ODE and num. math.: Roots of functions (N) [lectures]
c pHabala 2016
DEN: Roots of functions numerically
Definition.
By a root of a function f we mean any number r such that f (r) = 0.
Fact.
Let f be a function and r ∈ IR its root. Assume that there is a neighborhood U of r and m1 > 0 such
that f is differentiable on U and m1 ≤ |f 0 | on U .
Then for r̂ ∈ U we have the estimate |r − r̂| ≤ m11 |f (r̂)|.
Definition.
Consider a certain iterating method for finding a root of a function. We say that it is a method of
order q, or that it has error of order q, where q > 0, if it satisfies one of the following conditions:
• For every sequence {xk } generated by this method that converges to a root r, there is some K > 0
such that the inequality |r − xk+1 | ≤ K|r − xk |q is true for k sufficiently large.
• For every sequence {xk } generated by this method that converges to a root r there is a an upper
estimate ek for |r − xk | and K > 0 such that the inequality ek+1 ≤ K eqk is true for k sufficiently large.
The latter is sometimes called the R-order of convergence.
Fact.
Let f be a function on an interval [a, b]. If f (a) · f (b) < 0 (i.e. f (a) and f (b) have opposite signs) and
f is continuous on [a, b], then f must have a root in the interval [a, b].
Algorithm
(bisection method for finding root of f )
Given: a function f continuous on interval [a, b] and a tolerance ε.
Assumption: f (a) and f (b) have opposite signs.
0. Set x0 = a, y0 = b. Let k = 0.
1. Assumption: f (xk ) and f (yk ) have opposite signs.
Let mk = 21 (xk + yk ).
2. If f (mk ) = 0 or |yk − xk | < ε then algorithm stops, output is mk . Otherwise:
If f (xk ) and f (mk ) have opposite signs, set xk+1 = xk , yk+1 = mk , increase k by one
and go back to step 1.
If f (mk ) and f (yk ) have opposite signs, set xk+1 = mk , yk+1 = yk , increase k by one
and go back to step 1.
Theorem.
Assume that a function f continuous on [a, b] satisfies f (a)·f (b) < 0. Then the sequence {mk } generated
by the bisection method with starting values x0 = a, y0 = b converges to a root of f .
The bisection method is of linear order.
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ODE and num. math.: Roots of functions (N) [lectures]
Algorithm
c pHabala 2016
(regula falsi method for finding root of a function f )
Given: a function f continuous on interval [a, b] and a tolerance ε.
Assumption: f (a) and f (b) have opposite signs.
0. Set x0 = a, y0 = b. Let k = 0.
1. Assumption: f (xk ) and f (yk ) have opposite signs.
(yk )−yk f (xk )
Let rk = xk ff (y
.
k )−f (xk )
2. If |rk − rk−1 | < ε or |f (rk )| < ε then algorithm stops, output is rk . Otherwise:
If f (xk ) and f (rk ) have opposite signs, set xk+1 = xk , yk+1 = rk , increase k by one
and go back to step 1.
If f (rk ) and f (yk ) have opposite signs, set xk+1 = rk , yk+1 = yk , increase k by one
and go back to step 1.
Algorithm
(secant method for finding root of a function f )
Given: a function f continuous on IR and a tolerance ε.
0. Choose x0 , x1 . Let k = 1.
k )−xk f (xk−1 )
1. Let xk+1 = xk−1f f(x(xk )−f
.
(xk−1 )
If |xk=1 − xk | < ε or |f (xk+1 )| < ε then algorithm stops, output is xk+1 . Otherwise:
Otherwise increase k by one and go back to step 1.
Theorem.
Let f be a function continuous on an interval [a, b]. Assume that it is concave-up or concave-down on
[a, b] and f (a)f (b) < 0. Then for arbitrary values x0 , x1 chosen so that f (x0 ) > 0, f (x1 ) > 0 for the
concave-up case, or f (x0 ) < 0, f (x1 ) < 0 for the concave-down case, the sequence {xk } generated by
the secant method converges to a root r ∈ (a, b) of f .
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ODE and num. math.: Roots of functions (N) [lectures]
c pHabala 2016
Theorem.
Let f be a function twice continuously differentiable on some neighborhood U of its root r. Let {xk }
be a sequence generated by the secant method such that xk → r. If the root r is simple,
then there is a
√
1+ 5
α
constant K and N ∈ IN such that |r − xk+1 | ≤ K|r − xk | for k ≥ N , where α = 2 .
Algorithm
(Newton method for finding root of a function f )
Given: a function f continuous on IR and a tolerance ε.
0. Choose x0 . Let k = 0.
k)
1. Let xk+1 = xk − ff0(x
(xk ) .
If |xk+1 − xk | < ε or |f (xk+1 )| < ε the the algorithm stops, output is xk+1 .
Otherwise increase k by one and go back to step 1.
Theorem.
Let f be a function on an interval [a, b] such that f (a) · f (b) < 0. Assume that f is twice continuously
differentiable on (a, b) and f 0 6= 0, f 00 6= 0 on (a, b).
If x0 ∈ (a, b) is chosen so that f (x0 ) · f 00 (x0 ) > 0, then the sequence {xn } generated by the Newton
methodou converges to a root r ∈ [a, b] of f .
Theorem.
Let r be a root of a function f that is twice continuously differentiable on some neighborhood U of r
and for which there are m1 , M2 > 0 such that |f 0 | ≥ m1 and |f 00 | ≤ M2 on U . Consider a sequence {xk }
2
2
generated by the Newton method. Then for all xk , xk+1 ∈ U we have estimates |r − xk+1 | ≤ M
m1 |r − xk |
M2
and |r − xk+1 | ≤ 2m
|xk+1 − xk |2 .
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3
ODE and num. math.: Roots of functions (N) [lectures]
c pHabala 2016
Definition.
Let ϕ be a function. By a fixed point of ϕ we mean any number xf satisfying ϕ(xf ) = xf .
Theorem.
Let ϕ be a function, x0 ∈ IR and xk+1 = ϕ(xk ) for k ∈ IN . If xk → xf and ϕ is continuous at xf , then
ϕ(xf ) = xf .
Definition.
Let ϕ be a function on an interval I. We say that it is contractive there, or that it is a contraction,
if there exists q < 1 such that for all x, y ∈ I we have
|ϕ(x) − ϕ(y)| ≤ q · |x − y|.
Theorem. (Banach’s fixed-point theorem)
Let ϕ be a contractive function on I = [a, b] with coefficient q such that ϕ[I] ⊆ I. Then there exists
exactly one solution xf of the equation ϕ(x) = x in I. Moreover, for all choices of x0 ∈ I the sequence
given by xk+1 = ϕ(xk ) converges to xf and satisfies
|xf − xk+1 | ≤ q|xf − xk |
|xf − xk+1 | ≤
and
q
|xk+1 − xk |.
1−q
Theorem.
Assume that function ϕ defined on an interval I has a continuous derivative on the interior I O of I. If
there is q < 1 such that |ϕ0 (t)| ≤ q on I O , then ϕ is a contraction on I with coefficient q.
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