Finding the multiple zeros within a prescribed interval

I wish to solve the nonlinar function:

=0

within a prescribed interval, say (0,100] say, I'm aware of using an annonymous function and using fzero or fsolve, but how do I get say multiple solutions?

Answer 1

[Torsten]

deltax = 1e-4;
xright = 100;
n = floor(xright/pi);
fun = @(x)tan(x)-x;
for i=1:n
  left = (2*i-1)*pi/2.0 + deltax;
  right = (2*i+1)*pi/2.0 - deltax;    
  sol(i) = fzero(fun,[left right]);
end
sol
fun(sol)

[Matthew Hunt]

So basically you're splitting the domain up into smaller domains and checking for roots in each one?

[Torsten]

Yes, with the knowledge that there is exactly one root in the interval ((2*k-1)*pi/2:(2*k+1)*pi/2)

[Matthew Hunt]

That's assuming quite a bit. Can you do it if you take that assumption away?

[Torsten]

No, the strategy to find all zeros of a function in a specified interval will always depend on the behaviour of the function itself. So no general guideline can be given.

Of course you can start at x0=0.01, evaluate the function at x=x0+deltax, x=x0+2*deltax and so on until you get a sign change. Then you can call "fzero" with the interval of the last two x-values as search interval, save the root you get and continue with this strategy until you reach the right endpoint of the search interval. But this will be quite time-consuming, and - depending on deltax - it is still not ensured that you really find all zeros.

[Matthew Hunt]

I see where the error is. You can't tell where the solutions will be, so the use of fzero is limited. I'm beginning to understand now. Is there another way?

I know that the definition of tan is from -pi/2 tp pi/2 and is pi periodic, so I search on the intervals: [0,pi/2), (pi/2,3*pi/2), (3*pi/2, 5*pi/2) and so on?

[Matt J]

No, the strategy to find all zeros of a function in a specified interval will always depend on the behaviour of the function itself. So no general guideline can be given.

Imagine, for example, that you were instead trying to find all roots of contained in the interval [0,a]. No matter what you choose, there would always be infinite roots in the interval.

[Torsten]

@Matthew Hunt:

You know that tan(x) -x -> -Inf for x->2*(k-1)*pi/2 from the right and tan(x) - x -> +Inf for x->2*(k+1)*pi/2 from the left. So there must be a root in the interval 2*(k-1)*pi/2 : 2*(k+1)*pi/2. Plotting the function tan(x) - x you can see that there is exactly one root in this interval. This explains my code and the fact that it captures all roots in a specified interval.