function x=newton(Fs,FPs,P0,y,P,TOL);
 % NEWTON-RAPHSON ALGORITHM 2.3               
 %
 % To find a solution to f(x) = 0 given an
 % initial approximation p0:
 %
 % y,P can be used for additional info if the function
 % requires it.
 %
 % Initially written by Burden and Faires, revised by 
 % B.A. Broughton, 10/01/1999
 %
 % INPUT:   initial approximation p0; tolerance TOL;
 %          maximum number of iterations NO.
 %
 % OUTPUT:  approximate solution p or a message of failure

 TRUE = 1;
 FALSE = 0;
 F = inline(Fs,'x','P','y');
 FP = inline(FPs,'x','P');
 NO=150; 
 OUP = 1;
 F0 = F(P0,P,y);
 % STEP 1
 I = 1;
 OK = TRUE;        
% STEP 2
 while I <= NO & OK == TRUE   
        % STEP 3
        % compute P(I)
         FP0 = FP(P0,P);
         D = F0/FP0;
        % STEP 6
         P0 = P0 - D;  
         F0 = F(P0,P,y);
        % STEP 4
         if abs(D) < TOL 
                % procedure completed successfully
                % fprintf(OUP,'\nApproximate solution = %.10e\n',P0);
                % fprintf(OUP,'with F(P) = %.10e\n',F0);
                % fprintf(OUP,'Number of iterations = %d\n',I);
                % fprintf(OUP,'Tolerance = %.10e\n',TOL);
                 OK = FALSE;
        % STEP 5
         else
                I = I+1;
         end
 end
 if OK == TRUE 
% STEP 7
% procedure completed unsuccessfully
         fprintf(OUP,'\nIteration number %d',NO);
         fprintf(OUP,' gave approximation %.10e\n',P0);
         fprintf(OUP,'with F(P) = %.10e not within tolerance  %.10e\n',F0,TOL);
 end
 x = P0;