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سلام خدمت بچه های عزیز مهندسی شیمی این برنامه مطلب روش نیوتون ونصف کردن درمحاسبات عددی را براتون نوشتم امیدوارم که مورد استفاده قرار بگیره روشن نصف کردن function root=bisect(a0,b0,ep,max_iterate,index_f)
%
% function bisect(a0,b0,ep,max_iterate,index_f)
%
% This is the bisection method for solving an equation f(x)=0.
%
% The function f is defined below by the user. The function f is
% to be continuous on the interval [a0,b0], and it is to be of
% opposite signs at a0 and b0. The quantity ep is the error
% tolerance. The routine guarantees this as an error bound
% provided: (1) the restrictions on the initial interval are
% correct, and (2) ep is not too small when the machine epsilon
% is taken into account. Most of these conditions are not
% checked in the program! The parameter max_iterate is an upper
% limit on the number of iterates to be computed.
%
% For the given function f(x), an example of a calling sequence
% might be the following:
% root = bisect(1,1.5,1.0E-6,10,1)
% The parameter index_f specifies the function to be used.
%
% The following will print out for each iteration the values of
% count, a, b, c, f(c), (b-a)/2
% with c the current iterate and (b-a)/2 the error bound for c.
% The variable count is the index of the current interate. Tap
% the carriage return to continue with the iteration.
if a0 >= b0
disp('a0 < b0 is not true. Stop!')
return
end
format short e
a = a0; b = b0;
fa = f(a,index_f); fb = f(b,index_f);
if sign(fa)*sign(fb) > 0
disp('f(a0) and f(b0) are of the same sign. Stop!')
return
end
c = (a+b)/2;
it_count = 0;
while b-c > ep & it_count < max_iterate
it_count = it_count + 1;
fc = f(c,index_f);
% Internal print of bisection method. Tap the carriage
% return key to continue the computation.
iteration = [it_count a b c fc b-c]
if sign(fb)*sign(fc) <= 0
a = c;
fa = fc;
else
b = c;
fb = fc;
end
c = (a+b)/2;
pause
end
format long
root = c
format short e
error_bound = b-c
format short
it_count
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function value = f(x,index)
% function to define equation for rootfinding problem.
%===========================
%===========================
%===========================
switch index
case 1
value = x + 2*cos(x)-1;
%===========================
%===========================
%===========================
case 2
value = x^2;
end
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ادامه مطلب
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