From: boost (boost_at_[hidden])
Date: 2001-04-24 16:29:26
On Tuesday 24 April 2001 19:19, you wrote:
> > You will need just one function call to check if x has a
> > zero in the interval -1..1:
> > if f( interval<double>(-1,1) ) contains no zero then it is proven that
> > there is no zero.
> But isn't that just a matter of redefining the problem away?
> When you write f(interval<double>(-1, 1)), what you really mean
> is "the ordered pair (a, b) such that a is less than or equal
> to f(x) for all x in [-1, 1] and b is greater than or equal to
> f(x) for all x in [-1, 1]".
> Certainly if you find such an (a, b), and if you know that f
> is continuous, then you know whether or not f has a zero in
> [-1, 1]. But you've turned a root-finding problem into a
> minimax problem, which isn't necessarily easier.
o.k., I assumed that f is continuous, but I can just bisection
the inteval and check [-1,0] and [0,1] for zeros. Maybe the
example is not so great, but the method works even for
'ugly' function which have zeros due to some small dips.
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