帕德近似

帕德近似（英語：Padé approximant）是法国数学家亨利·帕德发明的有理多项式近似法。帕德近似往往比截断的泰勒級數准确，而且当泰勒级数不收敛时，帕德近似往往仍可行，所以多用于在计算机数学中。 例如 1 1 − x {\displaystyle {\frac {1}{1-x}}} 的泰勒级数 1 + x + x 2 + x 3 + ⋯ {\displaystyle 1+x+x^{2}+x^{3}+\cdots } 只有在 − 1 < x < 1 {\displaystyle -1<x<1} 时收敛，不如原函数广泛。 给定自然数m和正整数n, 函数 f ( x ) {\displaystyle f(x)} 的[m,n]阶帕德近似为 R ( x ) = ∑ j = 0 m a j x j 1 + ∑ k = 1 n b k x k = a 0 + a 1 x + a 2 x 2 + ⋯ + a m x m 1 + b 1 x + b 2 x 2 + ⋯ + b n x n {\displaystyle R(x)={\frac {\sum _{j=0}^{m}a_{j}x^{j}}{1+\sum _{k=1}^{n}b_{k}x^{k}}}={\frac {a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{m}x^{m}}{1+b_{1}x+b_{2}x^{2}+\cdots +b_{n}x^{n}}}} 并且 f ( 0 ) = R ( 0 ) f ′ ( 0 ) = R ′ ( 0 ) f ″ ( 0 ) = R ″ ( 0 ) ⋮ f ( m + n ) ( 0 ) = R ( m + n ) ( 0 ) {\displaystyle {\begin{array}{rcl}f(0)&=&R(0)\\f'(0)&=&R'(0)\\f''(0)&=&R''(0)\\&\vdots &\\f^{(m+n)}(0)&=&R^{(m+n)}(0)\end{array}}} 对于给定的 m , n {\displaystyle m,n} 函数 f ( x ) {\displaystyle f(x)} 的[m,n]阶帕德近似是唯一的。 函数 f ( x ) {\displaystyle f(x)} 的帕德近似记为 [ m / n ] f ( x ) . {\displaystyle [m/n]_{f}(x).\,} [ 6 / 6 ] sin ⁡ ( x ) = ( 12671 / 4363920 ) ∗ x 5 − ( 2363 / 18183 ) ∗ x 3 + x 1 + ( 445 / 12122 ) ∗ x 2 + ( 601 / 872784 ) ∗ x 4 + ( 121 / 16662240 ) ∗ x 6 {\displaystyle [6/6]_{\sin(x)}={\frac {(12671/4363920)*x^{5}-(2363/18183)*x^{3}+x}{1+(445/12122)*x^{2}+(601/872784)*x^{4}+(121/16662240)*x^{6}}}} [ 6 / 6 ] sin ⁡ ( x ) {\displaystyle [6/6]_{\sin(x)}} 的6+6=12阶泰勒级数展开为 x − ( 1 / 6 ) ∗ x 3 + ( 1 / 120 ) ∗ x 5 − ( 1 / 5040 ) ∗ x 7 + ( 1 / 362880 ) ∗ x 9 − ( 1 / 39916800 ) ∗ x 11 + O ( x 13 ) {\displaystyle {x-(1/6)*x^{3}+(1/120)*x^{5}-(1/5040)*x^{7}+(1/362880)*x^{9}-(1/39916800)*x^{11}+O(x^{13})}} 和 sin ⁡ ( x ) {\displaystyle \sin(x)} 的12阶泰勒级数全同： sin ⁡ ( x ) ≈ x − ( 1 / 6 ) ∗ x 3 + ( 1 / 120 ) ∗ x 5 − ( 1 / 5040 ) ∗ x 7 + ( 1 / 362880 ) ∗ x 9 − ( 1 / 39916800 ) ∗ x 11 + O ( x 13 ) {\displaystyle \sin(x)\approx {x-(1/6)*x^{3}+(1/120)*x^{5}-(1/5040)*x^{7}+(1/362880)*x^{9}-(1/39916800)*x^{11}+O(x^{13})}} [ 5 / 5 ] e x p ( x ) = 1 + ( 1 / 9 ) ∗ x 2 + ( 1 / 2 ) ∗ x + ( 1 / 72 ) ∗ x 3 + ( 1 / 1008 ) ∗ x 4 + ( 1 / 30240 ) ∗ x 5 1 + ( 1 / 9 ) ∗ x 2 − ( 1 / 2 ) ∗ x − ( 1 / 72 ) ∗ x 3 + ( 1 / 1008 ) ∗ x 4 − ( 1 / 30240 ) ∗ x 5 {\displaystyle [5/5]_{exp(x)}={\frac {1+(1/9)*x^{2}+(1/2)*x+(1/72)*x^{3}+(1/1008)*x^{4}+(1/30240)*x^{5}}{1+(1/9)*x^{2}-(1/2)*x-(1/72)*x^{3}+(1/1008)*x^{4}-(1/30240)*x^{5}}}} 其泰勒级数为 1 + x + ( 1 / 2 ) ∗ x 2 + ( 1 / 6 ) ∗ x 3 + ( 1 / 24 ) ∗ x 4 + ( 1 / 120 ) ∗ x 5 + ( 1 / 720 ) ∗ x 6 + ( 1 / 5040 ) ∗ x 7 + ( 1 / 40320 ) ∗ x 8 + ( 1 / 362880 ) ∗ x 9 + ( 1 / 3628800 ) ∗ x 10 + ( 23 / 914457600 ) ∗ x 11 + O ( x 12 ) {\displaystyle {1+x+(1/2)*x^{2}+(1/6)*x^{3}+(1/24)*x^{4}+(1/120)*x^{5}+(1/720)*x^{6}+(1/5040)*x^{7}+(1/40320)*x^{8}+(1/362880)*x^{9}+(1/3628800)*x^{10}+(23/914457600)*x^{11}+O(x^{12})}} 与exp(x)本身的泰勒级数展开的前10阶完全等同： 1 + x + ( 1 / 2 ) ∗ x 2 + ( 1 / 6 ) ∗ x 3 + ( 1 / 24 ) ∗ x 4 + ( 1 / 120 ) ∗ x 5 + ( 1 / 720 ) ∗ x 6 + ( 1 / 5040 ) ∗ x 7 + ( 1 / 40320 ) ∗ x 8 + ( 1 / 362880 ) ∗ x 9 + ( 1 / 3628800 ) ∗ x 10 + ( 1 / 39916800 ) ∗ x 11 + O ( x 12 ) {\displaystyle {1+x+(1/2)*x^{2}+(1/6)*x^{3}+(1/24)*x^{4}+(1/120)*x^{5}+(1/720)*x^{6}+(1/5040)*x^{7}+(1/40320)*x^{8}+(1/362880)*x^{9}+(1/3628800)*x^{10}+(1/39916800)*x^{11}+O(x^{12})}} 又如 f := 1 − cos ⁡ ( 2 ∗ x ) 2 1 + arctan ⁡ ( 3 ∗ x ) {\displaystyle f:={\frac {1-\cos(2*x)^{2}}{1+\arctan(3*x)}}} [ 3 / 3 ] f ( x ) = ( 64 / 75 ) ∗ x 3 + 4 ∗ x 2 1 + ( 241 / 75 ) ∗ x + ( 148 / 75 ) ∗ x 2 − ( 1061 / 225 ) ∗ x 3 {\displaystyle [3/3]_{f(x)}={\frac {(64/75)*x^{3}+4*x^{2}}{1+(241/75)*x+(148/75)*x^{2}-(1061/225)*x^{3}}}} − ( 9853969 / 39583665 ) ∗ z 5 − ( 1493060 / 2638911 ) ∗ z 3 + z 1 + ( 968375 / 879637 ) ∗ z 2 − ( 1167506 / 7916733 ) ∗ z 4 + ( 867043 / 2159109 ) ∗ z 6 {\displaystyle {\frac {-(9853969/39583665)*z^{5}-(1493060/2638911)*z^{3}+z}{1+(968375/879637)*z^{2}-(1167506/7916733)*z^{4}+(867043/2159109)*z^{6}}}} − ( 107 / 28416000 ) ∗ x 7 + ( 1 / 3840 ) ∗ x 5 1 + ( 151 / 5550 ) ∗ x 2 + ( 1453 / 3729600 ) ∗ x 4 + ( 1339 / 358041600 ) ∗ x 6 + ( 2767 / 120301977600 ) ∗ x 8 {\displaystyle {\frac {-(107/28416000)*x^{7}+(1/3840)*x^{5}}{1+(151/5550)*x^{2}+(1453/3729600)*x^{4}+(1339/358041600)*x^{6}+(2767/120301977600)*x^{8}}}} ( 2 / 15 ) ∗ ( 49140 ∗ x + 3570 ∗ x 3 + 739 ∗ x 5 ) ( 165 ∗ π ∗ x 4 + 1330 ∗ π ∗ x 2 + 3276 ∗ π ) {\displaystyle {\frac {(2/15)*(49140*x+3570*x^{3}+739*x^{5})}{(165*{\sqrt {\pi }}*x^{4}+1330*{\sqrt {\pi }}*x^{2}+3276*{\sqrt {\pi }})}}{}} ( 1 / 135 ) ∗ ( 990791 ∗ x 9 ∗ π 4 − 147189744 ∗ x 5 ∗ π 2 + 8714684160 ∗ x ) ( 1749 ∗ π 4 ∗ x 8 + 523536 ∗ π 2 ∗ x 4 + 64553216 ) {\displaystyle {\frac {(1/135)*(990791*x^{9}*\pi ^{4}-147189744*x^{5}*\pi ^{2}+8714684160*x)}{(1749*\pi ^{4}*x^{8}+523536*\pi ^{2}*x^{4}+64553216)}}} Maple中 pade(f(x),x,[m,n]); 其中 m,n 分别表示 分子、分母的级数； Baker, G. A., Jr.; and Graves-Morris, P. Padé Approximants. Cambridge U.P., 1996 Baker, G. A., Jr. Padé approximant （页面存档备份，存于互联网档案馆）, Scholarpedia （页面存档备份，存于互联网档案馆）, 7(6):9756. Brezinski, C.; and Redivo Zaglia, M. Extrapolation Methods.= Theory and Practice. North-Holland, 1991 Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP, Section 5.12 Padé Approximants, Numerical Recipes: The Art of Scientific Computing 3rd, New York: Cambridge University Press, 2007 [2015-02-24], ISBN 978-0-521-88068-8, （原始内容存档于2016-03-03）  Frobenius, G.; Ueber Relationen zwischem den Näherungsbrüchen von Potenzreihen, [Journal für die reine und angewandte Mathematik (Crelle's Journal)]. Volume 1881, Issue 90, Pages 1–17 Gragg, W.B.; The Pade Table and Its Relation to Certain Algorithms of Numerical Analysis [SIAM Review], Vol. 14, No. 1, 1972, pp. 1–62. Padé, H.; Sur la répresentation approchée d'une fonction par des fractions rationelles, Thesis, [Ann. \'Ecole Nor. (3), 9, 1892, pp. 1–93 supplement. 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