2.2 Secant method.ipynb

Kernel: Octave

# Approximating f(x) = sqrt(a)
# Using Newton 
# Author: Fabian with additional tweaks from CS154 classmates

# Parameters
tol = 1e-8;
x = 1.5;

# define the function
function y = f(x)
  y = (x.^2+log(x)-2);
endfunction

fun = @f;
root = fzero(fun, 1);

# define the first derivative of f’(x) = 2x here:
function y = ff(x)
  y = (2*x+(1/x));
endfunction

err = 3*tol;
i= 0;
while err > tol
  i += 1;
  z = f(x)/ff(x);
  err = abs(z);
  x = x - z;
  printf("current approx: %e \n", x)
  printf("At Iter [%d]: the rel. error is:%e \n",i,(root-x)/root)
endwhile
printf("the approximated value of x is: %e \n", x)

current approx: 1.321237e+00 
At Iter [1]: the rel. error is:-5.433378e-03 
current approx: 1.314107e+00 
At Iter [2]: the rel. error is:-8.131713e-06 
current approx: 1.314097e+00 
At Iter [3]: the rel. error is:-1.821527e-11 
current approx: 1.314097e+00 
At Iter [4]: the rel. error is:0.000000e+00 
the approximated value of x is: 1.314097e+00 

# Bisection method
# Edited by Mano, from Prof's notes

function s = fa(x)
  s = (x.^2+log(x)-2);
endfunction

lb = 1;
ub = 2;
tol = 1e-8;
maxit = 15;

function [c, v_err] = bisection2(lb, ub, tol, maxit)

# Checking the trap interval condition for the Bisection method
if (lb >= ub) 
  printf("Invalid trapping interval for bisection a[%10.8f]:b[%10.8f]  \n",     
          lb, ub);
else 
# We store the errors at every iteration to be plotted at the end.  
  v_err = zeros(1, maxit+1);
  v_err(1) = (ub - lb)/2;
  i = 0;
  printf("Initial approx. err:%e and input tolerance: %e\n",v_err(1),tol);

# To monitor the improvement per iteration, we will compute the approximate 
# order of convergence, gamma, described in the Holmes textbook Eq. 2.15 
# (Subsection 2.4.1). Gamma computes the ratio of the natural logarithms 
# of the current error and the error in the previous iteration.  We can
# only compute this after the first iteration.
  lnerr = log(v_err(1));
  while (v_err(i+1) > tol) && (i < maxit)
    i++;
    c = (lb + ub)/2;
    if fa(c) != 0 
      if (fa(lb) * fa(c)) < 0
          ub = c;
      else 
          lb = c ;
      endif
      v_err(i+1) = (ub - lb)/2;
      lnerr1 = log(v_err(i+1));
    endif
    c = (lb + ub)/2;
    printf("x_bar [ %d ] = %e and error = %e  \n",i,c, v_err(i+1));
    lnerr = lnerr1;  
  endwhile
endif
endfunction
bisection2(lb, ub, tol, maxit)

Initial approx. err:5.000000e-01 and input tolerance: 1.000000e-08
x_bar [ 1 ] = 1.250000e+00 and error = 2.500000e-01  
x_bar [ 2 ] = 1.375000e+00 and error = 1.250000e-01  
x_bar [ 3 ] = 1.312500e+00 and error = 6.250000e-02  
x_bar [ 4 ] = 1.343750e+00 and error = 3.125000e-02  
x_bar [ 5 ] = 1.328125e+00 and error = 1.562500e-02  
x_bar [ 6 ] = 1.320312e+00 and error = 7.812500e-03  
x_bar [ 7 ] = 1.316406e+00 and error = 3.906250e-03  
x_bar [ 8 ] = 1.314453e+00 and error = 1.953125e-03  
x_bar [ 9 ] = 1.313477e+00 and error = 9.765625e-04  
x_bar [ 10 ] = 1.313965e+00 and error = 4.882812e-04  
x_bar [ 11 ] = 1.314209e+00 and error = 2.441406e-04  
x_bar [ 12 ] = 1.314087e+00 and error = 1.220703e-04  
x_bar [ 13 ] = 1.314148e+00 and error = 6.103516e-05  
x_bar [ 14 ] = 1.314117e+00 and error = 3.051758e-05  
x_bar [ 15 ] = 1.314102e+00 and error = 1.525879e-05  
ans =  1.3141

#Secant method
#Mano

# Input parameters, incl. tolerance and original interval for secant line
tol = 1e-8;
a = 1;
b = 2;

# define the function
function y = f(x)
  y = (x.^2+log(x)-2);
endfunction

fun = @f;
root = fzero(fun, 1);

err = 3*tol;
i= 0;
while err > tol
  z = (f(a)*(a-b))/(f(a)-f(b));
  err = abs(z);
  b = a;
  a = a - z;
  printf("current approx: %e \n", a)
  i += 1;
  printf("At Iter [%d]: the rel. error is:%e \n", i, (root-a)/root)
endwhile
printf("the approximated value of a is: %e \n", a)

fun = @f
root =  1.3141
current approx: 1.270772e+00 
At Iter [1]: the rel. error is:3.296944e-02 
current approx: 1.316883e+00 
At Iter [2]: the rel. error is:-2.120151e-03 
current approx: 1.314072e+00 
At Iter [3]: the rel. error is:1.925183e-05 
current approx: 1.314097e+00 
At Iter [4]: the rel. error is:1.124360e-08 
current approx: 1.314097e+00 
At Iter [5]: the rel. error is:-5.964686e-14 
current approx: 1.314097e+00 
At Iter [6]: the rel. error is:0.000000e+00 
the approximated value of a is: 1.314097e+00