The roots for an equation containing tangent
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Mia Finit
le 4 Oct 2020
Réponse apportée : Milind Amga
le 8 Oct 2020
I'm going to find the roots of an equation containing tangent. like x*1.4 - atan(0.28*x)=37.5
I dont know how to write the code to find x in above equation.
1 commentaire
James Tursa
le 4 Oct 2020
You say "like". Is that your actual equation? Or is your actual equation something else?
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Image Analyst
le 5 Oct 2020
Assuming it's not your homework, try this:
x = linspace(0, 40, 10000);
y = x*1.4 - atan(0.28*x);
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
yline(37.5, 'LineWidth', 2, 'Color', 'r');
diffs = abs(y - 37.5);
[minDiff, index] = min(diffs)
xCrossing = x(index)
caption = sprintf('y crosses 37.5 at x = %f', xCrossing);
title(caption, 'FontSize', 18);
xlabel('x', 'FontSize', 18);
ylabel('y', 'FontSize', 18);
xline(xCrossing, 'LineWidth', 2, 'Color', 'g');
4 commentaires
Image Analyst
le 6 Oct 2020
Look:
x = 27.8147814781478
y = x*1.4 - atan(0.28*x)
and you'll see in the command window:
x =
27.8147814781478
y =
37.4975993926413
Didn't you want to know the value of x where y = 37.5? Because that's what we were all thinking. If not, explain it to someone else and let them write the question. Maybe it will be better understood by us if you do that.
Plus de réponses (3)
Alan Stevens
le 5 Oct 2020
If x*1.4 - atan(0.28*x)=37.5 is the equation then fixed point iteration will work.
Rewrite the equation as x(n+1) = (37.5+atan(0.28*x(n)))/1.4, use an initial guess for x, say, x(1) = 20, then use a while loop until x(n+1) and x(n) are the same (or within some tolerance).
3 commentaires
Alan Stevens
le 6 Oct 2020
The following is what I mean:
% Fixed point iteration
% If the equation is 1.4x - atan(0.28x) = 37.5 rearrange it as
% x = (37.5 + atan(0.28x))/1.4
tol = 1^-8; % Set desired tolerance
x = 20; % initial guess
flag = true; % Set to false when converged
while flag
xold = x;
x = (37.5 + atan(0.28*xold))/1.4;
if abs(x-xold)<tol
flag = false;
end
end
disp(x)
However, if you have a different equation in mind, you might have to manipulate it in a few different ways in order to find an arrangement that converges.
An alternative is to look up the Newton-Raphson method.
Bruno Luong
le 6 Oct 2020
>> fzero(@(x) x*1.4 - atan(0.28*x) - 37.5, 0)
ans =
27.8165
2 commentaires
Image Analyst
le 7 Oct 2020
I did it numerically rather than analytically or by using a function. So it's not exact but as close as you want to get. However there are optimization functions (seems like a lot of them) that may do the trick. I'm not very familiar with them, since I don't have the optimization toolboxes. There is a function fminsearch() you may want to study up on. Or lsqnonneg().
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