 then
end if;
end do:
evalf(x[i],5);
eps_step := 0.00001
eps_abs := 0.00001

                 f := x -> x^3-3*x^2+4*x-1

                     x[0] := 1.0

                     x[1] := 1.5

                        0.31767

