5.1 introduction problem : Given a function f(x), find a polynomial approximation p n (x).

Transcription

1 capter 5 : polynomial approximation and interpolation 5 introduction problem : Given a function f(x), find a polynomial approximation p n (x) Z b Z application : f(x)dx b p n(x)dx, a a one solution : Te Taylor polynomial of degree n about a point x = a is p n (x) =f(a)+f (a)(x a)+ f (a)(x a) + + n f (n) (a)(x a) n ex : f(x) = +x,a=,p n(x) =? In tis case we can find p n (x) witout computing f(a),f (a),,f (n) (a) recall te geometric series : r =+r + r +,convergesfor r < +x = ( x ) =+( x )+( x ) +,convergesfor x apple 5 5 p (x) = 5 p (x) = x Turs 3/ p 4 (x) = x +x p (x) = x +x 4 5x Te Taylor polynomial p n (x) is a good approximation to f(x) wenx is close to a, but in general we need to consider oter metods of approximation

2 5 polynomial interpolation tm : Assume f(x) isgivenandletx,x,,x n be n + distinct points Ten tere exists a unique polynomial p n (x) ofdegreeapple n wic interpolates f(x) at te given points, ie suc tat p n (x i )=f(x i )fori =:n pf :omit ex : n = ) x,x questions p (x) =f(x )+ ) f(x ) A(x x ) x x ceck : Wat is te form of p n (x) forn? ( deg p apple, p (x )=f(x ),p (x )=f(x ) Wat is te best coice of te interpolation points x,,x n? note :Teinterpolatingpolynomialp n (x) can be written in di erent forms standard form p n (x) =a + a x + + a n x n Newton s form note p n (x) =a + a (x x )+a (x x )(x x )+ + a n (x x ) (x x n ) Te example above wit n = used Newton s form for p (x) Te coe cients in eac form are di erent; ow can tey be computed? tm : Te coe a = f(x )=f[x ] cients in Newton s form of p n (x) canbecomputedasfollows a = f[x ] f[x ] x x = f[x,x ] : st divided di erence a = f[x,x ] f[x,x ] x x = f[x,x,x ] : nd divided di erence a n = f[x,,x n ] f[x,,x n ] x n x = f[x,,x n ] : nt divided di erence ok

3 3 pf : skip n =: okbecausep (x) =a,p (x )=f(x ) ) a = f(x ) n =:okbecausep (x) =a + a (x x ) and we sowed tat a = f(x ) f(x ) x x n =: needtowork p (x) =a + a (x x )+a (x x )(x x ) define g(x) = x x q (x)+ x x p (x) x x x x were p (x) =f[x ]+f[x,x ](x x ):interpolatesf(x) atx,x q (x) =f[x ]+f[x,x ](x x ):interpolatesf(x) atx,x ten g(x) astefollowingproperties deg g apple g(x )=p (x )=f(x ) g(x )= x x x x g(x )=q (x )=f(x ) q (x )+ x x x x p (x )= = f(x ) ten g(x) =p (x) forallx (by uniqueness teorem on polynomial interpolation) note : te coe cient of x in g(x) is f[x,x ] x x f[x,x ] x x te coe cient of x in p (x) isa ) a = f[x,x,x ]= f[x,x ] x x f[x,x ] x x as required n 3 : follows te same way ok

4 4 ex : n = ) x,x,x p (x) =a + a (x x )+a (x x )(x x ) = f[x ]+f[x,x ](x x )+f[x,x,x ](x x )(x x ) divided di erence table Te starred values are te coe cients in Newton s form of p (x) ex f(x) = +x,x =,x =,x = ) p (x) =? x i f(x i ) H H H H H H Turs 4/ p (x) = + (x ( )) (x ( ))(x ) : Newton s form = + (x +) (x +)x = x : standard form ceck : p ( ) =,p () =,p () = ok Z Z f(x)dx = dx +x = = 5 tan 5x = 5 tan 5=5494 Z p (x)dx = Z ( x )dx =(x 3 x3 ) =( 78 )= 78 =359 Hence p (x) isapoorapproximationtof(x) Can we do better?

5 5 53 optimal interpolation points Given f(x) for apple x apple, ow sould te interpolation points x,,x n be cosen? Consider two options uniform points : x i = +i, = n,i=:n Cebysev points : x i = cos i, i = i, = n,i=:n x uniform points, n= Cebysev points, n= note :TeCebysevpointsareclusterednearteendpointsofteinterval

6 ex : f(x) = +x, apple x apple solid line : f(x),givenfunction dased line : p n (x),interpolatingpolynomial 5 uniform points, n=4 5 Cebysev points, n= uniform points, n=8 5 Cebysev points, n= uniform points, n= 5 Cebysev points, n= Interpolation at te uniform points gives a good approximation near te center of te interval, but it gives a bad approximation near te endpoints Interpolation at te Cebysev points gives a good approximation on te entire interval

7 7 54 spline interpolation Let x <x < <x n following conditions <x n Acubicsplineis a function s(x) satisfyingte s(x) is a cubic polynomial on eac interval x i apple x apple x i+ s(x),s (x),s (x) arecontinuousatteinteriorpointsx,,x n ex : x =,x =,x = s(x) = s (x) = s (x) = (, apple x apple x 3, apple x apple (, apple x apple 3x, apple x apple (, apple x apple x, apple x apple ceck : s(x) satisfiesteconditionsrequiredtobeacubicspline problem : Given f(x) andx <x < <x n <x n,findtecubicspline s(x) tatinterpolatesf(x) attegivenpoints,ies(x i )=f(x i ),i=:n x i apple x apple x i+ ) s(x) =s i (x) =c + c x + c x + c 3 x 3,i=:n n +points ) n intervals ) 4n unknown coe interpolation conditions ) n equations cients continuity of s (x),s (x) atinteriorpoints ) (n ) equations Hence we can coose more conditions; a popular coice is s (x )=s (x n )=, wic gives te natural cubic spline interpolant ow to find s(x) ex : apple x apple,x i = +i, = n,i=,,n : uniform points step : nd derivative conditions s i (x) isalinearpolynomial ) s i (x) =a i x i+ x ) s i (x i )=a i,s i (x i+ )=a i+ + a i+ x x i, a i,a i+ :tobedetermined ) s i (x i )=a i = s i (x i ) ) s (x) iscontinuousatteinteriorpoints Turs 4/4

8 8 step : interpolation integrate twice s i (x) = a i(x i+ x) 3 + a i+(x x i ) 3 x i+ x + b i s i (x i )= a i + b a i i = f i ) b i = f i s i (x i+ )= a i+ a i+ + c i = f i+ ) c i = f i+ step 3 : st derivative conditions s i(x) = a i(x i+ x) s i(x i )= a i s i(x i+ )= a i+ we require s i (x i )=s i(x i ) ) a i ) a i f i + a i + a i + a i+(x x i ) a i + f i a i+ f i + a i + f i+ f i + a i + f i+ + + f i a i+ a i = a i + a i+ x x i + c i + f i+ f i + a i + f i+ = f i f i + f i+ ) a i +4a i + a i+ = (f i f i + f i+ ),i=:n a i+ a i+ step 4 : apply BC s (x )=a =,s n (x n )=a n = a C B C A a n = f f + f f n f n + f n C A A :symmetric,tridiagonal,positivedefinite

9 9 ex : natural cubic spline interpolation f(x) = +x, apple x apple,x i = +i, = n,i=:n solid line : f(x),givenfunction dased line : s(x),naturalcubicsplineinterpolant n= n=4 n=' n=8 error bound : f(x) s(x) apple max aapplexappleb f (4) (x) 4 : 4t order accurate Te natural cubic spline interpolant as inflection points at te endpoints of te interval, due to te boundary conditions s (x )=s (x n )=;tereare also inflection points in te interior of te interval wic are not present in te original f(x), and tese are problematic in some applications