how to solve tan(x)-x=0.4267?

May I ask how to solve tan(x)-x=0.4267 in matlab?

Answer 1

[John D'Errico]

To be honest, I'd use fzero, not fsolve. But simple iteration is entirely sufficient.

x0 = atan(x0 + 0.4267)

Put a loop around it.

[Titus Edelhofer]

And since the derivative is smaller than one in e.g. [0..2], it converges for any start point from [0..2]. Nice.

Titus

[dpb]

DOH! Headslap! Disconnect between intent and typed, fzero was what I really was thinking of for function, but wrote fsolve instead.

But, the iteration is gooder...

[John D'Errico]

A Titus points out, I had to choose a form for the iteration scheme that will converge.

Had I instead chosen to write it as

x = tan(x) - 0.4267

this would not converge so well or so stably for any set of starting values.

x = -5:5;
xprev = inf(size(x));
iter = 0;
while any(abs(x - xprev) > 1.e-12)
  iter = iter + 1;
  xprev = x;
  x = atan(x + 0.4267);
end

iter
iter =
    32

x
x =
      0.93864      0.93864      0.93864      0.93864      0.93864      0.93864      0.93864      0.93864      0.93864      0.93864      0.93864

So that worked pretty well, converging for essentially any value of x that I chose, and to 12 digits, after only 32 iterations. Some of those converged more rapidly, but I was being lazy when I wrote it.

x = -5:5;
xprev = inf(size(x));
iter = 0;
for iter = 1:100
  iter = iter + 1;
  xprev = x;
  x = tan(x) - 0.4267;
end

iter =
   101
x =
    -2.6874      -2.0619     -0.79099      -3.0194      -0.8124     -0.10595      -1.1468      -1.4749      -2.2183     -0.71188       1.5208

So, after 100 iterations with the "wrong" form, essentially none of the start points have converged. Even after 1000 iterations of the latter loop, none of them converged.

The trick is to choose a form that will robustly converge. In general, fzero is usually simpler. Let the code do the thinking for you.

But, if you want more precision, better yer is just...

syms x
vpasolve(tan(x)-x==0.4267,1)
ans =
0.93864428424672204276667386408592

[Titus Edelhofer]

Yes, you chose the right form. In fact, since the derivative of your iteration function is 1/((x+0.4267)^2 +1 ) is always between zero and one, your iteration will converge for any starting point.

Answer 2

[dpb]

Start with

 doc fsolve

[SUHAS GUPTA THUNUGUNTLA]

Can you please help me with the code to solve for this problem?

Thanks, Suhas

[Jan de Wilde]

rewrite the formula to tan(x)-x-0.4267 = 0 and solve with: fzero('tan(x)-x-0.4267',1)

[Titus Edelhofer]

Now that the full solution is there, one more suggestion is to use anonymous functions to call fzero:

fzero(@(x) tan(x)-x-0.4267, 1)

Titus