polynomial approximation and interpolation

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1 28. Wed /8 capter 5 : polynomial approximation and interpolation questions How can we approximate a given function f(x) by a polynomial p(x)? How can we interpolate a set of data values (x i,f i ) by a polynomial p(x)? recall A polynomial of degree n as te form p n (x) =a 0 + a x + + a n x n. Tere are oter forms in wic a polynomial may be written, as we sall see. application b a f(x)dx b a p n(x)dx,... Weierstrass Approximation Teorem Let f(x) be continuous for a x b. Ten for any ɛ> 0, tere exists a polynomial p(x) suc tat max f(x) p(x) ɛ. a x b pf : Mat 45 Given f(x), tere are many ways to find an approximating polynomial p(x). Taylor approximation Given f(x) and a point x = a, te Taylor polynomial of degree n about x = a is p n (x) =f(a)+f (a)(x a)+ 2 f (a)(x a) n f (n) (x a) n. Te Taylor polynomial satisfies te following conditions. p n (a) =f(a), p n(a) =f (a),..., p (n) n (a) =f (n) (a) f(x) =p n (x)+r n (x), r n (x) : remainder, error r n (x) = x a (x t) n n f (n+) (t)dt = (n+) f (n+) (ζ)(x a) n+ for some ζ [a, a + x]. f(x) p n (x) (n+) max f (n+) (t) x a n+ : error bound a t x

2 2 ex : f(x) = + 25x,a=0,p n(x) =? 2 In tis case tere is a way to find te Taylor polynomial witout computing derivatives, using te formula for te sum of a geometric series. if r < recall : + r + r 2 + r 3 + = r diverges if r + 25x = 2 ( 25x 2 ) = + ( 25x2 )+( 25x 2 ) 2 +( 25x 2 ) 3 + # % &'()*) # % + &'()*)+' + # # # # # # # # # %&,&'()*)+' + -.+', # %. &'()*)+' + -.+',.+'. # # # # # # # # 25x 2 < x < 5 lim n p n(x) =f(x), x 5 lim n p n(x) diverges explanation : f(x) as complex singularities at x = ± 5i, Mat 555 Hence te Taylor polynomial is a good approximation wen f(x) is sufficiently differentiable and x is close to a, but in general we need to consider oter metods of approximation.

3 29. Fri /20 3 polynomial interpolation tm : Assume f(x) is given and let x 0,x,..., x n be n + distinct points. Ten tere exists a unique polynomial p n (x), of degree less tan or equal to n, wic interpolates f(x) at te given points, i.e. suc tat p n (x i )=f(x i ) for i = 0 : n. pf : existence : soon, uniqueness : Fundamental Tm of Algebra, Mat 42 ex : n = x 0,x # questions. How can we construct p n (x) for larger n? 2. Is tere an efficient way to evaluate p n (x) for x x i? 3. How large is te error f(x) p n (x) for x x i? section 5. : Lagrange form of te interpolating polynomial def : L k (x) = n ( ) x xi,k= 0:n : Lagrange polynomial x k x i i=0 i k ex : n =2,x 0 =,x =0,x 2 = L 0 (x) = (x x )(x x 2 ) (x 0 x )(x 0 x 2 ) L (x) = (x x 0)(x x 2 ) (x x 0 )(x x 2 ) L 2 (x) = (x x 0)(x x ) (x 2 x 0 )(x 2 x ) = (x 0)(x ) ( 0)( ) = 2 x2 2 x = (x ( ))(x ) (0 ( ))(0 ) = x2 + = (x ( ))(x 0) ( ( ))( 0) = 2 x2 + 2 x properties. deg L k = n, 2. L k (x i )= { if i = k 0 if i k

4 4 Now let f(x) be given. Te Lagrange form of te interpolating polynomial is p n (x) =f(x 0 )L 0 (x)+f(x )L (x)+ + f(x n )L n (x) = n f(x k )L k (x). ceck. deg p n n, 2. p n (x i )= n f(x k )L k (x i )=f(x i ),i=0:n ex f(x) = k=0 + 25x 2,x 0 =,x =0,x 2 = p 2 (x) =f(x 0 )L 0 (x)+f(x )L (x)+f(x 2 )L 2 (x) = 2 ( 2 x2 2 x)+ ( x2 + ) + 2 ( 2 x2 + 2 x)= 25 2 x2 + ceck... ok Te Lagrange form is useful teoretically (to sow tat p n (x) exists), but it as some practical disadvantages.. It is expensive to evaluate p n (x) for x x i using tis form. 2. Adding an extra point x n+ means tat all te L k (x) must be recomputed. section 5.3 : k=0 Newton form of te interpolating polynomial p n (x) =a 0 + a (x x 0 )+a 2 (x x 0 )(x x )+ + a n (x x 0 ) (x x n ) ex : n =,x 0,x ( x x0 p (x) =f(x 0 ) x 0 x = f(x 0 )+ ) f(x ) f(x 0 ) x x 0 ( x x + f(x ) x x 0 ) (x x 0 ) : : Lagrange form Newton form For n 2, we need a metod to compute a 0,..., a n. Assume tat p n (x) is already known. Ten define p n (x) =p n (x)+a n (x x 0 ) (x x n ). We will derive a formula for a n. i =0:n p n (x i )=p n (x i )+a n (x i x 0 ) (x i x n )=f(x i ) i = n p n (x n )=p n (x n )+a n ( ) (x n x n )=f(x n ) a n = f(x n ) p n (x n ) ( ) (x n x n ) : Tis formula is correct, but tere is a more efficient way to compute a n. ok

5 30. Mon /23 5 First define some notation. a n = f[x 0,..., x n ] a 0 = f[x 0 ],a = f[x 0,x ],a 2 = f[x 0,x,x 2 ],... claim : f[x 0,..., x n ]= f[x,..., x n ] f[x 0,..., x n ] : divided difference pf p n (x) interpolates f(x) at x 0,..., x n, deg p n n q n (x) interpolates f(x) at x,..., x n, deg q n n ( ) ( ) x x0 xn x define g(x) = q n (x)+ p n (x) ten deg g n g(x 0 )=p n (x 0 )=f(x 0 ) g(x n )=q n (x n )=f(x n ) ( ) xi x 0 i = : n g(x i )= q n (x i )+ ( xn x i ) p n (x i )=f(x i ) g(x) =p n (x), now equate te coefficients of x n in tese two polynomials f[x,..., x n ] f[x 0,..., x n ] = f[x 0,..., x n ] ok Hence we can rewrite Newton s form of te interpolating polynomial as follows. p n (x) =f[x 0 ]+f[x 0,x ](x x 0 )+f[x 0,x,x 2 ](x x 0 )(x x ) + + f[x 0,..., x n ](x x 0 ) (x x n ) Te coefficients can be efficiently computed using a divided difference table. ex : n =2 # % % % # % # % # # % # # # Te starred values are te coefficients in Newton s form of p 2 (x). p 2 (x) =f[x 0 ]+f[x 0,x ](x x 0 )+f[x 0,x,x 2 ](x x 0 )(x x )

6 3. Wed /25 ex : f(x) = +25x 2,x 0 =,x =0,x 2 = p 2 (x) = (x ( )) 2 (x ( ))(x 0) = 25 2 x2 ok Using Newton s form for p n (x), if an extra interpolation point x n+ is added, ten te previous terms remain uncanged. evaluation of p n (x) p 2 (x) =a 0 + a (x x 0 )+a 2 (x x 0 )(x x ) : Newton form, 3 mults = a 0 +(x x 0 )(a + a 2 (x x )) : nested form, 2 mults general case p n (x) =a 0 + a (x x 0 )+a 2 (x x 0 )(x x )+ + a n (x x 0 ) (x x n ) = a 0 +(x x 0 )(a +(x x )(a a n (x x n )) ) operation count : Newton form = n(n+) 2 mults, nested form = n mults tm (te error in polynomial interpolation) f(x),x 0 <... < x n : given, p n (x) : interpolating polynomial Ten x 0 x x n f(x) =p n (x) + (n + ) f (n+) (ζ)(x x 0 ) (x x n ). pf : omit, but tat tis resembles te error in Taylor approximation application : n =,x 0 = a, x = b # a x b f(x) p (x) 8M b a 2, pf : w8 were M = max a x b f (x)

7 32. Mon /30 7 section 5.4 : optimal points for interpolation Given f(x) for x, ow sould we coose te interpolation points x 0,..., x n? Consider two options.. uniform points : x i = +i, = 2 n,i=0:n 2. Cebysev points : x i = cos θ i,θ i = i π n,i= 0 : n %&'()*+,-)'&./,0,&2 # # # # 3457/458,-)'&./,0,&2 # # # # x : Te Cebysev points are clustered near te endpoints of te interval.

8 8 ex : f(x) = + 25x 2, polynomial interpolation on te interval x # %&'()*+,-)'&./0,&2 # 3457/458,-)'&./,0,&2 # # # # # # # %&'()*+,-)'&./,0,&9 # 3457/458,-)'&./,0,&9 # # # # # #. Interpolation at te uniform points gives a good approximation near te center of te interval, but it gives a bad approximation near te endpoints. 2. Interpolation at te Cebysev points gives a good approximation on te entire interval.

9 33. Wed 2/2 9 section 5.5 : piecewise linear interpolation Given f(x),a x b, a = x 0 <x < <x n <x n = b. %% # # Te interpolating polynomial p n (x) may not be a good approximation to f(x) on te entire interval, so instead we may consider te piecewise linear interpolant, ded by q(x). x i x x i+ q(x) =f[x i ]+f[x i,x i+ ](x x i ) q(x i )=f(x i ),i= 0 : n error : f(x) q(x) 8 max a x b f (x) max i x i+ x i 2 : 2nd order accurate : q(x) is continuous, but it is not differentiable at x = x i. section 5. : piecewise cubic interpolation A cubic spline is a function s(x) satisfying te following conditions.. For eac interval x i x x i+, s(x) is a cubic polynomial. 2. s(x),s (x),s (x) are continuous at te interior points x,..., x n ex : x 0 =,x =0,x 2 = s(x) = s (x) = s (x) = { 0, x 0 x 3, 0 x { 0, x 0 3x 2, 0 x { 0, x 0 x, 0 x Hence s(x) satisfies te conditions required to be a cubic spline.

10 0 problem Given f(x) and a = x 0 <x < <x n <x n = b. Find a cubic spline s(x) wic interpolates f(x) at te given points, i.e. s(x i )=f(x i ),i= 0 : n. x i x x i+ s(x) =s i (x) =c 0 + c x + c 2 x 2 + c 3 x 3,i= 0 : n #% # # # & #% # # n + points n intervals 4n unknown coefficients interpolation conditions 2n equations continuity of s (x),s (x) at interior points 2(n ) equations We can coose 2 more equations. A popular coice is to set s (x 0 )=s (x n ) = 0, wic gives te natural cubic spline interpolant. ow to find s(x) ex : x,x i = +i, = 2 n step : 2nd derivative conditions ( s i (x) is a linear polynomial s xi+ x i (x) =a i,i=0,..., n : uniform points s i (x i )=a i,s i (x i+ )=a i+ s i (x i )=a i = s i (x i ) Hence s (x) is continuous at te interior points. step 2 : integrate twice interpolation s i (x) = a i(x i+ x) 3 + a i+(x x i ) 3 ) ( ) xi+ x + b i s i (x i )= a i 2 + b i = f i b i = f i a i 2 s i (x i+ )= a i+ 2 + c i = f i+ c i = f i+ a i+ 2 s i (x) = a i(x i+ x) 3 + f i a i 2 + a i+(x x i ) 3 ( ) xi+ x + f i+ a i+ 2 ( ) x xi + a i+ ( ) x xi + c i ( x xi )

11 34. Fri 2/4 step 3 : st derivative conditions s i(x) = a i(x i+ x) f i a i a i+(x x i ) 2 2 f i+ a i+ 2 s i(x i )= a i 2 f i + a i + f i+ s i(x i+ )= a i+ 2 we require s i (x i )=s i(x i ) a i 2 f i a i + a i + a i ( f i + a i + f i+ a i+ a i+ + f i a i = a i 2 f i + a i + f i+ ) + a i+ = f i 2f i + f i+ a i+ a i +4a i + a i+ = 2 (f i 2f i + f i+ ),i= : n step 4 : BC s 0(x 0 )=0 a 0 =0,s n (x n )=0 a n = a... a n = 2 A : tridiagonal, symmetric, positive definite f 0 2f + f 2... f n 2 2f n + f n

12 2 ex : f(x) = natural cubic spline interpolation + 25x 2, x,x i = +i, = 2 n,i= 0 : n % %& # # # # %' %( # # # #. f(x) s(x) max a x b f (4) (x) 4 : 4t order accurate, pf : omit 2. Te natural cubic spline interpolant as inflection points at te endpoints of te interval, due to te boundary conditions s (x 0 )=s (x n ) = 0. In fact, tere are additional inflection points in te interior of te interval, wic are problematic in some applications.