亨利·帕德 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}}}
1
+
x
+
x
2
+
x
3
+
⋯
{\displaystyle 1+x+x^{2}+x^{3}+\cdots }
−
1
<
x
<
1
{\displaystyle -1<x<1}
给定自然数 m和正整数n, 函数
f
(
x
)
{\displaystyle f(x)}
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}
f
(
x
)
{\displaystyle f(x)}
函数
f
(
x
)
{\displaystyle f(x)}
[
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)}}
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}}}}
雅可比橢圓函數
sn
(
x
;
3
)
{\displaystyle \operatorname {sn} (x;3)}
[ 编辑 ]
−
(
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}}}}
第一類 5 階貝塞爾函數
J
5
(
x
)
{\displaystyle J_{5}(x)}
[ 编辑 ]
−
(
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 }})}}{}}
菲涅耳積分
C
(
x
)
{\displaystyle C(x)}
[ 编辑 ]
(
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计算 [ 编辑 ] Maple中
pade(f(x),x,[m,n]);
其中 m,n 分别表示 分子、分母的级数；
参考文献 [ 编辑 ] 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.
Wynn, P. , Upon systems of recursions which obtain among the quotients of the Padé table, Numerische Mathematik, 1966, 8 (3): 264–269, doi:10.1007/BF02162562 
![{\displaystyle [m/n]_{f}(x).\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9fafa682ed8fe334d93506eca1b70639b29ed3e)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66704d8fd2d861021c55cb84a9fb623e37f25f00)
![{\displaystyle [6/6]_{\sin(x)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d7e5964ced96191cf94d20880d88ce39f629540)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51174c9dceb2ee02fdd342eb8e78a332fc91d6a1)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb2833ba76aaa09f44f30d46739787b20d0795fd)
