# The roots for an equation containing tangent

3 visualizaciones (últimos 30 días)
Mia Finit el 4 de Oct. de 2020
Respondida: Milind Amga el 8 de Oct. de 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 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
James Tursa el 4 de Oct. de 2020
You say "like". Is that your actual equation? Or is your actual equation something else?

Iniciar sesión para comentar.

Image Analyst el 5 de Oct. de 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 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
Image Analyst el 6 de Oct. de 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.
Mia Finit el 6 de Oct. de 2020
Thank you so much for your support. That is not the answer of my question. but if someone wants to help me out, please dont give me the exact answer of that. I asked you that to help me out in order to improve my programming skill. So, please just tell me the way by which Ill be able to code myself.
Thank all of you.

Iniciar sesión para comentar.

### Más respuestas (3)

Alan Stevens el 5 de Oct. de 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 comentariosMostrar 1 comentario más antiguoOcultar 1 comentario más antiguo
Mia Finit el 5 de Oct. de 2020
or maybe I havent got what you mean...
Alan Stevens el 6 de Oct. de 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.

Iniciar sesión para comentar.

Bruno Luong el 6 de Oct. de 2020
>> fzero(@(x) x*1.4 - atan(0.28*x) - 37.5, 0)
ans =
27.8165
##### 2 comentariosMostrar NingunoOcultar Ninguno
Mia Finit el 6 de Oct. de 2020
Yeah. I did the same as you. But I was wondering that if it is not a correcct answer... So, i was looking for another straight forward way by which I can get an answer. But it seems it is a proper one... cause the other answer given by our friend (the one with a graph) is the same.... however, my equation is much more complicated and I was not sure of that. But now, I am sure that the way I pass through was the best.
Thank you for your support guys.
Of course I am looking forward for the answer given by other guys as well.
Thanks all.
Image Analyst el 7 de Oct. de 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().

Iniciar sesión para comentar.

Milind Amga el 8 de Oct. de 2020
@Image Analyst
Could you please have a look at my code too?
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Iniciar sesión para comentar.

### Categorías

Más información sobre Linear Least Squares en Help Center y File Exchange.

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by