option nolet
print "-----------------------------------------------------------"
print "- muller method : (Solve Nonlinear Equation),version 1.0  -"
print "- peter.vlasschaert@gmail.com,21/08/2020                  -"
print "-----------------------------------------------------------"
! filename :TRUEBASICMULLERMETHODVERSION1NOSUB
!----------------------------------------------------------
print 
print" 3 value : start 'method,see : h' "
print
print "---------------------------------------------"
print " solve nonlinear : equation , Newton Raphson "
print " equation (solve: x) => f(x)=x^2-2 => f(x)=0 " 
print "---------------------------------------------"
print
print
print " f(x) = x^2-2,see below : function"
print 
print "input => "
print
xr = 0.01
h  = 0.001
eps = 0.0001
maxit = 20
print " start value            = ";xr
print " h : xr ,xr(1+h),xr(1-h)= ";h
print " error                  = ";eps
print " max number of iteration= ";maxit
print
x2 =xr       ! 1e value
x1 =xr*(1+h) ! 2e value
x0 =xr*(1-h) ! 3e value
declare function f
! algorithm : mullermethod ( Do while structure)
print "output => "
print
print " iter ,xr"
print "---------"
do
 iter = iter + 1
 h0 = x1-x0
 h1 = x2-x1
 d0 = (f(x1)-f(x0))/h0
 d1 = (f(x2)-f(x1))/h1
 a=(d1-d0)/(h1+h0)
 b=a*h1+d1
 c=f(x2)
 rat = sqr(b^2-4*a*c)  
!*****************************************
! othermethod to find sign before +/-'rat'
!*****************************************
 if abs(b+rat) > abs(b-rat) then
   den = b + rat
 else
   den = b - rat
 end if
   dxr  = -2*c/den
   xr = x2 +dxr
 print  iter , xr
if abs(dxr) < eps*xr or iter => maxit then
   exit do
end if
 x0=x1
 x1=x2
 x2=xr
loop
print
print "final solution f(x) = 0 =>"
print
print "                                 number iteration   xr"
print " final: iter ,x value f(x) = 0 =>  ";iter,xr
end

Function f(x)
f=x^2-2
! other f = x^3-1
end function