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However, most root-finding algorithms do not guarantee that they will find all roots of a function, and if such an algorithm does not find any root, that does not necessarily mean that no root exists. Most numerical root-finding methods are iterative methods, producing a sequence of numbers that ideally converges towards a root as a limit.
The first complete root-isolation procedure results of Sturm's theorem (1829), which expresses the number of real roots in an interval in terms of the number of sign variations of the values of a sequence of polynomials, called Sturm's sequence, at the ends of the interval.
For example, when testing if the given interval [40 ,60) overlaps the intervals in the tree shown above, we see that it does not overlap the interval [20, 36) in the root, but since the root's low value (20) is less than the sought high value (60), we must search the right subtree. The left subtree's maximum high of 41 exceeds the sought low ...
A better way is to the divide the range into intervals halfway between the squares. So any number between 25 and halfway to 36, which is 30.5, estimate 5; any number greater than 30.5 up to 36, estimate 6. [Note 3] The procedure only requires a little arithmetic to find a boundary number in the middle of two products from the multiplication ...
Given real numbers x and y, integers m and n and the set of integers, floor and ceiling may be defined by the equations ⌊ ⌋ = {}, ⌈ ⌉ = {}. Since there is exactly one integer in a half-open interval of length one, for any real number x, there are unique integers m and n satisfying the equation
If the multiplicity m of the root is finite then g(x) = f(x) / f ′ (x) will have a root at the same location with multiplicity 1. Applying Newton's method to find the root of g(x) recovers quadratic convergence in many cases although it generally involves the second derivative of f(x).
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b k is the current iterate, i.e., the current guess for the root of f. a k is the "contrapoint," i.e., a point such that f(a k) and f(b k) have opposite signs, so the interval [a k, b k] contains the solution. Furthermore, |f(b k)| should be less than or equal to |f(a k)|, so that b k is a better guess for the unknown solution than a k.