An Interval Newton Method

6.3.3 An Interval Newton Method

Sometimes it is desired to compute a tiny interval that is guaranteed to enclose a real simple root x ∗ of f (x), even when rounding errors are taken into account. This can be done using an adaptation of Newton’s method to interval arithmetic method due to Moore [271].

Suppose that the function f (x) is continuously differentiable. Using the notation in Sec. 2.5.3, let f ′ ( [x 0 ]) denote an interval containing f ′ (x) for all x in a finite interval [x 0 ] := [a, b]. Define the Newton operator on N[x] by

f (m)

N( [x]) := m − ′

f ( [x])

where m = mid ([x]) = 1

2 (a + b).

Theorem 6.3.7.

If α ∈ [x] is a zero of f (x), then α ∈ N([x]). If N([x]) ⊆ [x], then f (x) has one and only one zero in N([x]).

Proof. Suppose α is a zero of f (x) in [x]. If 0 ∈ f ′ ( [x]), then N([x]) = [−∞, ∞]. Otherwise, by the mean value theorem

0 = f (α) = f (m) + f ′ (ξ )(α − m), ξ ∈ int [α, m] ⊆ [x].

This implies that α = m − f (m)/f ′ (ξ ) ⊆ N([x]), which proves the first statement. If N([x]) ⊆ [x], then f ′ (

and ξ 2 in [x] such that (m − f (m)/f ′ (ξ 1 )) − a = −f (a)/f ′ (ξ 1 ),

b − (m − f (m)/f ′ (ξ 2 )) = f (b)/f ′ (ξ 2 ).

Because N([x]) ⊆ [a, b], the product of the left sides is positive. But since f ′ (ξ 1 ) and

f ′ (ξ 2 ) have the same sign this means that f (a)f (b) < 0, and f has therefore a zero in [x]. Finally, there cannot be two or more zeros in [x], because then we would have f ′ (c) =

0 for some c ∈ [x]. In the interval Newton method, a starting interval [x 0 ] is chosen, and we compute for

k = 0, 1, 2, . . . a sequence of intervals [x k +1 ] given by

If N([x k ]) ⊂ [x k ] we set [x k +1 ] = N([x k ]) ∩ [x k ]. Otherwise, if N([x k ]) ∩ [x k ] is empty, we know that [x k ] does not contain a root and stop. In neither condition holds we stop,

subdivide the initial interval, and start again. It can be shown that if [x 0 ] does not contain a root, then after a finite number of steps the iteration will stop with an empty interval.

If we are close enough to a zero, then the length of the intervals [x k ] will converge quadratically to zero, just as with the standard Newton method.

6.3. Methods Using Derivatives 647

Example 6.3.4.

Take f (x) = x 2 − 2 and [x 0 ] = [1, 2]. Using the interval Newton method

+1 ] = N([x k ]) ∩ [x k 2 [x ],

, [x k

we obtain the sequence of intervals

[x 1 ] = N([x 0 ]) = 1.5 −

45 ( 45) 2 − 2(32) 2 [x 2 ] = N([x 1 ]) =

32 − 128 [22, 23] The quadratic convergence of the radius of the intervals is evident:

The interval Newton method is well suited to determine all zeros in a given interval. Divide the given interval into subintervals and for each subinterval [x] check whether the

condition N([x]) ⊆ [x] in Theorem 6.3.7 holds. If this is the case, we continue the interval Newton iterations, and if we are close enough the iterations converge toward a root. If the

condition is not satisfied but N([x]) ∩ [x] is empty, then there is no zero in the subinterval and this can be discarded. If the condition fails but N([x])∩[x] is not empty, then subdivide the interval and try again. The calculations can be organized so that we have a queue of intervals waiting to be processed. Intervals may be added or removed from the queue. When the queue is empty we are done.

The above procedure may not always work. Its performance will depend on, among other things, the sharpness of the inclusion of the derivative f ′ ( [x]). The algorithm will fail, for example, in the case of multiple roots where N([x]) = [−∞, ∞].