Pub Date : 1981-12-31DOI: 10.1002/ZAMM.19810610304
H. E. Salzer
In rational interpolation of ƒ(x) by (a0 + a1x + … + anxn)/(1 + b1x + … + bmxm) at xi, i = 1(1)n + m + 1, instead of solving an (n + m + 1)-th order linear system for ai and bi, we solve only an m-th order linear system to obtain all n + m + 1 coefficients of 1/(x - xi) in the “incomplete barycentric form” of the numerator and denominator. The extension to r-th order osculatory interpolation by (a0 + a1x + … + anr - 1xnr - 1)/(1 + b1x + … + bmrxmr) requires the solution of an rm-th, instead of the usual (rm + rn)-th order linear system. In similar treatment of non-osculatory odd-point rational trigonometric interpolation by a quotient of trigonometric sums, we solve a 2m-th, instead of (2n + 2m + 1)-th order linear system for the coefficients of 1/sin 1/2 (x - xi). Bei der rationalen Interpolation von ƒ(x) durch (a0 + a1x + … + anxn)/(1 + b1x + … + bmxm) in xi, i = 1(1) n + m + 1 losen wir anstatt eines linearen Systems (n + m + 1)-ter Ordnung Fur ai und bi nur ein System m-ter Ordnung, um samtliche Koeffizienten von 1/(x - xi) in der „unvollstandigen baryzentrischen Form” von Zahler und Nenner zu erhalten. Die Ausdehnung auf osculatorische Interpolation r-ter Ordnung durch (a0 + a1x + … + anr - 1xnr - 1)/(1 + b1x + … + bmrxmr) erfordert die Losung eines Systems rm-ter Ordnung anstelle des gewohnlichen linearen Systems (rm + rn)-ter Ordnung. In ahnlicher Behandlung nicht-osculatorischer rationaler trigonometrischer Interpolation uber eine ungerade Punktanzahl durch den Quotienten trigonometrischer Summen losen wir ein lineares System 2m-ter anstelle (2n + 2m + 1)-ter Ordnung Fur die Koeffizienten von 1/sin 1/2 (x - xi).
在(a0 + a1x +…+ anxn)/(1 + b1x +…+ bmxm)在xi, i = 1(1)n + m + 1处的有理插值中，我们不求解ai和bi的(n + m + 1)阶线性系统，而只求解一个m阶线性系统，得到1/(x - xi)的所有n + m + 1个系数，以分子和分母的“不完全质心形式”。由(a0 + a1x +…+ anr - 1xnr - 1)/(1 + b1x +…+ bmrxmr)扩展到r阶共轭插值需要一个rm-th的解，而不是通常的(rm + rn)-th阶线性系统的解。在类似的治疗non-osculatory odd-point理性三角插值系数的三角,我们解决2 m,而不是(2 n + 2 m + 1) th阶线性系统的系数1 /罪1/2 (x - xi)。贝der rationalen插值冯ƒ(x)的军队(a0 + a1x +…+ anxn) / (1 + b1x +……+ bmxm)在xi, i = 1 (1) n + m + 1 losen我们anstatt进行linearen系统(n + m + 1) - t好毛皮ai和bi努尔静脉系统的m - t好,1/(x - xi) in der " unvollstandigen baryzentrischen Form " von Zahler and Nenner zu erhalten。Die Ausdehnung audehnung auto - ordinung durch (a0 + a1x +…+ anr - 1xnr - 1)/(1 + b1x +…+ bmrxmr) /(1 + b1x +…+ bmrxmr) /(1 + b1x +…+ bmrxmr) / Die Losung eines Systems (rm + rn)- terordnung。In ahnlicher Behandlung nicht-osculatorischer - rationer trigonometrischer插值与直线ungerade Punktanzahl durch den Quotienten trigonometrischer Summen losen wir In lineares System 2m-ter anstelle (2n + 2m + 1)-ter Ordnung Fur die Koeffizienten von 1/ sin1 /2 (x - xi)。
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Pub Date : 1981-12-31DOI: 10.1515/9783112524121-011
A. Chyliński, T. Radoszewski
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Pub Date : 1981-12-31DOI: 10.1002/ZAMM.19810610309
N. Anderson, A. Arthurs
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