(*  Computing the "exact value" of the solution of 
        sin(x)+x=1 ; 
     you have to run this part to initialize 
      the variable exactvalue used later for comparison.
       BE PATIENT, OBTAINING exactvalue MAY TAKE A WHILE!  *)
p=N[1,10000];  f[x_] := Sin[x]+x-1;
For[i=1, i<=20, i++, { p=p-f[p]/f'[p], }]
exactvalue=p;
Print[N[exactvalue,50]]

(*  Newton's method for computing the solution of  sin(x)+x=1  *)
p=N[1,10000];  f[x_] := Sin[x]+x-1;
For[i=1, i<=13, i++, 
  { p=p-f[p]/f'[p], 
    error = Abs[p - exactvalue], 
    Print[i,"   ",Log[1.0*error]]
    }
  ]

(*  Secant method for computing the solution of  sin(x)+x=1  *)
p0=N[0,10000];  p1 = N[1,10000];  f[x_] := Sin[x]+x-1;
For[i=1, i<=16, i++, 
  {p=p1-f[p1]*(p1-p0)/(f[p1]-f[p0]), 
    error = Abs[p - exactvalue], 
    Print[i,"   ",Log[1.0*error]], 
    p0 = p1, p1 = p
    }
  ]

(*  Fixed point iteration for computing the solution of sin(x)+x=1 *)
p=N[1,10000];  g[x_] := 1-Sin[x];
For[i=1, i<=50, i++, 
  {p=g[p], 
    error = Abs[p - exactvalue],  
    Print[i,"   ",Log[1.0*error]]
    }
  ]

(*  Aitken's extrapolation for iterative solution of  sin(x)+x=1  *)
p=N[1,10000];  g[x_] := 1-Sin[x];
p1 = g[p];  p2=g[p1];
For[i=1, i<=50, i++, 
  {pa=p-(p1-p)^2/(p2-2*p1+p), 
    error=Abs[pa-exactvalue],
    Print[i,"   ",Log[1.0*error]],
    p=p1, p1=p2, p2=g[p1], 
    }
  ]

(*  Steffensen's method for iterative solution of  sin(x)+x=1  *)
p=N[1,10000];  g[x_] :=  1-Sin[x];
For[i=1, i<=12, i++, 
  {p1 = g[p];  p2=g[p1];
    pa=p-(p1-p)^2/(p2-2*p1+p), 
    error=Abs[pa-exactvalue],
    Print[i,"   ",Log[1.0*error]],
    p=pa; 
    }
  ]