Shampine, Allen, Pruess - Fundamentals of Numerical Computing (523185), страница 24
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In general, the curvature K(X) of a functionf(x) isIf’W I( ) = (1 + (fyx))2)3’2K xand in this theory the expression is linearized to K(X) E If’(x)I. When (S’(X))~ < 1,the quantity J’F (S”)2 dx can be regarded as a measure of the curvature of the splineS(x). We prove now that in this measure, any smooth interpolating function satisfyingthe type (1) end conditions must have a curvature at least as large as that of S(x). Thisis sometimes referred to as the minimum curvature property of the complete cubicspline. The same result is true for the natural cubic spline when the requirement thatthe interpolant satisfy the type (1) end conditions is dropped.If g is any C2[x1 ,XN] function that interpolates f over (xl, .
. .,Theorem 3.5.xN ) and satisjes the type (1) end conditions, thenJ” (sy2 dx 2 J” (g”)2dx,XlXlwhere S(x) is the complete cubic interpolatory spline. The same inequality holds forg that do not necessarily satisfy the type (1) end conditions when S(x) is the naturalcubic interpolatory spline.Proo$ First observe thatxN(g/’ -S//)2& =JXlxN(g~~)2& _ 2 xN (g/I - S”)S”dt - 1” (s”)2 dt.Jx1JXlXlIf it can be shown that the second integral on the right is zero, thenJ’“Y’)zdt = JxN(g//)2dt-JxN~g”_Sii)2dt < JxN(g”)2dr,XlXlXlXlsince the integral of a nonnegative function is always nonnegative, and we are finished.l-r\ aotohl;ch that thn Ano;rmA ;ntmn+ol ;c ~P,-T\ nntm th,st3.5 SPLINE INTERPOLATION115Figure 3.11 S(v) from Example 3.11.n7.41, and S(9.5) = 9.74. They are in good agreement with the actual values.Example 3.12.rocket aret I >0.00T0.00.35t21.5T>0.70tT16.0Observed values for the thrust (T) versus time (t) curve of a model>0.051.0>0.4016.50.7516.0>0.105.00.4516.0>0.8016.0>0.1515.0>0.5016.0>0.8516.0>0.2033.50.5516.0>0.906.00.2538.0>0.6016.0>0.952.0>0.3033.00.6516.0>1.000.0The 15 values indicated by (>) were used as data.
The resulting complete cubic splineS(x) is graphed in Figure 3.12. Note that the large curvatures near t = 0.40 and t = 0.85are difficult to handle. Values of S at some reserved data points are S(0.25) = 39.1,S(0.35) = 23.6, and S(0.65) = 16.1.nEXERCISES3.20 Derive equation (3.44) from the end condition f'(x n ) = S'(x n ).3.21 If the end conditions S"(x1) = f“(x1) and S"(xN) = f"(xN) are used, what equations should replace (3.43) and (3.44)?3.22 The vapor pressure P of water (in bars) as a function of temperature T (0 C ) is116CHAPTER 3INTERPOLATIONFigure 3.12 S(t) from Example 3.12.0Tp mpk)100.012277500.123380.70112 900.006107400.0737740.47364 80200.023378600.199241.01325 100300.042433700.31166Interpolate these data with the cubic spline S(x).
It is also known that P(5) =0.008721, P(45) = 0.095848, and P(95) = 0.84528. How well does S(x) do atthese points?3.23 The following data give the absorbance of light (A) as a function of wavelength(3L) for vanadyl D-tartrate dimer.hA@)h4 vhA(U> 31250.70038750.769> 45000.170> 32500.572> 40000.83646250.14433750.40041250.750> 47500.183> 35000.382> 42500.53048750.25236250.44943750.315> 50000.350> 37500.560Use the cubic spline S(x) to interpolate the nine indicated (>) data points. Explorethe effects of scaling and shifting the independent variable (x = wavelength) witheach of the following.(a) The data as is.(b) Replace x by x/ 1000 for all inputs.(c) Replace x by (x - 4000)/1000 for all inputs.For each case evaluate S(x) at the remaining noninterpolated wavelengths.
HowIf f is in C4[x1, xN], and S(x) is the complete cubic interpolator-yTheorem 3.6.spline for f with knots {XI < - - - < XN), then for any x in [xl ,XN]If($--WI I &h4M43If’(x) - S’(x) 1 L hh M41 f”(x) - S”(x)1 < ;h2M4,where M4 = max If t4) (x) I for x in [xl ,xN].In contrast to polynomial interpolation, S(x) does converge to f(x) as N + 00 asI - - -_ ^I -r \q - t -I - 1 , , &, , , l_^ _ _ _1I , , , . , & : , . , ,, cC L , ,, , I : - -, l , ,- , . .
- _ * - - _ ^c -C L ,116CHAPTER 3INTERPOLATIONFigure 3.12 S(t) from Example 3.12.TP(T)TP(T)TP(T)o0.006107400.073774800.47364100.012277500.12338900.70112200.023378600.199241001.01325300.042433700.31166Interpolate these data with the cubic spline S(x). It is also known that P(5)=0.008721, P(45) = 0.095848, and P(95) = 0.84528. How well does S(x) do atthese points?3.23 The following data give the absorbance of light (A) as a function of wavelength(λ) for vanadyl D-tartrate dimer.λΑ(λ)λΑ(λ)λΑ(λ)>31250.70038750.769>4500o.170>32500.572>40000.83646250.14433750.40041250.750>47500.183>35000.382>42500.53048750.25236250.44943750.315>50000.350>37500.560Use the cubic spline S(x) to interpolate the nine indicated (>) data points.
Explorethe effects of scaling and shifting the independent variable (x = wavelength) witheach of the following.(a) The data as is.(b) Replace x by x/ 1000 for all inputs.(c) Replace x by (x – 4000)/ 1000 for all inputs.For each case evaluate S(x) at the remaining noninterpolated wavelengths. How3.5 SPLINE INTERPOLATION117well do these values compare with the known absorbances? Does shifting and/orscaling affect the accuracy of S(X)?3.24 Repeat Exercise 3.23 except use P9(x) instead of S(X). Use the method suggestedin Exercise 3.4. What effect does the scaling have on COND?3.25 The absorption of sound (at 20°C, 40% humidity) as a function of frequency, f, isUse the cubic spline S(X) to interpolate the nine indicated (>) points in the following two ways.(a) The data as is.(b) log f versus logA(f)Which seems to be better?3.26 The following table gives values for a property of titanium as a function of temperature T.Compute and plot the cubic spline S(T) for these data (use about 15 interpolatingpoints).
How well does it do?3.27 In performing potentiometric titrations one obtains a potential difference curveplotted against volume of titrant added. The following table gives the measurements for the potentiometric titration of Fe2+ solution with 0.1095N Ce4+ solution using platinum and calomel electrodes.Compute the cubic spline S(X) for these data (use about 15 interpolating points).Graph S(x) or x in [20, 24]. How well does it behave? The physical problem hasexactly one inflection point. Is this true for S(X)?3.28 The potential energy of two or more interacting molecules is called van der Waal’sinteraction energy. A theoretical calculation for two interacting helium atoms hasthe set of energies V(r) for various values of the internuclear distance r givenbelow.
The energy exhibits repulsion (V > 0) for small r and attraction (V < 0)118CHAPTER 3INTERPOLATIONfor larger values of r.Compute the cubic spline S(x) using about 12 interpolating points. How well doesit work?3.29 Modify the routine SVALUE or Spline_value to return S´(x) and S´´(X) as well asS(x).3.30 In [5] a method is given for deducing the diffusion coefficient D for chloroform inpolystyrene from uptake measurements.
Using several assumptions, they arrive atthe quantitywhich can be measured for a number of C0 values. A differentiation with respectto C0 gives an expression for D in terms of the quantityUsing the dataapproximate D for each C0 value by differentiating the appropriate spline fit.3.31 Show that the cubic spline S(X) has a critical point z in [x n ,x n + 1 ], that is, S´(z) = 0,if and only if the following are true:Why is it not sufficient merely to use (i) and the test bnbn+1 = S´(xn)S´(xn+1) < 0?3.32 Show that the cubic spline S(X) has an inflection point z in ( x n ,x n+1 ), that is,S´´(z) = 0, if and only if cncn+1 < 0, in which case z = xn - cn /(3dn).3.33 Use the formula in Exercise 3.31 to find all local minima for the data in Exercise 3.23.3.34 Use the formula in Exercise 3.31 to find the local maximum (near T = 905) forthe data in Exercise 3.26.3.6 INTERPOLATION IN THE PLANE119Figure 3.13 Scattered data in the plane.3.35 For the data in Exercise 3.28 the global minimum at r = re corresponds to stableequilibrium (V’ = 0).
There is also an inflection point (where V´´ = 0) at r = ri .What does S(x) yield for re and ri ? Are the answers reasonable?3.36 Use the formulas in Exercises 3.31 and 3.32 to find the local maximum (nearv = 4.5) and all inflection points for the data in Example 3.5.3.6 INTERPOLATION IN THE PLANEIn this section a few of the ideas involved in interpolating functions of several variablesare taken up. Although the ideas of the case of one variable generalize, there are newdifficulties arising from geometrical considerations. To be specific, only the case oftwo independent variables will be considered.Suppose we have values fi given at distinct points pi for i = 1,.
. . , N in a regionin the x-y plane (see Figure 3.13), and we seek a polynomial in the two variables x andy that interpolates the data. This is easily accomplished in a way similar to Lagrangianan interpolating polynomial is giveninterpolation. If p = (x,y) is a general point inbyprovided that120CHAPTER 3INTERPOLATIONand each φi is a polynomial in x and y. It is easy to verify thatsatisfies the requirements. Thus, it is easy to construct a polynomial in two variables,which interpolates given values at any set of distinct points in the plane.The interpolating polynomial given is not closely analogous to that for one variable because the degree is much higher than the number of nodes.
Unfortunately, thefacts are simply different when there is more than one independent variable. This canbe seen by considering a general quadratic polynomial in two variables (x, y):P(x, y) = a0 + a1x + a2y + a3x2 + a4xy + a5y2.There are six parameters ai . The analog of the result for one variable would be thatthere is a unique polynomial interpolating at six distinct points (xi,yi) in the plane. Foreach node the interpolation condition isInterpolation at six nodes provides six linear equations for the six parameters. Supposethat five of the nodes are {(0,0),(1,0),(0,-1),(-1,0),(0,1)} and the sixth is ( α,β) .In the equation corresponding to each of the first five nodes, the coefficient of a4 iszero.
It is also zero in the sixth equation if α = 0 or if β = 0. For any node ( α,β)of this kind, the system amounts to six equations in only five unknowns. The systemmay or may not have a solution, but if there is a solution, it cannot be unique becausea4 can have any value. In several dimensions the placement of the nodes has a role inquestions of existence and uniqueness that we did not see in one dimension.If we had a choice about the points of interpolation, we might like to work on arectangular grid. By this is meant there are n x-coordinates, x1 < x2 < ··· < xn, andm y-coordinates, y1 < y2 < ··· < ym and the n × m points of interpolation consist ofthe pairs (x i ,y j) for 1 < i < n, 1 < j < m; see Figure 3.14.
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