LECTURE 5 HERMITE INTERPOLATING POLYNOMIALS

1 CE 30125 - LECTURE 5p. 5 HERMITE INTERPOLATING POLYNOMIALS So far we have considered Lagrange interpolation schemes which fit an degreepolynomial to data or interpolation points All these Lagrange interpolation methods discussed had the general form: Fitting the data points meant requiring the INTERPOLATING polynomial to be equal to thefunctional values at the data points:, NthN1+ f1 f0 fN xN x0 f2 x1 x2 xgx aixii0=N =gx aoa1xa2x2a3x3 aNxN++ + ++=gxi fi=i0N =CE 30125 - LECTURE 5p. HERMITE interpolation : Develop an INTERPOLATING polynomial which equals the func-tion and its derivatives up to order at data points. Therefore we require that constraints constraints : constraints We have a total of constraints We need to set up a general polynomial which is of degree (number of constraints must equal the number of unknowns in the interpolatingpolynomial).

2 PthN1+ f0 , f0 ,.. ,f0 xN x0 x1 x(1)(P) f1 , f1 ,.. ,f1 (1)(P) fN , fN ,.. ,fN (1)(P)gxi fi=i0N =N1+ g1 xi fi1 =i0N =N1+ gp xi fip =i0N =N1+ p1+ N1+ p1+ N1+ 1 CE 30125 - LECTURE 5p. Setting up a polynomial with a total of unknowns: Procedure to develop HERMITE interpolation : Set up the INTERPOLATING polynomial Implement constraints Solve for unknown coefficients, , , Note that Lagrange interpolation is a special case of HERMITE interpolation (, derivatives are matched). It is also possible to set up specialized HERMITE interpolation functions which do notinclude all functional and/or derivative values at all nodes There may be some missing functional or derivative values at certain nodes This lowers the degree of the INTERPOLATING + N1+ gx aixii0=p1+ N1+ 1 =aii0=p1+ N1+ 1 p0=CE 30125 - LECTURE 5p. HERMITE interpolation Develop a two data point HERMITE interpolation function which passes through the func-tion and its first derivative for the interval [0, 1].

3 Therefore and . We must impose constraint equations (match function and its derivativeat two data points). Therefore we require a 3rd degree xofofo1 x1+1f1f11 p1=N1+2=11+ 2 4=gx aoa1xa2x2a3x3++ +=g1 x a12a2x3a3x2++=CE 30125 - LECTURE 5p. Application of constraints Constraint equations may be written in matrix form:g0 fo=aofo=g1 f1=aoa1a2a3+++f1=g1 0 fo1 =a1fo1 =g1 1 f11 =a12a23a3++f11 =1 0 0 01 1 1 10 1 0 00 1 2 3aoa1a2a3fof1fo1 f11 =CE 30125 - LECTURE 5p. Solve for Therefore Checking to ensure that the constraints are satisfied:aoa1a2a3 aofo=a1fo1 =a23f13fo f11 2fo1 =a32f1 2fofo1 f11 +++=gx fofo1 x3f13fof11 2fo1 x22f1 2fofo1 f11 +++ x3+++=gx g0 fo=g1 f1=g1 0 fo1 =g1 1 f11 =CE 30125 - LECTURE 5p. We note that can be re-written such that the functional and derivative values arefactored out: can be expressed in generic form as: Each basis function is a third degree polynomial associated with the function at data point associated with the function at data point associated with the first derivative at data point associated with the first derivative at data point gx gx fo2x33x2 1+ f12x3 3x2+ fo1 ++x32x2 x+ f11 x3x2 +=gx gx fo ox f1 1x fo1 ox f11 ++ + 1x = ox 2x33x2 1+ xo 1x 2x3 3x2+ x1 ox x32x2 x+ xo 1x x3x2 x1CE 30125 - LECTURE 5p.

4 The cubic HERMITE basis functions vary with x as: Therefore we can define 2 separate functions associated with each data point. Each is athird degree polynomial. NOW WE NEED 2 NODES2 FUNCTIONS PER NODE4 DEGREES OFFREEDOM PER FUNCTION = 16 CONSTRAINTS. Each of these functions satisfies the following constraints 0 (x) x0 x1 x0 x1 1 (x) x 0 (x) 1 (x) x CE 30125 - LECTURE 5p. 8 constraints on the functions themselves for to match the specified functionalvalues 8 constraints on the derivatives of the functions for to match the specifiedderivative values 1000010000100001gx gx fo ox f1 1x fo1 ox f11 ++ + 1x =x ox 1x ox 1x xo0=x11=g1 x g1 x fo o1 x f1 11 x fo1 o1 x f11 ++ + 11 x =x o1 x 11 x o1 x 11 x xo0=x11=CE 30125 - LECTURE 5p. Mathematically these 16 constraints can be expressed as Kronecker Delta Therefore an alternative method for setting up is to associate a basis functionwith each functional value and the various derivative values at each data point.

5 Each of these basis functions is a polynomial of degree . There will be basis functions. We must set up constraints. ixj ij= ixj 0=ij 01 = i1 xj 0= i1 xj ij=ij 01 = ij0 ij 1 i = j =gx p1+ N1+ 1 p1+ N1+ p1+ N1+ 2CE 30125 - LECTURE 5p. HERMITE interpolation Using Basis Functions In general, HERMITE interpolation can be set up as: We must satisfy the constraints: .. gx ix fii0=N ix fi1 i0=N ix fip i0=N +++=gxj fj=j0N =g1 xj fj1 =j0N =gp xj fjp =j0N =CE 30125 - LECTURE 5p. In order to satisfy We require the following constraints: .. gxj fj= ixj fii0=N ixj fi1 i0=N ixj fip i0=N +++fj= ixj ij0 ij 1 i = j ==ij 0N = ixj 0=ij 0N = ixj 0=ij 0N =CE 30125 - LECTURE 5p. In order to satisfy: We require the following constraints.

6 G1 xj fj1 =j0N = i1 i0=N xj fi i1 xj fi1 i1 xj fip i0=N ++i0=N +fj1 = i1 xj 0=ij 0N = i1 xj ij=ij 0N = i1 xj 0=ij 0N =CE 30125 - LECTURE 5p. In order to satisfy the pth derivative conditions: We require the following constraints: .. gp xj fjp =j0N = ip i0=N xj fi ip xj fi1 ip xj fip i0=N ++i0=N +fjp = ip xj 0=ij 0N = ip xj 0=ij 0N = ip xj ij=ij 0N =CE 30125 - LECTURE 5p. Each set of basis functions has the general form .. EXTRAPOLATION Use an INTERPOLATING function outside of the range within which the data points lie Extrapolation must always be used with caution How was interpolation established? What is the behavior of the function? How far away from the interval is the point extrapolated? ix aijxjj0=p1+ N1+ 1 =i0N = ix bijxjj0=p1+ N1+ 1 =i0N = ix tijxjj0=p1+ N1+ 1 =i0N =CE 30125 - LECTURE 5p.

7 OF LECTURE 5 Normalized Chebyshev POLYNOMIALS are polynomial functions whose maximum ampli-tude is minimized over a given interval. If we select the roots of the degree Chebyshev polynomial as data (or interpola-tion) points for a degree polynomial interpolation formula ( Lagrange), we willhave minimized the maximum error over the interval as far as we can. If Chebyshev roots are used as data points: The polynomial terms in the error expression actually equal the degreeChebyshev polynomial and its maximum value on the interval is a minimum ascompared to using any other set of interpolation points. degree polynomial interpolation error is also more evenly distributed whenChebyshev roots are used as data points as opposed to using evenly spaced pointssince in the latter case the error is increased near the ends of the interval anddecreased within the + thNthecx 1N1+ ------------------ N1+fN1+ =N1th+NthCE 30125 - LECTURE 5p.

8 HERMITE interpolation passes through the function and its first derivatives at datapoints. This results in a polynomial function of degree . Extrapolation is the use of an INTERPOLATING formula for locations which do not lie withinthe +p1+ N1+ 1