How can I input values?

Question

Melchor 2023-5-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1961844-how-can-i-input-values

回答： Sulaymon Eshkabilov 2023-5-11

syms x;
f = input('Type the function expression, f: ');
xi = input('Enter the initial estimate, x0: ');
tol = input('Tolerable error: ');
disp('               Results of Newton-Raphson Iteration')
disp('=================================================================')
fprintf('k\txi\t\txr\t\tf(xi)\t\tder(xi)\n');
i = 1000;
k = 1;
g = diff(f,x);
fa = eval(subs(f,x,xi));
while abs(fa)> tol
    fa = eval(subs(f,x,xi));
    ga = eval(subs(g,x,xi));
    
    b = xi - fa/ga;
    fprintf('%d\t%f\t%f\t%f\t%f\n',k,xi,b,fa,ga);
    xi = b;
   
    k = k + 1;
end
fprintf('Root is %f\n', xi);
f=@(x)exp(x)+23*x;
xint=[2 5];
tol=1e-6;
perf='max';
[iter,x,fval]=goldensection(f,xint,tol,perf)
-----------------------------
function [iter,root]=falsepos(f,x,tol)
format short g
if(f(x(1))*f(x(2)))>0
    error('Your initial values are incorrect! Cannot find root within the interval.')
else 
    k=1;
    xr=(x(1)+x(2))/2;
    iter(1,1:7)=[k x xr f(x(1)) f(x(2)) f(xr)];
    while(abs(iter(k,7))>tol)
        k=k+1;
        if (iter(k-1,7)*iter(k-1,5))<0
            iter(k,3)=iter(k-1,4);
            iter(k,6)=iter(k-1,7);
            iter(k,2)=iter(k-1,2);
            iter(k,5)=iter(k-1,5);
        else
            iter(k,2)=iter(k-1,4);
            iter(k,5)=iter(k-1,7);
            iter(k,3)=iter(k-1,3);
            iter(k,6)=iter(k-1,6);
        end
        iter(k,4)=(iter(k,2)+iter(k,3))/2;
        iter(k,7)=f(iter(k,4));
        iter(k,1)=k;
    end
    root=iter(k,4);
end
end
f=@(x) exp(x)-3*x;
x=[0 1];
tol = 1e-5;
[iter,root]=falsepos(f,x,tol)
or (bisect)
k=1;
xr=(x(1)+x(2))/2;
iter(1,1:7)=[k x xr f(x(1)) f(x(2)) f(xr)];
while(abs(iter(k,7))>tol)
k=k+1;
if (iter(k-1,7)*iter(k-1,5))<0
iter(k,3)=iter(k-1,4);
iter(k,6)=iter(k-1,7);
iter(k,2)=iter(k-1,2);
iter(k,5)=iter(k-1,5);
else
iter(k,2)=iter(k-1,4);
iter(k,5)=iter(k-1,7);
iter(k,3)=iter(k-1,3);
iter(k,6)=iter(k-1,6);
end
iter(k,4)=(iter(k,2)+iter(k,3))/2;
iter(k,7)=f(iter(k,4));
iter(k,1)=k;
end
root=iter(k,4);
end
end
f=@(x) exp(x)-3*x;
x=[0 1];
tol = 1e-5;
[iter,root]=bisection(f,x,tol)

2 个评论
显示无隐藏无

Walter Roberson 2023-5-11

You show a function named falsepos but you call a function named goldensection ??

Walter Roberson 2023-5-11

fa = eval(subs(f,x,xi));

You should never eval() the result of subs(). MATLAB has no documented meaning for eval() of a symbolic expression. You should either use vpa() or double() instead of eval(), depending what you are trying to do.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Sulaymon Eshkabilov 2023-5-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1961844-how-can-i-input-values#answer_1233429

在 MATLAB Online 中打开

You code contains several crucial errors including wrong call of fcn, wrong placements of commands after the fcn file, wrong input variables and output variables, calling inexistance or not yet defined fcns (bisection), etc. Here is the corrected code (maybe still a few edits are necessary). It runs ok, computes the solutions and shows all iteration values. It can be tested with e.g.:

Input 1. ... f: @ @(x) exp(x)-3*x

Input 2. ... : [0, 1]

Input 3. ... : 1e-1

syms x;
f = input('Type the function expression, f: ');
xi = input('Enter the initial estimate, x0: ');
tol = input('Tolerable error: ');
disp('               Results of Newton-Raphson Iteration')
disp('=================================================================')
fprintf('k\txi\t\txr\t\tf(xi)\t\tder(xi)\n');
i = 1000;
k = 1;
g = diff(f,x);
fa = double(f(xi));
f=@(x)exp(x)-3*x;
x=[0 1];
tol = 1e-5;
[iter,root]=falsepos(f,x,tol)
f=@(x) exp(x)-3*x;
x=[0 1];
tol = 1e-5;
%[iter,root]=bisection(f,x,tol)
while abs(fa)> tol
    fa = double(f(xi));
    ga = double(subs(g,xi));
    
    b = xi - fa/ga;
    fprintf('%d\t%f\t%f\t%f\t%f\n',k,xi,b,fa,ga);
    xi = b;
   
    k = k + 1;
end
fprintf('Root is %f\n', xi);
xr=(x(1)+x(2))/2;
iter(1,1:7)=[k x xr f(x(1)) f(x(2)) f(xr)];
k=1;
while(abs(iter(k,7))>tol)
k=k+1;
if (iter(k-1,7)*iter(k-1,5))<0
iter(k,3)=iter(k-1,4);
iter(k,6)=iter(k-1,7);
iter(k,2)=iter(k-1,2);
iter(k,5)=iter(k-1,5);
else
iter(k,2)=iter(k-1,4);
iter(k,5)=iter(k-1,7);
iter(k,3)=iter(k-1,3);
iter(k,6)=iter(k-1,6);
end
iter(k,4)=(iter(k,2)+iter(k,3))/2;
iter(k,7)=f(iter(k,4));
iter(k,1)=k;
end
root=iter(k,4);
f=@(x)exp(x)+23*x;
xint=[2 5];
tol=1e-6;
perf='max';
[iter,x]=falsepos(f,xint,tol)
%-----------------------------
function [iter,root]=falsepos(f,x,tol)
format short g
if(f(x(1))*f(x(2)))>0
    error('Your initial values are incorrect! Cannot find root within the interval.')
else 
    k=1;
    xr=(x(1)+x(2))/2;
    iter(1,1:7)=[k x xr f(x(1)) f(x(2)) f(xr)];
    while(abs(iter(k,7))>tol)
        k=k+1;
        if (iter(k-1,7)*iter(k-1,5))<0
            iter(k,3)=iter(k-1,4);
            iter(k,6)=iter(k-1,7);
            iter(k,2)=iter(k-1,2);
            iter(k,5)=iter(k-1,5);
        else
            iter(k,2)=iter(k-1,4);
            iter(k,5)=iter(k-1,7);
            iter(k,3)=iter(k-1,3);
            iter(k,6)=iter(k-1,6);
        end
        iter(k,4)=(iter(k,2)+iter(k,3))/2;
        iter(k,7)=f(iter(k,4));
        iter(k,1)=k;
    end
    root=iter(k,4);
end
end