There is a unique function s(x) that has the required properties. It turns out to also satisfy

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1 Numerical Analysis Grinshpan Natural Cubic Spline Let,, n be given nodes (strictly increasing) and let y,, y n be given values (arbitrary) Our goal is to produce a function s() with the following properties: s( ) y,,, n, s() is twice continuously differentiable on [, n ], n s () d is as small as possible There is a unique function s() that has the required properties It turns out to also satisfy s() restricted to [, + ] is a cubic,,, n, 5 s() is almost linear at the endpoints, s ( ) s ( n ) 0 This function is called a natural cubic spline It is arbitrarily smooth on every open subinterval (, + ) and a) s ( ) eists (s ( ) s ( +)),,, n, b) s ( ) eists (s ( ) s ( +)),,, n Construction of s() We will first construct a unique function s() that satisfies properties,,, 5, and then show that it also satisfies property An auiliary formula is needed Lemma Let u() be a cubic polynomial on [0, ], and let u(0) a, u() b, u (0) A, and u () B Then u() has the form u() A ( ( ) + a A ) ( ) + B ( + b B ) Proof of lemma It is straightforward to chec that the formula wors Alternatively, the second derivative u () is a linear function with u (0) A and u () B Hence u () A( ) + B Integrating u () twice, we obtain u() A ( ) + B + c( ) + d To satisfy u(0) a and u() b we must tae c a A and d b B If the interval [0, ] is replaced by [, + ], with δ +, the formula changes as follows: u() A ( a ( + ) + Aδ ) ( + )+ B ( b ( ) + Bδ ) ( ) δ δ δ δ

2 For each,, n, define s() M ( + ) + δ [, + ] ( y δ ( + ) δ + M + δ ( ) + ( y+ δ +δ ( ), where M M n 0 and parameters M,, M n are to be determined By construction, s() restricted to [, + ] is a cubic polynomial and s( ) y,,, n, which maes it continuous The first derivative s () restricted to (, + ) is a quadratic polynomial, s () M ( ( + ) y δ + M ( + ( ) y+ + δ δ δ (, + ) We are going to choose M to mae s () continuous at each interior joint δ +δ The second derivative s () restricted to (, + ) is linear, s () M ( + ) + M + ( ) δ δ (, + ) Since s ( ) s ( +) M,,, n, the line segments connect continuously Hence s () is continuous as long as s () is To mae sure that s () is continuous, ie, that condition a) holds, eamine two abutting intervals [, ], [, + ] and equate one-sided derivatives s ( +) δ M δ M + + y + y + This gives s ( ) δ M + δ M + y y δ M + δ + δ M + δ M + y + y + y y Multiplying by and dividing by δ + δ +, we have δ δ + δ M + M + where () δ are the second divided differences () M + (),,, n, δ + δ y + y + y y +

3 We thus obtain a linear system of (n ) equations in (n ) unnowns M,, M n In matri form, δ δ +δ M δ δ δ +δ δ +δ δ δ δ +δ M δ +δ M δ δ +δ 5 M 5 δ n δ n +δ n M n M n δ n δ n +δ n () () () () 5 () n () n The values of M,, M n are uniquely determined by these equations: the tridiagonal coefficient matri is invertible because it is diagonally dominant, δ δ +δ + δ δ +δ < M s ( ) are sometimes called moments If all subintervals [, + ] are of equal length δ, we have simplification: M + + M + M δ (y + y + y ),,, n, or M M M M 5 M n M n () () () () 5 () n () n Etremum property Let f() be an interpolant to the values of y at points, and let f () eist and be continuous The difference () s() f() is then zero at each Eamine the following integration: s () ()d s () () s () ()d s () ()d (because s ( ) s ( n ) 0) n + c ()d (because s () [, + ] n [ c (+ ) ( ) ] 0 (because ( ) 0) is constant)

4 It follows that f () d (s () ()) d s () d s () () + () d s () d + s () d () d Thus s() minimizes the integral n f () d over all twice continuously differentiable functions f() with f( ) y Equality in f () d s () d holds if and only if () 0 on [, n ], ie, () is linear on [, n ] But, since ( ) 0, this can only happen if () 0 or f() s() We have proved: THEOREM Among all twice continuously differentiable functions f() satisfying the interpolation conditions f( ) y,,, n, the minimum value of the integral f () d is realized by the natural cubic spline s() and only by s() Error estimate Let f() be a four times continuously differentiable function on [, n ] with f( ) y and f ( ) f ( n ) 0 Let δ ma be the maimum of δ and δ min be the minimum of δ Then f() s() C ma [, n] f () () δ ma, where C δ ma /δ min In particular, if all subintervals [, + ] are of equal length δ, we have: f() s() ma [, n] f () () δ I m leaving out a proof of the error estimate We may later come bac to this topic and supply justification

5 5 Appendi: diagonal dominance A matri A (a ij ) is diagonally dominant if a ii > j i a ij for each i LEMMA Diagonal dominance implies invertibility Proof If v (v i ) 0 but Av 0, choose v m of maimum modulus Then 0 j a mjv j and so a mm v m j m a mj v j j m a mj v m, a contradiction

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