Chapter 3 Interpolation methods Peter­Wolfgang Gr¨aber Systems Analysis in Water Management CHAPTER 3. INTERPOLATION METHODS Problem: Some measured values (dependent variable) are dependent on independent variables in one -, two -, three- or four-dimensional space measurement, and generally they are represented by the three space coordinates (depending upon coordinate system e.g. xn� yn� zn orr n� αn� zn orr n� αn� ϑn (see 2 vector analysis)) and the timet n. We have a discontinuous value tables in this case. For the one dimensional case e.g.: independent dependend variable variable x0 y0 f(x 0) x1 y1 f(x 1) . xn yn f(x n) The pointsx 0� x1� ...� xn are called supporting points, and the pointsy 0� y1� ...� yn are the basic or supporting values. If we are loo#ing for function values, whose arguments lie within the range (x0� xn), we name it interpolation In contrast if the function values we are looking for lie outside the range (x0� xn) we call this extrapolation. $y interpolation or extrapolation we try to &nd a continuous functionw p(x), which reflects the orig inal functiony n f(x n) as exactly as possible (see &gure (.)). It is always assumed that the interpolation function only matches the original function on the supporting points. !he accuracy for the interval in between, e.g. the matching of both functions, depends on the number and the distribution of the supporting points. According to the sampling theorem the quantisation error increases proportionally to the rise of the function. Note: No interpolation algorithm can be used as replacement for an enlargement ot the measured value density. By means of an interpolation algorithm one receives in each case approximated values. ,- Figure 3.): representation of the discontinuous measured data acquisition ,) CHAPTER 3. INTERPOLATION METHODS Example for application of interpolation: The pollutant concentrationC(x), which runs from a refuse dump, is measured at the pointsx 0� x1� x2 (see &gure (.2). The pollutant concentration at the pointx F l, which cause the danger by 'owing in the river, is to be estimated by interpolation. * conclusion is to be given whether this value exceeds the limiting value. Figure 3.2: representation of an interpolation problem x0 C0 f(x 0) x1 C1 f(x 1) xF l / x2 C2 f(x 2) For the solution of this problem an interpolation functionw p(x) is to seek for as 0replacement0 for the functionC n f(x n). This function should ful&l the following conditions: wi p(x i) C i i )� . � n (3.)) ∀ 72 i.e. w0 p(x 0) C 0 w1 p(x 1) C 1 (3.2) . wn p(x n) C n Then it is supposed that the intermediate values of the functionw p(x) are a good appro%imation of the intermediate values of the functionC n f(x n). For the determination of the functionw p(x) different interpolation methods can be used. We differentiate thereby one- and multi-dimensional procedures. !he mul- tidimensional methods play an important role in connection with the geographical information systems �GIS) and are also often applied in connection with geostatis tics. In the following some methods will be introduced in connection with water economical .uestions. 2olynomial interpolation • Spline interpolation (peace wise polynomial interpolation) • Kriging method • ,( 3.1 Polynomial interpolation In this methodp(x) has the form of an alg ebraic polynomial of ordern: 2 n w p(x) a 0 4a 1x4a 2x 4 ...4a nx (3.() The advantage of this method is that the intermediate values can be computed as easily as possible. Based on a value table withn4) pairs of variates ma ximally ann-th order polynomial can be exactly determined: n k y : p(x) ak x (3.5) · k �=0 with the property: n k y(xi) p(x i) ak x w i (3.6) ≈ · i k �=0 This polynomial is the interpolation polynomial to the given system of interpolation supporting points. Normally we loo# for polynomials of lower order (n (), which &t together t he pairs ≤ of values at least by pairs: p(x) a 0 4a 1x linear interpolation 2 p(x) a 0 4a 1x4a 2x .uadratic interpolation 2 3 p(x) a 0 4a 1x4a 2x 4a 3x cubic interpolation The application of polynomials with higher order ma#es the calculations more di7cult and leads to very large 'uctuations. From the different display formats for the polynomials follow di1erent interpolation methods for the determination of the coefficientsa i of ann-th order polynomial. * ll this different methods lead to the same polynomial. Systems Analysis in Water Management Peter­Wolfgang Gr¨aber 3.1. Polynomial interpolation This interpolation methods are: analytical power function • Lagrange • Aiken • Newton • 3.1.1 Analytical power function This method assumes that for each supporting point the polynomialw p(x) ful&ls the conditiony(x i) p(x i). In this case we get for then4) supporting points a system ofn 4 ) equations withn 4 ) un#nown coefficien tsa 0� . � an. 2 n a0 4a 1x0 4a 2x0 4 ...4a nx0 y 0 (3.8) 2 n a0 4a 1x1 4a 2x1 4 ...4a nx1 y 1 . 2 n a0 4a 1xn 4a 2xn 4 ...4a nxn y n This equation system can be written as a matri% equation: ! " # · with the matrices: 2 n )x 0 x0 x 0 a0 y0 � ··· � � 2 n )x 1 x1 x 1 a1 y1 ! ··· " # . .. . . . . . 2 n )x n xn x n an yn ··· ,6 CHAPTER 3. INTERPOLATION METHODS The matri%! and the vector# on the right side repre sent the #nown coefficients, whereby" is the solution vector. ! he linear equation system can be solved by all the #nown methods (see section 1.(, solution of equations systems, page 22) The determinant of this linear equation system is: (x1 x 0) (x 2 x 0) (x 3 x 0) ... (x n x 0) � − · − · − · · − · x x n x x x x ... x x ) 0 0 ( 2 1) ( 3 1) ( n 1) � ··· � · − · − · · − · � � � n� �)x 1 x 1 � (x3 x 2) ... (x n x 2) D � ··· � · − · · − · (3.,) � � �. .. � . �. � . � � � � � n� �)x n x n� (xn 1 x n 2) (x n x n 2) � ··· � · − − − · − − · � � � � � � (xn x n 1) · − − and is named the andermond determinant. Since all supporting points are (must be) pairwise different, isD - and the linear � equation system is explicit solvable. There is only one polynomial of the ordern which ful&ls the propert yy i f(x i) p(x i) with the coefficients (see section ).2.( determinants, page )9): Da0 Da1 Dan a � a � �a n (3.:) 0 D 1 D ··· D With this coefficients the interpolation polynomial is: 2 n y(x) p(x) a 4a x4a x 4 4a n x ≈ 0 1 · 2 · ··· · The interpolation value at the placex P is: 2 n y(xP ) p(x P ) a 4a x P 4a x 4 4a n x ≈ 0 1 · 2 · P ··· · P Although the beginning of this method is very simple, the &nal determination of the interpolation polynomial requires a relative large computation, particularly if a great number of supporting points are to be ta#en in account. ,8 3.1. Polynomial interpolation Example for the interpolation $ith the analytical po$er function method: Find a .uadratic polynomial by using the values of the following table and calculate 1 1 the valuey f( 2 ) at the placex 2 . x - ) 2 y - ) - Since only three supporting points are given, the polynomial can only be a second order polynomial. * .uadratic polynomial has the form 2 p(x) a 0 4a 1x4a 2x It must be: yi p(x i) 2 yi a 0 4a 1xi 4a 2xi p(-) - a 4a -4a - - a - ⇒ 0 1 · 2 · ⇒ 0 p()) ) a 4a )4a ) 2 ) a 4a ) ⇒ 0 1 · 2 · ⇒ 1 2 p(2) - a 4a 24a 2 2 - 2a 4 5a - ⇒ 0 1 · 2 · ⇒ 1 2 From this three equations follows: a0 - a1 2 a ) 2 − Thus the interpolation polynomial is: p(x) 2x x 2 − 1 With this function the value at the placex 2 can be computed: ) ) ) ) 2 ( f p 2 2 ≈ 2 · 2 − 2 5 � � � � � � ,, CHAPTER 3. INTERPOLATION METHODS 3.1.2 Lagrange interpolation formula Lagrange wrote the interpolation in the following form: y(xP ) p(x P ) L (xP ) y 4L (xP ) y 4...4L n(xP ) y n (3.9) ≈ 0 · 0 1 · 1 · With the Lagrange interpolation no closed analytical functions are computed, but only single valuesp(x P ) for each interpolation pointx P . !hereby the coe7cientsL i(x) of the interpolation valuesy i (fori -�)� . � n) aren-th order polynomials o fx P . These are computed from the supporting pointsx i and are called the Lagrange polynomials. The Lagrange polynomials ofn-th order are: (xP x 1)(xP x 2) (x P x n) L0(x) − − ··· − (x x )(x x ) (x x n) 0 − 1 0 − 2 ··· 0 − (xP x 0)(xP x 2) (x P x n) L1(x) − − ··· − (x x )(x x ) (x x n) 1 − 0 1 − 2 ··· 1 − . (3.)-) (xP x 0)(xP x 1) (x P x i 1)(xP x i+1) (x P x n) Li(x) − − ··· − − − ··· − (xi x 0)(xi x 1) (x i x i 1)(xi x i+1) (x i x n) − − ··· − − − ··· − . (xP x 0)(xP x 1)(xP x 2) (x P x n 1) Ln(x) − − − ··· − − (xn x 0)(xn x 1) (x n x n 1) − − ··· − − Thus the Lagrange interpolation polynomial is: y f(x P ) p(x P ) L (xP )y 4L (xP )y 4...4L n(xP )yn ≈ 0 0 1 1 (xP x 1)(xP x 2) (x P x n) y − − ··· − y0 (x x )(x x ) (x x n) 0 − 1 0 − 2 ··· 0 − (xP x 0)(xP x 2) (x P x n) 4 − − ··· − y1 (3.))) (x x )(x x ) (x x n) 1 − 0 1 − 2 ··· 1 − .