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It has some nonvanishing coefficients. For if they were it would follow that N(x)e<9^ m _ 1 > ( .. However, since N(x) is surely bounded, we would conclude that N(x) = 0, which contradicts our assumption. 20), then cp(x; t 0 ) is an eigenspline of ,5^m_ 1 > r . However, all eigenspline solutions of 5(x + 1) = tS(x) have been determined 56 LECTURE 5 and found to correspond to the zeros A 1 ? 20) has a zero t0 in common with n 2 m-i,r( x )» ifl contradiction with our Assumption 1. 19) has a nonvanishing root to.
Some authors call them Hermite splines. If r = m = 1 we have the familiar case of linear spline interpolation. 2. The main results. Let y ^ 0. We recall the class of sequences Let S(x)e^2m-i,r- An easv application of Markov's theorem on bounds for the derivatives of a bounded polynomial (see [59, Chap. 2], or ) will show that if S(x) = 0(\x\y) as x -» ±00, then also 5(s)(x) = O(\x\y) as x -» ±00, and this for r = 1, 2, • • • , 2m — 1. Accordingly we consider the class Our main result is the following theorem.
The periodic function Qn(x) of Lemma 1 admits the Fourier series expansion This result has a number of corollaries (see [4, §§ 8,9]). One such is the following. COROLLARY 5. Iff is not a negative real then The question of convergence for negative t, in which case we must assume that t 7* A v , v = 1, • • • , n — 1, is more delicate. If n is odd, we may choose t = — 1 (which in this case is not an eigenvalue) and find that Sn(x; — 1) -> cos nx, which limit is certainly different from tx = einx. However, if we choose t = — e (e is the base of natural logarithms), then surely t ^ A v , because e is transcendental, while all Av are algebraic numbers.