Complex Analysis in Polynomial Interpolation

4.3.2 Complex Analysis in Polynomial Interpolation

We shall encounter multivalued functions: the logarithm and the square root. For each of these we choose that branch which is positive for large positive values of the argument z. They will appear in such contexts that we can then keep them nonambiguous by forbidding z to cross the interval [−1, 1]. (We can, however, allow z to approach that interval.)

We first consider the general problem of interpolation of an analytic function, at an arbitrary sequence of nodes u 1 ,u 2 ,...,u n in C. Multiple nodes are allowed. Set

M n (z) = (z − u 1 )(z −u 2 ) · · · (z − u n ), z, u j ∈ C. Let D be a simply connected open domain in C that contains the point u and the nodes.

The interpolation problem is to find the polynomial p ∗ ∈P n , that is determined by the

386Chapter 4. Interpolation and Approximation conditions p ∗ (u j ) = f (u j ) , j = 1 : n, or the appropriate Hermite interpolation problem in

the case of multiple nodes. We know that p ∗ depends linearly on f . In other words, there exists a linear mapping L n from some appropriate function space so that p ∗ =L n f . Assume that f is an analytic function in the closure of D, perhaps except for a finite number of poles p. A pole must not be a node. Recall the elementary identity (4.2.20) in Example 4.2.3,

, (4.3.14) z −u

M n (z)(z − u)

j =1

which is valid also for multiple nodes. Introduce the linear operator K n ,

1 M n (u)f (z)

(K n f )(u) = dz, (4.3.15)

2πi ∂D M n (z)(z − u)

multiply the above identity by f (z)/(2πi), and integrate along the boundary of D to get

dz + (K n f )(u) . (4.3.16)

First, assume that f has no poles in D. By applying the residue theorem to the first two integrals, we note that the equation has the same structure as Newton’s unique interpolation formula with exact remainder (4.2.6),

f (u) = M j −1 (u) [u 1 ,...,u j ]f + M n (u) [u 1 ,u 2 ,...,u n ,u ]f.

j =1

Matching terms in the last two formulas we obtain

(f −L n f )(u) = (K n f )(u),

and, if f is analytic in D, also the formula for the divided-difference,

1 f (z) dz

[u 1 ,u 2 ,...,u j ]f =

. (4.3.17) 2πi ∂D (z −u 1 ) · · · (z − u j )

From this we obtain an upper bound for the divided-difference (

([u |f (z)|

max z

1 ,u 2 ,...,u j

]f ∈∂D (=

, (4.3.18) 2π min z ∈∂D |(z − u 1 ) · · · (z − u j ) |

where L = ∂D |dz|. If there are poles p ∈ D with residues res f (p) , we must add p res f (p)/(z − p) to the left-hand side of (4.3.17) and p res f (p) j M j −1 (u)/M j (p) to the right-hand side. By (4.3.14) this is, however, equivalent to subtracting

res f (p)M n (u)/M n (p)(p − u)

from the right-hand side. This yields the following theorem.

4.3. Generalizations and Applications 387

Theorem 4.3.3.

Assume that f is analytic in the closure of the open region D, perhaps except for a finite number of poles, p ∈ D, with residues res f (p). D also contains the interpolation

points u 1 ,u 2 ,...,u n , as well as the point u. Multiple nodes are allowed. The point u and Then the interpolation error can be expressed as a complex integral,

res f (p) (f −L n f )(u) = (K n f )(u) −M n (u)

M n (p)(p − u) where K n is defined by (

4.3.15) and the sum (which may be void) is extended over the poles of f in D.

This theorem is valid when the interpolation points u j are in the complex plane, although we shall here mainly apply it to the case when they are located in the interval

[−1, 1]. An important feature is that this expression for the interpolation error requires no knowledge of f (n) (z) .

For the case of distinct nodes we have, by the residue theorem,

n (u)

(K n f )(u) =

f (u j ) + f (u), (4.3.20)

(u j − u)M n (u j )

where the sum, with reversed sign, is Lagrange’s form of the interpolation polynomial L n f (u) .

Example 4.3.4 (Chebyshev Interpolation). In Chebyshev interpolation on the interval [−1, 1], the nodes are the zeros of the

Chebyshev polynomials T n (u) , and

M n (u) =2 1−n T n (u).

Recall the notation and results in the discussion of Chebyshev expansions in Sec. 3.2.3. In this example we assume that f (u) is analytic in D. We shall, by means of the integral K n f , show that it yields almost as accurate an approximation as the first n terms of the Chebyshev expansion of the same function. Let D = E R , where E R is the ellipse with foci at ±1, and where R is the sum of the semiaxis. Let z ∈ ∂E R and u ∈ [−1, 1]. Then |T n (u) | ≤ 1, and it can be shown (Problem 4.3.13) that

|dz| ≤ 2πa R .

∂E R

If we assume that |f (z)| ≤ M R for z ∈ ∂E R , a straightforward estimation of the line integral (4.3.20) gives

1 2 1−n M R 2πa R |f (u) − (L n f )(u) | = |(K n f )(u) |≤ 2π

2 1−n 1 2 (R n −R −n )(a R − 1)

2M R a R R −n

. (4.3.21) ( 1−R −2n )(a R − 1)

388 Chapter 4. Interpolation and Approximation We see that the interpolation error converges at the same exponential rate as the truncation

error of the Chebyshev expansion (see Sec. 3.2.3). If f (z) has a singularity arbitrarily close to the interval [−1, 1], then R ≈ 1, and the exponential convergence rate will be very poor.

This above analysis can be extended to the case of multiple poles by including higher derivatives of f . This yields the general Lagrange formula valid also when there are interpolation points of multiplicity greater than one:

Clearly, when r i = 1 for all i = 1 : m, this formula equals the usual Lagrange interpolation formula. It can be written in the simpler form

n r i −1

p(u) =

f (k) (u i )L ik (u),

i =1 k =0

where L ik (u) are generalized Lagrange polynomials. These can be defined starting from the auxiliary polynomials

r (u j −u

i ) & u −u j

l ik (u) = , i = 1 : m, k =0:r i − 1. k !

u i −u j

j =1

Set L i,r i −1 =l i,r i −1 , i = 1 : m, and form recursively

r i −1

L (ν) ik (u) =l ik (u) − l ik (u i )L i,ν (u), k =r i − 2 : −1 : 0.

ν =k+1

It can be shown by induction that

L (σ ) ik (u j ) = ! 1 if i = j and k = σ, 0 otherwise.

Hence the L ik are indeed the appropriate polynomials.

Example 4.3.5.

When r i = 2, i = 1 : n, the Hermite interpolating polynomial is the polynomial of degree less than n = 2m, which agrees with f (u) and f ′ (u) at u = u i , i = 1 : m. We have

p(u) = (f (u i )L i 0 (u) +f ′ (u i )L i 1 (u)),

i =1

where L ik (u) can be written in the form L

i 1 (u) = (u − u i )l i (u) , L i 0 (u) = (1 − 2l ′ i (u i )(u −u i ))l i (u) ,

4.3. Generalizations and Applications 389 and l i (u) , i = 1 : m, are the elementary Lagrange polynomials:

r (u j −u

i ) k & u −u j

l ik (u) = , i = 1 : n, k =0:r i − 1. k !

u i −u j

j =1