The roots for an equation containing tangent

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.

[James Tursa]

You say "like". Is that your actual equation? Or is your actual equation something else?

Answer 1

[Image Analyst]

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');

[Image Analyst]

By the way, that equation is NOT a polynomial.

[Mia Finit]

no. thanks. that is not my answer again. maybe i could not say what i am looking for.

[Image Analyst]

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]

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.

Answer 2

[Alan Stevens]

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).

[Mia Finit]

I am afraid, it didnt work.

[Mia Finit]

or maybe I havent got what you mean...

[Alan Stevens]

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.

Answer 3

[Bruno Luong]

>> fzero(@(x) x*1.4 - atan(0.28*x) - 37.5, 0)

ans =

   27.8165

[Mia Finit]

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]

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().

Answer 4

[Milind Amga]

@Image Analyst

Could you please have a look at my code too?

Edge detection using sobel method