Counting zeros

Given a function f(x), Is there a simple way to get MATLAB to determine the number of zeros it has in some arbitrary interval [a,b], without necessarily finding them explicitly? Perhaps counting the number of sign changes or whatnot?

Thanks.

Answer 1

[Andrew Newell]

In general, no. You can try doing it by counting sign changes, but there is no guarantee you'll catch all the answers. However, we might be able to suggest a better approach if you post the code for f(x).

[Richard]

Thanks, I am just wondering about it, I don't really have a particular function in mind. Suppose we proceed by counting sign changes, how might we make MATLAB scan the interval? Also, is it possible for MATLAB to locate zeros in ascending order given an interval? Thanks again!

[Andrew Newell]

Not knowing anything about the function, I'd suggest picking some regularly spaced points in [a,b] and counting sign changes, then add more points and see if anything changes.

Answer 2

[Dr. Seis]

I know you don't need to find the roots explicitly, but this will work:

p = [1 -6 -72 -27]; % polynomial coeffs for f(x) = x^3 - 6*x^2 - 72*x - 27
a = -20; % lower interval bound
b = 0; % upper interval bound
r = roots(p); % roots
n = sum((a <= r).*(r <= b)); % number of roots in interval [a,b]
disp(n);

If you really want to search, then:

f = @(x)x.^3 - 6*x.^2 - 72*x - 27; % Define function
n = 0; % Intialize n to 0
dx = 0.01; % set increment in x
for i = a+dx : dx : b
  if sign(f(i)) ~= sign(f(i-dx))
      n=n+1;
  end
end
disp(n);

Though, I am sure there must be a much more savvy way of doing this.

Answer 3

[Walter Roberson]

For arbitrary functions, there is no way of doing this, not even by counting sign changes. If f(x) is negative and f(x+t) is positive then the best you can say is that f might be discontinuous or might have at least one zero in the interval x to x+t .

For certain classes of functions, such as polynomials, if you are given the symbolic form, it becomes possible to find the roots within the limits of round-off.