Interpolation.pdf - Chapter 3 Interpolation 3.1 Polynomial...

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Chapter 3 Interpolation 3.1 Polynomial Interpolation One of the basic tools for approximating a function or a given data set is interpolation. In this chapter we focus on polynomial and piece-wise poly- nomial interpolation. The polynomial interpolation problem can be stated as follows: Given n + 1 distinct points, x 0 , x 1 , ..., x n called nodes and corresponding values f ( x 0 ) , f ( x 1 ) , ..., f ( x n ), find a polynomial of degree at most n , p n , which sat- isfies (the interpolation property) p n ( x 0 ) = f ( x 0 ) , p n ( x 1 ) = f ( x 1 ) , . . . p n ( x n ) = f ( x n ) . Let us represent such polynomial as p n ( x ) = a 0 + a 1 x + · · · + a n x n . Then, the interpolation property implies a 0 + a 1 x 0 + · · · + a n x n 0 = f ( x 0 ) , a 0 + a 1 x 1 + · · · + a n x n 1 = f ( x 1 ) , . . . a 0 + a 1 x n + · · · + a n x n n = f ( x n ) . 31
32 CHAPTER 3. INTERPOLATION This is a linear system of n +1 equations in n +1 unknowns (the polynomial coe ffi cients a 0 , a 1 , . . . , a n ). In matrix form: 2 6 6 6 4 1 x 0 x 2 0 · · · x n 0 1 x 1 x 2 1 · · · x n 1 . . . 1 x n x 2 n · · · x n n 3 7 7 7 5 2 6 6 6 4 a 0 a 1 . . . a n 3 7 7 7 5 = 2 6 6 6 4 f ( x 0 ) f ( x 1 ) . . . f ( x n ) . 3 7 7 7 5 (3.1) Does this linear system have a solution? Is this solution unique? The answer is yes to both. Here is a simple proof. Take f 0, then p n ( x j ) = 0,for j = 0 , 1 , ..., n but p n is a polynomial of degree at most n , it cannot have n + 1 zeros unless p n 0, which implies a 0 = a 1 = · · · = a n = 0. That is, the homogenous problem associated with (3.1) has only the trivial solution. Therefore, (3.1) has a unique solution. In general, the values to interpolate might not come from a function. They can be regarded as just data supplied to us. We will often write ( x 0 , f 0 ) , ( x 1 , f 1 ), etc., to emphasize this more general setting. Example 5. As an illustration let us consider interpolation by a linear poly- nomial, p 1 . Suppose we are given ( x 0 , f 0 ) and ( x 1 , f 1 ) . We have written p 1 explicitly in the Introduction. We write it now in a di erent form: p 1 ( x ) = x - x 1 x 0 - x 1 f 0 + x - x 0 x 1 - x 0 f 1 . (3.2) Clearly, this polynomial has degree at most 1 and satisfies the interpolation property: p 1 ( x 0 ) = f 0 , (3.3) p 1 ( x 1 ) = f 1 . (3.4) Example 6. Given ( x 0 , f 0 ) , ( x 1 , f 1 ) , ( x 2 , f 2 ) let us construct p 2 , the polyno- mial of degree at most 2 which interpolates these points. The way we have written p 1 in (3.2) is suggestive of how to explicitly write p 2 : p 2 ( x ) = ( x - x 1 )( x - x 2 ) ( x 0 - x 1 )( x 0 - x 2 ) f 0 + ( x - x 0 )( x - x 2 ) ( x 1 - x 0 )( x 1 - x 2 ) f 1 + ( x - x 0 )( x - x 1 ) ( x 2 - x 0 )( x 1 - x 1 ) f 2 .
3.1. POLYNOMIAL INTERPOLATION 33 If we define l 0 ( x ) = ( x - x 1 )( x - x 2 ) ( x 0 - x 1 )( x 0 - x 2 ) , (3.5) l 1 ( x ) = ( x - x 0 )( x - x 2 ) ( x 1 - x 0 )( x 1 - x 2 ) , (3.6) l 2 ( x ) = ( x - x 0 )( x - x 1 ) ( x 2 - x 0 )( x 1 - x 1 ) , (3.7) then we simply have p 2 ( x ) = l 0 ( x ) f 0 + l 1 ( x ) f 1 + l 2 ( x ) f 2 . (3.8) Note that each of the polynomials (3.5), (3.6), and (3.7) are exactly of degree 2 and they satisfy l j ( x k ) = δ jk 1 . Therefore, it follows that p 2 given by (3.8) satisfies the interpolation property p 2 ( x 0 ) = f 0 , (3.9) p 2 ( x 1 ) = f 1 , (3.10) p 2 ( x 2 ) = f 2 . (3.11) We can now write down the polynomial (of degree at most n ) which interpolates n + 1 given values, ( x 0 , f 0 ) , . . . , ( x n , f n ), where the interpolation nodes x 0 , . . . , x n are assumed distinct.

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