2.  3.3 Hermite Interpolation Chapter 3 Interpolation and Polynomial Approximation -- Hermite Interpolation Find the osculating polynomial P(x) such that P(x i ) = f (x i ), P’ (x i ) = f ’ (x i ), …, P (m_i) (x i ) = f (m_i) (x i ) for all i = 0, 1, …, n. 1/12 Note: Given N conditions (and hence N equations), a polynomial of degree can be determined.  Given N conditions (and hence N equations), a polynomial of degree can be determined. N  1N  1N  1N  1 The osculating polynomial that agrees with f and all its derivatives of order  m 0 at one point x 0 is just the Taylor polynomial  The osculating polynomial that agrees with f and all its derivatives of order  m 0 at one point x 0 is just the Taylor polynomial with remainder The case when m i = 1 for each i = 0, 1, …, n gives the Hermite polynomials.  The case when m i = 1 for each i = 0, 1, …, n gives the Hermite polynomials.

3. Chapter 3 Interpolation and Polynomial Approximation -- Hermite Interpolation Example: Suppose x 0  x 1  x 2. Given f(x 0 ), f(x 1 ), f(x 2 ) and f ’(x 1 ), find the polynomial P(x) such that P(x i ) = f (x i ) ， i = 0, 1, 2, and P’(x 1 ) = f ’(x 1 ). Analyze the errors. Solution: First of all, the degree of P(x) must be … 3. Similar to Lagrange polynomials, we may assume the form of Hermite polynomial as    2 13 )()()()()(  0 i ii xhx1x1 f ’xhxfxP  where h i (x j ) =  ij, h i ’(x 1 ) = 0, (x i ) = 0, ’(x 1 ) = 1  h1h1  h1h1 h0(x)h0(x) Has roots x 1, x 2, and h 0 ’(x 1 ) = 0  x 1 is a multiple root. )()()( 2 2 100 xxxxCxh  h 0 (x 0 ) = 1  C 0 h2(x)h2(x) Similar to h 0 (x). h1(x)h1(x) Has roots x 0, x 2  ))( ()( 201 xxxxBAxAxxh  A and B can be solved with h 1 (x 1 ) = 1 and h 1 ’(x 1 ) = 0. (x)  h1h1 Has roots x 0, x 1, x 2   h1h1 ))( ()( 2101 xxxxxxCx   h1h1 ’(x 1 ) = 1  C 1 can be solved. Similar to Lagrange error analysis 2/12

4. Chapter 3 Interpolation and Polynomial Approximation -- Hermite Interpolation In general, given x 0, …, x n ; y 0, …, y n and y 0 ’, …, y n ’. The Hermite polynomial H 2n+1 (x) satisfies H 2n+1 (x i ) = y i and H’ 2n+1 (x i ) = y i ’ for all i.    n i )()( )(  0 i i xhxh yiyi x H 2n+1   n  0 i yi’yi’ where h i (x j ) =  ij, h i ’(x j ) = 0, (x j ) = 0, ’(x j ) =  ij  hihi  hihi hi(x)hi(x) x 0, …, x i, …, x n are all roots with multiplicity 2   )()()(x L2L2 BxAxh n, iiii  A i and B i can be solved by h i (x i ) = 1 and h i ’(x i ) = 0  (x)  hihi All the roots x 0, …, x n have multiplicity 2 except x i   hihi )()( ii L n, i 2 (x) xxCx   hihi ’(x i ) = 1  C i = 1  hihi )(x)( i L n, i 2 (x) xx  If, then Solution: Let Such a Hermite interpolating polynomial is unique 3/12

5. Chapter 3 Interpolation and Polynomial Approximation -- Hermite Interpolation Quiz: Given x i = i +1, i = 0, 1, 2, 3, 4, 5. Which one is h 2 (x)?  x 0 - -1 0.5 123456 y x y 0 - - -1 123456 slope=1 HW: p.140 #7 4/12

6. Chapter 3 Interpolation and Polynomial Approximation -- Cubic Spline Interpolation  3.4 Cubic Spline Interpolation Remember what I have said? Increasing the degree of interpolating polynomial will NOT guarantee a good result, since high-degree polynomials are oscillating. Example: Consider the Lagrange polynomial P n (x) of on [  5, 5]. Take P n (x)  f (x)  Piecewise polynomial approximation 5/12

7. Chapter 3 Interpolation and Polynomial Approximation -- Cubic Spline Interpolation  Piecewise linear interpolation Approximate f (x) by linear polynomials on each subinterval : Let. Then as uniform No longer smooth.  Hermite piecewise polynomials Given Construct the Hermite polynomial of degree 3 with y and y’ on the two endpoints of. It is not easy to obtain the derivatives. How can we make a smooth interpolation without asking too much from f ? Headache … 6/12

8. Chapter 3 Interpolation and Polynomial Approximation -- Cubic Spline Interpolation  Cubic Spline Definition: Given a function f defined on [a, b] and a set of nodes a = x 0 < x 1 < … < x n = b, a cubic spline interpolant S for f is a function that satisfies the following conditions: a.S(x) is a cubic polynomial, denoted S i (x), on the subinterval [ x i, x i+1 ] for each i = 0, 1, …, n – 1; b.S(x i ) = f (x i ) for each i = 0, 1, …, n; c.S i+1 (x i+1 ) = S i (x i+1 ) for each i = 0, 1, …, n – 2; d.S’ i+1 (x i+1 ) = S’ i (x i+1 ) for each i = 0, 1, …, n – 2; e.S” i+1 (x i+1 ) = S” i (x i+1 ) for each i = 0, 1, …, n – 2; f(x)f(x) H(x)H(x) S(x)S(x) 7/12

9. Chapter 3 Interpolation and Polynomial Approximation -- Cubic Spline Interpolation  Method of Bending Moment Then S j ”(x) is a polynomial of degree, which can be determined by the values of f on nodes. 1 2 Assume S j ”(x j  1 ) = M j  1 ， S j ”(x j ) = M j. Then for all x  [ x j – 1, x j ], S j ”(x) = j j j j j j h xx M h xx M 1 1      Integrate S j ”(x) twice, we obtain S j ’(x) and S j (x) ： j j j j j j j A h xx M h xx M        2 )( 2 )( 2 1 1 2 1 S j ’(x) = jj j j j j j j BxA h xx M h xx M       6 )( 6 )( 3 1 3 1 S j (x) = Can be solved by S j (x j  1 ) = y j  1 S j (x j ) = y j Let h j = x j – x j – 1 and S(x) = S j (x) for x  [ x j – 1, x j ]. For each j it is a cubic polynomial The bending moment 8/12

10. Chapter 3 Interpolation and Polynomial Approximation -- Cubic Spline Interpolationj jj j jj j h MM h yy A 6 11      j j j j j j j j j jjj h xx h M y h xx h M yBxA 1 22 1 1 ) 6 () 6 (       Now solve for M j ： Since S’ is continuous at x j [x j  1, x j ]: S j ’(x) = j jj jj j j j j j j h MM xxf h xx M h xx M 6 ],[ 2 )( 2 )( 1 1 2 1 2 1           1 1 1 1 2 1 1 2 1 6 ],[ 2 )( 2 )(              j jj jj j j j j j j h MM xxf h xx M h xx M [x j, x j+1 ]: S j+1 ’(x) = From S j ’(x j ) = S j+1 ’(x j ), we can combine the coefficients for M j  1, M j, and M j+1. Define,, and 1 1 jj j j hh h     1 jj  ]),[],[( 6 11 1 jjjj jj j xxfxxf hh g      for  j  1 n1n1 That is, we have unknowns but only equations. n1n1 n+1n+1 Extra 2 boundary conditions are needed. 2 11 gMMM jjjjjj    9/12

11. Chapter 3 Interpolation and Polynomial Approximation -- Cubic Spline Interpolation  S’(a) = y 0 ’ ， S’(b) = y n ’ /* Clamped boundary */ [ a, x 1 ]: S 1 ’(x) = 1 01 10 1 2 1 1 2 1 0 6 ],[ 2 )( 2 )( h MM xxf h ax M h xx M       010 1 10 )],[( 6 2g y0’y0’ xxf h MM  nnn n nn gxxf yn’yn’ h MM     ]),[ ( 6 2 1 1 Similar for S n ’(x) on [ x n  1, b ]  S”(a) = y 0 ” = M 0 ， S”(b) = y n ” = M n Then The case when M 0 = M n = 0 is called a free boundary, and when that occurs, the spline is called a Natural Spline.  Periodic boundary: If f is a periodic function, that is, y n = y 0 and S’(a + ) = S’(b  )  M 0 = M n 10/12

12. Chapter 3 Interpolation and Polynomial Approximation -- Cubic Spline InterpolationNote: Cubic Spline can be uniquely determined by its boundary conditions as long as the coefficient matrix is strictly diagonally dominant.  Cubic Spline can be uniquely determined by its boundary conditions as long as the coefficient matrix is strictly diagonally dominant. If f  C[ a, b ] and. Then  If f  C[ a, b ] and. Then as max h i  0. as max h i  0. That is, the accuracy of approximation can be improved by adding more nodes without increasing the degree of the splines. Sketch of the Algorithm: Cubic Spline ① Compute  j, j, g j ; ② Solve for M j ; ③ Find the subinterval which contains x (i.e. find the corresponding j ); ④ Approximate f(x) by S j (x). HW: p.153-154 #9, 17 11/12

13. Chapter 3 Interpolation and Polynomial Approximation -- Cubic Spline Interpolation Lab 06. Cubic Spline Time Limit: 1 second; Points: 4 Construct the cubic spline interpolant S for the function f, defined at points x 0 < x 1 < … < x n, satisfying some given boundary conditions. Partition an given interval into m equal- length subintervals, and approximate the function values at the endpoints of these subintervals. 12/12