2 Interpolation

Chapter 2 InterpolationConsider a family of functions of a single variable x , 0(;,,),nx a a φ having n +1 parameters 0,,n a a ,whose values characterize the individual functions in this family. The interpolation problem for φ consists of determining these parameters i a so that for n +1 given real or complex pairs of numbers(,),0,,,i i x f i n = with i k x x ≠ for i k ≠,0(;,,)n x a a φ =i f , 0,,,i n =holds. We will call the pairs (,)i i x f support points, the locations i x support abscissas, and the values i f support ordinates. Occasionally, the values of derivatives of φ are also prescribed.The above is a linear interpolation problem if φ depends linearly on the parameters i a :0(;,,)n x a a φ ≡0011()()()n n a x a x a x φφφ+++ ,This class of problems includes the classical one of polynomial interpolation,0(;,,)n x a a φ ≡2012nn a a x a x a x ++++ ,as well as trigonometric interpolation,0(;,,)n x a a φ ≡2012 xi xi nxi n a a e a ea e ++++ (21i =-). In the past, polynomial interpolation was frequently used to interpolate function values gathered from tables. The availability of modern computing machinery has reduced the need for extensive table lookups. However, polynomial interpolation is also important as the basis of several types of numerical integration formulas in general use. In a more modern development, polynomial and rational interpolation are employed in the construc tion of “extrapolation methods” for integration, differential equations,and related problems .Trigonometric interpolation is used extensively for the numerical Fourier analysis of time series and cyclic phenomena in general. In this context, the so-cal led “fast Fourier transforms” are particularly important and successful.The class of linear interpolation problems also contains spline interpolation. In the special case of cubic splines, the functions φ are assumed to be twice continuously differentiable for 0[,]n x x x ∈and to coincide with some cubic polynomial on every subinterval 1[,]i i x x + of a given partition 01n x x x <<< .Spline interpolation is a fairly new development growing importance. It provides a valuable tool for representing empirical curves and for approximating complicated mathematical functions. It is increasingly used when dealing with ordinary or partial differential equations.Two nonlinear interpolation schemes are of importance: rational interpolation,00(;,,,,,)n m x a a b b φ ≡0101n n mm a a x a x b b x b x++++++ ,and exponential interpolation,00(;,,,,,)n n x a a φλλ ≡0101 n xxxn a ea ea eλλλ+++ .Rational interpolation plays a role in the process of best approximating a given function by one which is readily evaluated on a digital computer. Exponential interpolation is used, for instance, in the analysis of radioactive decay .Interpolation is a basic tool for the approximation of given function. For a comprehensive discussion of these and related topics consult Davis(1965).2.1 Interpolation by polynomials2.1.1 Theoretical Foundation: The interpolation Formula of LagrangeIn what follows, we denote by n ∏ the set of all real or complex polynomials p whose degrees do not exceed n:P(x )= 2012n n a a x a x a x ++++ ,(2.1.1.1) Theorem For n +1 arbitrary support points(,),0,,,i i x f i n =i k x x ≠ for i k ≠There exists a unique polynomial n P ∈∏ with()i i P x f =, i =0, 1, …, n .PROOF . Uniqueness: For any two polynomials P 1, P 2∈n ∏ with12()()i i i P x P x f ==, i =0, 1, …, n .the polynomial P :=P 1-P 2∈n ∏ has degree at most n , and it has at least n +1 different zeros, namely x i , i =0,1,…,n . P must therefore vanish identically, and P 1=P 2.Existence: We will construct the interpolating polynomial P explicitly with the help of polynomials L i ∈n ∏, i =0, 1, …, n , for which (2.1.1.2)1()0i k ik if i kL x if i k δ⋅⋅=⎧==⎨⋅⋅≠⎩The following Lagrange polynomials satisfy the above conditions: (2.1.1.3) 011011()()()()():()()()()i i n i i i i i i i n x x x x x x x x L x x x x x x x x x -+-+----≡----≡()()'()i i x x x x ωω- with 0():()ni i x x x ω=≡-∏.Note that our proof so far shows that the Lagrange polynomials are uniquely determined by (2.1.1.2).The solution P of the interpolation problem can now be expressed directly in terms of the polynomials L i , leading to the Lagrange interpolation formula:(2.1.1.4) 0()()nnnk i i i i i k i i kk x x P x f L x f x x ==≠=-≡≡-∑∑∏The above interpolation formula shows that that the coefficients of P depend linearly on the support ordinates f i .While theoretically important, Lagrange’s formulais ,in general, not as suitable for actual calculations as some other methods to be described below, particularly for large numbers n of support points. Lagrange’s formula may , however, be useful in some situations in which many interpolation problems are to be solved for the same support abscissas x i , i =0,1,…, n , but different sets of support ordinates f i , i =0,1,…, n .EXAMPLE . Give for n =2:Wanted: P (2),where P ∈2∏, P (x i )=f i for i =0, 1, 2. Solution: 0(1)(3)()(01)(03)x x L x --≡-- , 1(1)(3)()(01)(03)x x L x --≡--, 2(0)(1)()(30)(31)x x L x --≡--.0121110(2)1(2)3(2)2(2)1312333P L L L -=⋅+⋅+⋅=⋅+⋅+⋅=.2.1.3 Newton’s Interpolation Formula: Divided DifferencesNeville’s algor ithm is geared towards determining interpolating values rather than polynomials. If the interpolating polynomial itself is needed, or if one wants to find interpolating values for several arguments j ξ simultaneously , then Newton’s interpolation formula is to be preferred. Here we write the interpolating polynomial P ∈n ∏, P (x i )=f i , i =0,1, …, n , in the form (2.1.3.1)010********()()()()()()()()n n n P x P x a a x x a x x x x a x x x x x x -≡≡+-+--++---Note that the evaluation of [2.1.3] for x =ξ may be done recursively as indicated by the following expression:112100()((())())()n n n n P a x a x a x a ξξξξ---=-+-++-+ .This requires fewer operations than evaluating (2.1.3.1) term by term. It correspondsto the so-called Horner scheme for evaluating polynomials which are given in the usual form, i.e. in the representation (2.1.3.1) is well suited for evaluation.It remains to determine the coefficients are in (2.1.3.1).In principle, they can be calculated successively from000()f P x a ==,110110()()f P x a a x x ==+-,22012022021()()()()f P x a a x x a x x x x ==+-+--,This can be done with n divisions and n(n-1) multiplications. These is, however, a better way , which requires only n(n+1)/2 divisions and which produces useful intermediate results.Observe that the two polynomials 01()ki i i P x and 011()k i i i P x - differ by apolynomial of degree k with k zeros 0i x ,1i x , ,1k ix -,since cot h polynomialsinterpolate the corresponding support points. Therefore there exists a unique coefficient(2.1.3.2) 01ki i i f , k =0, 1, …,n ,such that(2.1.3.3) 010110101()()()()()kkk kii ii i iiii ii i P x P x fxx x x x x --≡+---From this and from the identity 0()i i P x f ≡ it follows immediately that (2.1.3.4) is aNewton representation of the partially interpolating polynomial 011()k i i i P x - . Thecoefficients (2.1.3.2) are called k th divided differences.The recursion (2.1.2.1) for the partially interpolating polynomials translates into the recursion(2.1.3.5) 0101101kk kk i i i i i i i i i i i f f f x x --=-for the divided differences, since by (2.1.3.3), 1ki i f and 011k i i i f - are the coefficientsof the highest terms of the polynomials 0i f and 011k i i i P - ,respectively . The aboverecursion starts for k =0 with the given support ordinates i f , i =0, …, n . It can be used in various ways for calculating divided differences 0i f ,01ki i i p , ,01ni i i f , which thencharacterize the desired interpolating polynomial 0101kkjj j i i i f f = .Because the polynomial 01ki i i p is uniquely determined by the support points it interpolates [Theorem(2.1.1.1)],the polynomial is invariant to any permutation of the indices 0i ,1i , ,k i ,and so is its coefficient 01ki i i f of k x .Thus:(2.1.3.6) The divided differences 01ki i i f are invariant to permutations of the indices0i , 1i , , k i : If(0j ,1j , k j )=(0s i ,sk i , ,sk i )is a permutation of the indices 0i ,1i , ,k i ,then0101k kj j j i i i f f =If we choose to calculate the divided differences in analogy to Neville’s method —instead of, say , Aitken’s method --then we are led to the following tableau, called the divided-difference scheme: (2.1.3.7)The entries in the second column are of the form100110f f f x x -=- , 211221f f f x x -=- ,Those in the third column,120101220f f f x x -=- , 231212331f f f x x -=- ,Clearly ,010********()()()()()()n n n P x P x f f x x f x x x x x x -≡≡+-++---is the desired solution to the interpolation problem at hand. The coefficients of the above expansion are found in the top descending diagonal of the divided-difference scheme (2.1.3.7).EXAMPLE . With the numbers of the example in section 2.1.1 and 2.1.2 , we have000111012122201251316232x f f x f f f x f ======-=-==0120125()12(0)(0)(1),6510(2)((21)2)(20)1.63P x x x x P =+----=--+-+=Instead of building the divided-difference scheme column by column, one might want to start with the upper left corner and add successive ascending diagonal rows. This amounts to adding new support points one at a time after having interpolated the previous ones. In the following ALGOL procedure, the entries in an ascending diagonal of (2.1.3.7) are found ,after each increase of i ,in the top portion of array t , and the first i coefficients 01i f are found in array a. for i:=0 step 1 until n do begin t[i]:=f[i];for j:=i-1 step -1 until 0 do t[j]:=(t[j+1]-t[j])/(x[i]-x[j]); a[i]:=t[0] end ;Afterwards, the interpolating polynomial (2.1.3.1) may be evaluated for any desiredargument z : p:=a[n];for i:=n-1 step -1 until 0 do p:=p×(z-x[i])+a[i];Some Newton representations of the same polynomial are numerically more trustworthy to evaluate than others. Choosing the permutation so that1,0,1,,1,ik ik x x k n ξξ--≥-=-Dampens the error (see section 1.3) during the Horner evaluation of (2.1.3.8) 0001000()()()()().nn ni i ii iiiii i p x p x f f xx fx x x x -≡≡+-++--All Newton representations of the above kind can be found in the single divided-difference scheme which arises if the support arguments x i , i =0, …, n , are ordered by size:1i i x x +< for i =0, …, n -1. Then the preferred sequence of indices01,,,k i i iis such that each index k i is “adjacent” to some previous index. Moreprecisely, either k i =min{l i |0≤l<k }-1 or k i =max{l i |0≤l<k }+1. Therefore the coefficients of (2.1.3.8) are found along a zigzag path —instead of the upper descending diagonal —of the divided-difference scheme. Starting with 0i f ,the pathproceeds to the upper right neighbor if 1k k i i -<, or to the lower right neighbor if1k k i i ->.EXAMPLE . In the previous example, a preferred sequence for 2ξ= is0121,2,0i i i ===.The corresponding path in the divided difference scheme is indicated below:000111012122201251316232x f f x f f f x f ======-=-==The desired Newton representation is :12015()3(1)(1)(3),26p x x x x ≡-----1205110(2)((23))(21)3.623p =----+=Frequently , the support ordinates i f are the values ()i i f x f = of a given function f (x ), which one wants to approximate by interpolation. In this case, the decided differences may be considered as multivariate functions of the support arguments i x , and are historically written as0[,,].k i i f x xThese functions satisfy (2.1.3.5). For instance, 00[](),f x f x ≡1010011010[][]()()[,]f x f x f x f x f x x x x x x --≡≡--,120121012002101220102021[,][,]()()()()()()[,,]()()()f x x f x x f x x x f x x x f x x x f x x x x x x x x x x x ----+-≡≡----1011010[,,][,,,][,,,]k k k k f x x f x x x f x x x x x --≡-Also, (2.1.3.6) gives immediately:(2.1.3.9)Theorem , The divided differences 0,,ki i x x are symmetric functions oftheir arguments, i.e., they are invariant to permutations of the 0,,ki i x x .If the function f(x) is itself a polynomial, then we have (2.1.3.10) Theorem , if f (x ) is a polynomial of degree N , then01[,,,]0k f x x x =for k >N .PROOF . Because of the unique solvability of the interpolation problem (Theorem 2.1.1.1),,,()()o k p x f x ≡ for k N ≥.The coefficient of k x in ,,()o k p x musttherefore vanish for k N >. This coefficient, however, is given by 01[,,,]k f x x x according to (2.1.3.3).EXAMPLE . 2()f x x =.If the function f(x) is sufficiently often differentiable, then its divided differences0[,,]k f x x can also be defined if some of the arguments ix coincide. for instance,if f(x) has a derivative at 0x , then it makes sense for certain purposes to define000[,]:'().f x x f x =For a corresponding modification of the divided-difference scheme (2.1.3.7) see Section 2.1.5 on Hermite interpolation.2.1.4 The error in Polynomial InterpolationOnce again we consider a given function f(x) and certain of its values (),0,1,,i i f f x i n==which are to be interpolated. We wish to ask how well the interpolating polynomial0()()n p x p x ≡with(),0,1,,i i p x f i n ==Reproduces ()f x for arguments different from the support arguments i x . The error()()f x p x -,Where ,0,1,i x x i ≠=can clearly become arbitrarily large for suitable functions f unless some restrictions are imposed on f .Under certain conditions, however, it is possible to bound the error. We have, for instance:(2.1.4.1) Theorem . If the function f has an (n +1)st derivative, then for every argument x there exists a number ξ in the smallest interval 0[,,,]n I x x x which contains x and all support abscissas i x ,satisfying(1)01()()()(),(1)!n n x ff x p x n ωξ+-=+where01()()()().n x x x x x x x ω≡---PROOF . Let 01():()n p x p x ≡ be the polynomial which interpolates the function ati x , i =0, 1, …, n, and suppose i x x ≠(for i x x = there is nothing to show ). We canfind a constant K such that the function():()()()F x f x p x K x ω≡--vanishes for x x =:()0.F x =Consequently , F (x ) has at least the n +2 zeros0,,,n x x xin the interval 0[,,,]n I x x x . By Rolle’s theorem, applied repeatedly , F ’(x ) has at least n +1 zeros in the above interval, F ’’(x) at least n zeros, and finally (1)()n F x + at least one zero 0[,,,].n I x x x ξ∈ Since (1)()0.n p x +≡(1)(1)()()(1)!0n n FfK n ξξ++=-+=or (1)()(1)!n fK n ξ+=+.This proves the proposition(1)()()()()().(1)!n x f x p x K x fn ωωξ+-==+A different error term can be derived from Newton’s interpolation formula (see Section 2.1.3):010*******()()[][,]()[,,,]()().n n n p x p x f x f x x x x f x x x x x x x -≡≡+-++--Here 01[,,,]k f x x x are the divided differences of the given function f . If in addition to the n +1 support points(,):(),i i i i x f f f x =0,1,,,i n = we introduce an (n +2)nd support point1111(,)::,:(),n n n n x f x x f f x ++++==where ,0,,,i x x i n ≠= then by Newton’s formula0100()()()[,,,](),n n n f x p x p x f x x x x ω+==+or(2.1.4.2) 00()()()[,,,].n n f x p x x f x x x ω-=The difference on the left-hand side appears in Theorem (2.1.4.1), and since()0x ω≠, we must have(1)0()[,,,](1)!n n ff x x x n ξ+=+ for some 0[,,,].n I x x x ξ∈This also yields(2.1.4.3) ()0()[,,]!n n ff x x n ξ=for some 0[,,,].n I x x x ξ∈which relates derivatives and divided differences.EXAMPLE .()sin f x x =: ,0,1,2,3,4,5,510i ix i n π===015sin sin ()()()(),720x p x x x x x x x ξ--≡---015()1sin ()()()().720720x x p x x x x x x x ω-≤---=We end this section with two brief warnings, one against trusting the interpolating polynomial outside of 0[,,]n I x x and one against expecting too much of polynomial interpolation inside 0[,,]n I x x .In the exterior of the interval 0[,,]n I x x , the value of ()x ω in Theorem (2.1.4.1) grows very fast. The use of the interpolation polynomial p for approximating f at some location outside the interval 0[,,]n I x x --called extrapolation —should be avoided if possible.Within 0[,,]n I x x on the other hand, it should not be assumed that finer and finer and finer samplings of the function f will lead to better and better approximations through interpolation.Consider a real function f which is infinitely often differentiable in a given interval [a,b]. To every interval partition 01{}n a x x x b ==<<<= there exists an interpolating polynomial n p ∈∏ with ()i i p x f = for i x ∈ .A sequence o finterval partitions()()()1{}m m m m m n a x x x b ==<<<=Gives rise to a sequence of interpolating polynomials mp .One might expect the polynomials mp to converge toward f if the fineness()()1:m ax m m m i iix x +=-Of the partitions tends to 0 as .m →∞In general this is not true. For example, it has been shown for the functions21(),[,][5,5],()[.][0,1],1f x a b or f x a b x≡=-≡=+That the polynomials mp do not converge pointwise to f for arbitrarily fineuniform partitions m ,()()/,0,,.m ia ib a m i m x=+-=2.1.5 Hermite InterpolationConsider the real numbers i ξ,()k iy,0,1,,1,0,1,,i k n i m =-= ,with01m ξξξ<<< .The Hermite interpolation problem for these data consists of determining a polynomial P whose degree does not exceed n , wheren +1:=0mi i n =∑,and which satisfies the following interpolation conditions: (2.1.5.1) ()()()k k i ipyξ=, 0,1,,1,0,1,,i k n i m =-= .This problem differs from the usual interpolation problem for polynomials in that it prescribes at each support abscissa i ξ not only the value but also the first1i n -derivatives of the desired polynomial. The polynomial interpolation of Section2.1.1 is the special case 1,0,1,,.i n i m == There are exactly1inn =+∑ conditions (2.1.5.1) for the n +1 coefficients of theinterpolating polynomials, leading us to expect that the Hermit interpolation problem can be solved uniquely:(2.1.5.2) Thoerem . For arbitrary numbers 01m ξξξ<<< ,()k i y ,0,1,,1i k n =- ,0,1,,i m= , there exists precisely one polynomialnp ∈∏, n+1:=0mi i n =∑,which satisfies (2.1.5.1).PROOF . We first show uniqueness. Consider the difference polynomial12():()()Q x p x p x =-of two polynomials 12,n p p ∈∏for which (2.1.5.1) holds.Since()()0,k i Qξ= 0,1,,1,0,1,ik n i m =-=,i ξ is at least an i n --fold root of Q ,so that Q has altogether1inn =+∑ roots, eachcounted according to its multiplicity . Thus Q must vanish identically, since its degree is less than n +1.Existence is a consequence of uniqueness: For (2.1.5.1) is a system of n linear equations for n unknown coefficients j c of 01().n n p x c c x c x =+++ The matrix of this system is not singular, because of the uniqueness of its solutions. Hence the linear system (2.1.5.1) has a unique solution for arbitrary right-hand sides()k iy.Hermite interpolating polynomials can be given explicitly in a form analogo us to the interpolation formula of Lagrange (2.1.1.4) .The polynomial n p ∈∏given by(2.1.5.3) 1()00()()i n mk ik ii k p x L x -===∑∑satisfies (2.1.5.1). The polynomials ik n L ∈∏are generalized Lagrange polynomials. They are defined as follows: Staring with the auxiliary polynomials 01():(),0,0!j k mnji ik i j i jj x x l i m k n k ξξξξ=≠--=≤≤≤<-∏[compare (2.1.1.3)], put.1.1():(),0,1,,.i i i n i n L x l x i m --==and recursively for 2,3,,0,i i k n n =--1()1():()()().i n v ik k i iv ikv k L x l x L x lξ-=+=-∑By induction()1,()0ik j if i j and k L otherw iseσσξ==⎧=⎨⎩Thus P in (2.1.5.3) is indeed the desired Hermite interpolating polynomial.In order to describe alternative methods for determining P , it will be useful to represent the data (),,0,1,,,0,1,,1,k i i i i m k n y ξ==- in a somewhat different formas a sequence 0,,{(,)}n i i i n F x f == of n +1 pairs of numbers. The pairs0000001111(,),(,),,(,),(,),,(,)n n n n n n x f x f x f x f x f -- .of n F denote consecutively the pairs0(1)(1)(0)(1)(0)00000011(,),(,),,(,),(,),,(,).mn n m m y y y y y ξξξξξ--Note that 01n x x x ≤≤≤ and that the number i ξ occurs exactly i n times in the sequence 0,,{}.i i n x =EXAMPLE 1.Suppose 011,2,3m n n === and(0)(1)000,1,2;y y ξ==-=-(0)(1)(2)11111,0,1040;yyyξ====This problem is described by the sequence 40,,4{(,)}i i i F x f == :001122(,)(0,1),(,)(0,2),(,)(1,0),x f x f x f =-=-=3344(,)(1,10),(,)(1,40).x f x f ==Given Hermite interpolation problem, it uniquely determines a sequence n F , as above. Conversely, every sequence 0,,{(,)}n i i i n F x f == of n+1 pairs of numbers with01nx x x ≤≤≤ determines a Hermite interpolation problem, which will be referredto simply as n F .It also will be convenient to denote by0jx x -The polynomials(2.1.5.4) 00:1,x x -=0011:()()()jj x x x x x x x x --=---of degree j .Our next goal is to represent the polynomial P which interpolates n F in Newton form [compare (2.1.3.1)]:(2.1.5.5) 2010200()nn p x a a x x a x x a x x =+-+-+++-And to determine the coefficients i a which the help again of divided differences (2.1.5.6) 01[,,,],k k a f x x x = 0,1,,k n =However, the recursive definition (2.1.3.5) of the divided differences has to be modified because there may be repetitions among the support abscissas01nx x x ≤≤≤ . For instance, if 01x x =,then the divided difference 01[,]f x x canno longer be defined as 0110([][])/()f x f x x x --.The extension of the definition of divided differences to the case of repeats arguments involves transition to a limit .To this end, let 01n ζζζ<<<Be mutually distinct support abscissas, and consider the divided differences[,,]i i k f ζζ+which belong to the function ():()f x p x =,where the polynomial p isthe solution of the Hermite interpolation problem n F .These divided differences are now well defined by the recursion (2.1.3.5),if we let :()i i f p ζ= initially. Therefore, and by (2.1.3.5),(2.1.5.7a) 0010(),:[,,,],njj j j j p x a x a f ζζζζ==-=∑(2.1.5.7b) [](),i i f p ζζ= (2.1.5.7c)12111[,,,][,,,][,,,],i i i k i i i k i i i k i k if f f ζζζζζζζζζζζ+++++-+++-=-for 0,1,,,1,,.i n k n i ==- Since 01n x x x ≤≤≤ , all limits11[,,,]:lim[,,,]j j i i i k i i i k x i j i kf x x x f ζζζζ++++→≤≤+=exist provided they exist for indices i , k with 1i i i k x x x ++=== . The latter follows from (2.1.4.3) , which yields(2.1.5.8) ()11lim [,,,]()!j jk i i i k i x i j i kf px k ζζζζ++→≤≤+=if 1i i i k x x x ++=== . We now denote by ()0r r i =≥ the smallest index such that1r r i x x x +=== .Then due to the interpolation properties of P which respect to n F .()()()()k k i r r kpx px f +==.So that by (2.1.5.8)()1[,,,]!r i k i i i k f f x x x k +++=if 1i i i k x x x ++===In the limit j j x ζ→,(2.1.5.7) becomes(2.1.5.9a) 0010(),:[,,,]njj j j j j p x a x a f x x x ζ==-=∑(2.1.5.9b) ()1[,,,]!r i k i i i k f f x x x k +++=if i i k x x +=(2.1.5.9c) 12111[,,,][,,,][,,,]:,i i i k i i i k i i i k i k if x x x f x x x f x x x x x +++++-+++-=-otherwise. (Note that 01n x x x ≤≤≤ has been assumed.) These formulas now permit a recursive calculation of the divided differences and thereby the coefficients j a of the interpolating polynomial P in Newton form,EXAMPLE 2. We illustrate the calculation of the divided differences with the data ofExample 1(m =1,012,3n n ==):{}4(0,1),(0,2),(1,0),(1,10),(1,40),F =--.The following difference scheme results:00x =01[]f x *-= 012[,]f x x *-=10x =11[]f x *-= 0123[,,]f x x x = 121[,]f x x = 036[,,]f x x =21x =20[]f x *= 1239[,,]f x x x = 045[,,]f x x = 2310[,]f x x *= 1411[,,]f x x =31x =30[]f x *=23420[,,]f x x x *= 3410[,]f x x *=41x =40[]f x *= The entries marked * have been calculated using (2.1.5.9b) rather than (2.1.5.9c) .The coefficients of the Hermite interpolating polynomial can be found in the upper diagonal of the difference scheme:23400002222()123651236(1)5(1).p x x x x x x x x x x x x x x x =---+-+-+-=--++-+-The interpolation error which is incurred by Hermite interpolation can be estimated in the same fashion as for the usual interpolation by polynomials. In particular, the proof of the following theorem is entirely analogous to the proof of Theorem (2.1.4.1):(2.1.5.10) Theorem . Let the real function f be n +1 times differentiable on the interval [a , b ] ,and consider m +1 support abscissas [,],i a b ξ∈01m ξξξ<< .If the polynomial p (x ) is of degree at most n ,1mii nn ==+∑And satisfies the interpolation conditions()()()(),0,1,,1,0,1,,,k k i i i pfk n i m ξξ==-=Then to every [,]x a b ∈ there exists 0[,,,]m I x ξξξ∈ such that(1)()()()(),(1)!n x ff x p x n ωξ+-=+where101():()()()mn n n m x x x x ωξξξ=--- .Hermite interpolation is frequently used to approximate a given real function f by a piecewise polynomial function ϕ .Given a partition01m a bξξξ=<<<=Of an interval [a , b ], the corresponding Hermite function space ()v H is defined as consisting of all functions :[,]a b ϕ→ with the following properties: (2.1.5.11)(a) 1[,]:v C a b ϕ-∈The (v -1)st derivative of ϕ exists and is continuous on [a , b ]. (b) 21|:i v I ϕ-∈∏On each subinterval 1:[,]i i i I ξξ+=, 0,1,,1i m =- , ϕ agrees with a polynomial of degree at most 2v -1.Thus the function ϕ consists of polynomial pieces of degree 2v -1 or less which are v -1 times differentiable at the “Knots” i ξ,in order to approximate a given real function1[,]v f Ca b -∈ by a function ()v H ϕ∈ ,we choose the componentpolynomials |i i p I ϕ= of ϕ so that 21i v p -∈∏ and so that the Hermiteinterpolation conditions()()()()11()(),()(),0,1,,1,k k k k ii i ii i p fp fk v ξξξξ++===-are satisfied.Under the more stringent condition 2[,]v f C a b ∈, Theorem (2.1.5.10) provides a bound to the interpolation error for i x I ∈ which arises if the component polynomiali preplaces f :1(2)21(2)2()()()()m ax ()(2)!)m ax ().2(2)!iivi i v i I vi i v vI x x f x p x fv fv ξξξξξξξξ+∈+∈---≤-≤⋅Combining these results for 0,1,,i m = gives for the function ()v H ϕ∈ , which was defined earlier,(2.1.5.13)2(2)2[,]1:max ()(),2(2)!vv vx a b f f x x fv ϕϕ∞∞∈-=-≤where 101max i i i m ξξ+≤≤-=- i s the “fineness” of the partition .The approximation error goes to zero with the 2v th power of the fineness j if we consider a sequence of partitions j of the interval [a , b ] with 0j → .Contrast this with the case of ordinary polynomial interpolation, where the approximation error not necessarily go to zero as 0j → (Section 2.1.4).Ciarlet, Schultz, and V arga (1967) were able to show that also the first v derivatives of ϕ are good approximations to the corresponding derivatives of f : (2.1.5.14) 1()()(2)1()()()()()m a x ()!(22)!iv ki i k k kv ii iI x x fx p x f k v k ξξξξξξ-++∈---≤-- For all ,0,1,,,0,1,,1,i x I k v i m ∈==- and therefore (2.1.5.15)2()()(2)222!(22)!v kk k v v kf fk v k ϕ--∞∞-≤-For 0,1,,.k v =2.4 Interpolation by Spline Functions。