Homework #6 Brief Solutions 2012

Transcription

1 Homework #6 Brief Solutions %page 95 problem 4 data=[-,;-,;,;4,] data = xk=data(:,);yk=data(:,);s=csfit(xk,yk,-,) %Using the program to find the coefficients S = S=.456*(x+).^ -.456*(x+).^-*(x+)+; S= -.5*(x+).^+.9*(x+).^ *(x+); S=.*(x-).^ -.58*(x-).^+.87*(x-)+; x=-:.:4; y=spline_eval(s(:,4),s(:,),s(:,),s(:,),xk,x); plot(xk,yk, o,x,y) % see plot below %p95 problem 5 Using function [a,b,c,d]=natspline_coeff(knots,data); %% function [a,b,c,d]=natspline_coeff(knots,data); %% Modification of Cheney Kincaid pseudo-code %% The natural spline for interpolating data at the knots a=x<...<x_n=b; %% System of equations is solved using the tridiagonal nature %% WARNING: THIS PROGRAM REQUIRES AT LEAST 4 KNOTS %% INPUT: %% knots - distinct points (t_j) of interpolation as a row vector %% data - data at the interpolation points as a row vector %% OUTPUT: %% a - column vector of constant terms for the spline on [t_j,t_(j+)] %% b - column vector of coefficients of x-t_j for the spline on [t_j,t_(j+)] %% c - column vector of coefficients of (x-t_j)^ for the spline on [t_j,t_(j+)] %% d - column vector of coefficients of (x-t_j)^ for the spline on [t_j,t_(j+)] % determine lengths of the intervals See web page for program, %applying this we obtain the coefficients for the polynomial pieces of the spline [a,b,c,d]=natspline_coeff(xk,yk ) a = b =

2 c = d = % Thus the spline is S=.47*(x+).^ -.47*(x+).^-.47*(x+)+; S= -.47*(x+).^+.48*(x+).^ -.575*(x+); S=.856*(x-).^-.77*(x-).^+.876*(x-)+; naty=spline_eval(a,b,c,d,xk,x); plot(xk,yk, o,x,naty) Figure : The clamped spline of problem 4 p95 on the left and the natural spline of problem 5 p95 on the right. %p95 # (a) Clamped Spline f=@(x) cos(x.^); df=@(x) -*x*sin(x.^); xk=[,sqrt(pi/),sqrt(*pi/),sqrt(5*pi/)]; yk=f(xk); dfa=df(xk());dfb=df(xk(4)); S=csfit(xk,yk,dfa,dfb); S = xx=:.: sqrt(5*pi/); yy=spline_eval(s(:,4),s(:,),s(:,),s(:,),xk,xx); plot(xk,yk, o,xx,yy,xx,f(xx)); % (b) Natural spline

3 [a,b,c,d]=natspline_coeff(xk,yk ) a =.. -. b = c = d = natyy=spline_eval(a,b,c,d,xk,xx); plot(xk,yk, o,xx,natyy,xx,f(xx)); % see below Figure : The clamped spline of problem a p95 on the left and the natural spline of problem b p95 on the right. 5 (a) Using the nodes x = and x =, show that f(x) = x x is its own clamped spline on the interval [, ]. There are several ways to answer this. First, if we assume that the algorithm has shown (which it has) that there is a unique piecewise cubic polynomial, S(x), that satisfies the interpolation conditions S( ) = f( ), S() = f(), S ( ) = f ( ), and S () = f (), then since f itself is a piecewise cubic, it must be equal S. A second way is to essentially reprove everything using the representation of a cubic spline and the given interpolation requirements. The clamped spline has the representation S(x) = s + s (x + ) + s (x + ) + s 4 (x + ). Determine the coefficients by S( ) = f( ) = 8 + = 6 and S ( ) = f ( ) = ( ) = and S() = f() = and S () = f () = : S( ) = 6 gives 6 = s, while S ( ) = gives = s. Thus, S(x) = 6+(x+)+s (x+) +s (x+). Then, S() = yields = 6+ +s 4+s 8 or 4 s + 8 s = 6, while S () = yields = + s + s or 4s + s =. Subtracting 4s + s = from 4 s + 8 s = 6 gives 4s = 4 or s =. Then s = 6.

4 Hence the clamped spline has the form S(x) = 6+(x+) 6(x+) +(x+), or 6+x+ 6x 4x 4+x +6x +x+8 = x+x = f(x). 5 (b) With nodes x =, x =, x = show that f(x) = x x is its own clamped spline on [, ]. Again there are several ways. Using the fact that a cubic polynomial is by definition a piecewise cubic polynomial for any way of breaking the interval in pieces, and that it satisfies the requirements of the clamped spline for f, then by the uniqueness of the clamped spline interpolant we must have f = S. Another way is to reprove this in the special case. This time the clamped spline has two cubic pieces and is of the form S (x) = s, + s, (x + ) + s, (x + ) + s, (x + ) () S (x) = s, + s, x + s, x + s, x () The endpoint conditions at give S( ) = 6 and S ( ) = which as before give s, = 6 and s, =. The interpolation condition S() = f() = gives = S () = s, 4 + s, 8 or s, + s, = 4 () and = S () = s,. The endpoint conditions S() = f() = 8 = 6 yields S () = 6 = s, + 4s, + 8s, or s, + s, + 4s, = (4) and S () = f () = 4 = yields S () = = s, + 4s, + s,, or Putting what we know into () and () gives s, + 4s, + s, = (5) S (x) = 6 + (x + ) + s, (x + ) + s, (x + ), (6) S (x) = s, x + s, x + s, x (7) We still have the spline conditions S () = S () and S () = S () which yield the equations + 4s, + s, = s, (8) s, + s, = s, (9) Equations (), (4) (5), (8) and (9) give you five equations in the 5 remaining unknowns s,, s,, s,, s,, s,. The coefficient matrix and RHS for this system are = This system has solution s, = 6, s, =, s, =, s, =,s, =. Hence S (x) = 6 + (x + ) 6(x + ) + (x + ) = x xas before () S (x) = x + x + x = x x () 4

5 5 (c) This should be the statement that a cubic polynomial is by definition a piecewise cubic polynomial for any way of breaking the interval in pieces, and that it satisfies the requirements of the clamped spline for f, then by the uniqueness of the clamped spline interpolant we must have f = S. You may also do it by comparing the polynomial on [x k, x k+ ] with the B-spline representaton. SplineProblem: Since the spline problem was laid out for you in Matlab, here is how it should have looked with the figures. xti=585+[:49]*; yti=[ ]; yti=[ yti ]; %Plotting the points, see below plot(xti,yti,o) %Computing the interpolating polynomial xx=xti():.:xti(49); ypoly=int_poly(yti,xti,xx); %see int_poly.m for how data is input plot(xx,ypoly,xti,yti,o) %plot below % It is way off near the ends %Computing the natural spline interpolant [a,b,c,d]=spline_coeff(xti,yti);% see spline_coeff.m ynatspl=spline_eval(a,b,c,d,xti,xx); % see spline_coeff.m plot(xx,ynatspl,xti,yti,o) %plot below MUCH BETTER %A comparison where we interpolate at fewer points knots=xti(:6:49); data=yti(:6:49); yyintpoly=int_poly(data,knots,xx); [a,b,c,d]=spline_coeff(knots,data); yynatspl=spline_eval(a,b,c,d,knots,xx); plot(xx,yyintpoly,xti,yti,o,xx,yynatspl) %See plot below %Again, the spline is very much better. %Computing error estimates max(abs(yti-int_poly(data,knots,xti))) ans = max(abs(yti-spline_eval(a,b,c,d,knots,xti))) ans = % and the least squares error sum(abs(yti-int_poly(data,knots,xti)).^) ans = sum(abs(yti-spline_eval(a,b,c,d,knots,xti)).^) ans =

6 % The spline is superior in both cases. % Looking for a true least squares approximation by splines % First discover the B-spline basis of independent functions % Setting up the knots knots=[knots(),knots(),knots(),knots(),knots()]; knots=[knots(),knots(),knots(),knots(),knots()]; knots=[knots(),knots(),knots(),knots(),knots(4)]; knots4=[knots(),knots(),knots(),knots(4),knots(5)]; knots5=[knots(),knots(),knots(4),knots(5),knots(6)]; knots6=[knots(),knots(4),knots(5),knots(6),knots(7)]; knots7=[knots(4),knots(5),knots(6),knots(7),knots(8)]; knots8=[knots(5),knots(6),knots(7),knots(8),knots(9)]; knots9=[knots(6),knots(7),knots(8),knots(9),knots(9)]; knots=[knots(7),knots(8),knots(9),knots(9),knots(9)]; knots=[knots(8),knots(9),knots(9),knots(9),knots(9)]; %Using the program cubicbspl that gives the B-spline based on the knots f=@(x) cubicbspl(knots,x); f=@(x) cubicbspl(knots,x); f=@(x) cubicbspl(knots,x); f4=@(x) cubicbspl(knots4,x); f5=@(x) cubicbspl(knots5,x); f6=@(x) cubicbspl(knots6,x); f7=@(x) cubicbspl(knots7,x); f8=@(x) cubicbspl(knots8,x); f9=@(x) cubicbspl(knots9,x); f=@(x) cubicbspl(knots,x); f=@(x) cubicbspl(knots,x); % Plot the B-splines to get a better feel for them plot(xx,f(xx),xx,f(xx),xx,f(xx),xx,f4(xx),xx,f5(xx),xx,f6(xx),xx,f7(xx),xx,f8(xx),xx,f9(xx),xx,f(xx % see below %Plot the sum to show that the sum of the B-splines is identically (not much to see!). sumbsplines= f(xx)+f(xx)+f(xx)+f4(xx)+f5(xx)+f6(xx)+f7(xx)+f8(xx)+f9(xx)+f(xx)+f(xx); plot(xx,sumbsplines) max(sumbsplines) ans =.8 %small round off error min(sumbsplines) ans = %well, not quite % Doing the least squares fit using the B-splines as the linearly independent functions F=@(x) [f(x);f(x);f(x);f4(x);f5(x);f6(x);f7(x);f8(x);f9(x);f(x);f(x)]; FC=[]; for j=:49; y=f(xti(j));fc=[fc;y]; end; FC %This gives the matrix called F in the book and F_C in class. FR=FC; %its transpose B=FR*yti; %The right hand side of the matrix form of the system of normal equations for least squares ap 6

7 FRFC=FR*FC %The coefficient matrix in the matrix form of the system of normal equations for least square %Notice that it has a banded property. This is because of the short support of the B-splines. FRFC = Columns through Columns 5 through Columns 9 through C=FRFC\B %This uses Matlab to solve the system C =

8 C()*f(x)+C()*f(x)+C()*f(x)+C(4)*f4(x)+C(5)*f5(x)+C(6)*f6(x)+C(7)*f7(x)+... C(8)*f8(x)+C(9)*f9(x)+C()*f(x)+C()*f(x); % Making the best linear combination of the basis plot(xx,bestlsspline(xx),xti,yti,o) %plot the best least square spline fit with the data See below %%Plotting the natural spline and the best least squares spline and comparing errors plot(xx,bestlsspline(xx),xti,yti,o,xx,yynatspl) spl_intxti=spline_eval(a,b,c,d,knots,xti); %evluate the natural spline interpolant at the data points ls_spl_xti=bestlsspline(xti); %evaluate the best least squares spline at all the data points sum((spl_intxti-yti).^) %least squares error of the natural spline interpolant ans = sum((ls_spl_xti-yti).^) %least squares error of the best least squares spline (its better of course:) ans = max(abs(spl_intxti-yti)) % max abs value error for the natural spline interpolant ans = max(abs(ls_spl_xti-yti)) %max abs value error for the least squares spline (again better) ans =

9 . x Figure : The titanium data on the left and the same data with its interpolating polynomial on the right Figure 4: The natural spline interpolation of the titanium data on the left. The natural spline and the polynomial that interpolates the data at every 6th data point on the right. 9

10 Figure 5: The cubic B-splines on the left the left and their sum on the interval on the right Figure 6: The best least squares approximation to the titanium data using the B-splines on the left. The natural spline interpolant on every 6th point plotted with the best least squares approximation by cubic splines on the same knot set on the right.