False Position Method (Plot)

Question

Brain Adams 2021-3-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/780752-false-position-method-plot

评论： Alan Stevens 2021-3-23

Hi everyone, I wrote a code that finds the root of the equation using False Position Method. I would like to ask that, how can I plot the root as a function of iteration number and approximate error as a function of itteration number? Thanks in advance to all who want to help!

f = @(x) 1000*x^3 + 3000*x - 15000; 
x_l = 0; 
x_u = 4; 
if f(x_l)*f(x_u) > 0
    fprintf('There is no solution in the given interval'); 
    return
elseif f(x_l) == 0
    fprintf('%f is the solution',x_l); 
elseif f(x_u) == 0
    fprintf('%f is the solution', x_u); 
end
fprintf('i     xl    xu      xm\n'); 
for i = 1:1000 
    
    xm = x_u - (x_l-x_u)*f(x_u)/(f(x_l)-f(x_u)); 
    
    fprintf('%i %f %f %f\n',i,x_l,x_u,xm) 
    
    if abs(f(xm)) < 0.0001 
        
        return
    end
 if f(x_l)*f(xm) < 0
     x_u = xm;
     
 elseif f(x_u)*f(xm) < 0
     x_l = xm;
 end 
     
end

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Answer 1

Alan Stevens 2021-3-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/780752-false-position-method-plot#answer_655647

在 MATLAB Online 中打开

Like this:

f = @(x) 1000*x^3 + 3000*x - 15000; 
x_l = 0; 
x_u = 4; 
if f(x_l)*f(x_u) > 0
    fprintf('There is no solution in the given interval'); 
    return
elseif f(x_l) == 0
    fprintf('%f is the solution',x_l); 
elseif f(x_u) == 0
    fprintf('%f is the solution', x_u); 
end
fprintf('i     xl    xu      xm\n'); 
for i = 1:100 
    
    xm(i) = x_u - (x_l-x_u)*f(x_u)/(f(x_l)-f(x_u)); 
    
    %fprintf('%i %f %f %f\n',i,x_l,x_u,xm) 
    
    if abs(f(xm(i))) < 0.0001 
        break
    end
 if f(x_l)*f(xm(i)) < 0
     x_u = xm(i);
     
 elseif f(x_u)*f(xm(i)) < 0
     x_l = xm(i);
 end 
     
end
i = 1:numel(xm);
plot(i,xm,'o'),grid
xlabel('iteration number')
ylabel('root approximation')

2 个评论
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Brain Adams 2021-3-23

Thanks! It is much more clear now. I am trying to plot other graphic. Unfortuanetly, I couldn't plot approximate error as a function of itteration number. Can you help me on that too?

Alan Stevens 2021-3-23

在 MATLAB Online 中打开

Depends on what you think of as the approximate error. If you mean the change in xm from one iteration to the next, then try the following:

f = @(x) 1000*x^3 + 3000*x - 15000; 
x_l = 0; 
x_u = 4; 
if f(x_l)*f(x_u) > 0
    fprintf('There is no solution in the given interval'); 
    return
elseif f(x_l) == 0
    fprintf('%f is the solution',x_l); 
elseif f(x_u) == 0
    fprintf('%f is the solution', x_u); 
end
xmold = (x_l + x_u)/2;
 %fprintf('i     xl    xu      xm\n'); 
for i = 1:100 
    
    xm(i) = x_u - (x_l-x_u)*f(x_u)/(f(x_l)-f(x_u)); 
    deltax(i) = abs(xm(i)-xmold);
    %fprintf('%i %f %f %f\n',i,x_l,x_u,xm) 
    
    if abs(f(xm(i))) < 0.0001 
        break
    end
 if f(x_l)*f(xm(i)) < 0
     x_u = xm(i);
     
 elseif f(x_u)*f(xm(i)) < 0
     x_l = xm(i);
 end 
     xmold = xm(i);
end
i = 1:numel(xm);
plot(i,xm,'o'),grid
xlabel('iteration number')
ylabel('root approximation')
figure
semilogy(i,deltax,'o'),grid
xlabel('iteration number')
ylabel('error')