The pade command finds a rational expression which agrees
with a function up to a given order.
pade takes 4 arguments
expr, an expression.
x, the variable name.
n, an integer or R, a polynomial.
p, an integer.
pade(expr,x,n,p) or
pade(expr,x,P,p)
returns a rational function P/Q such that
degree(P)<p and P/Q=expr (mod xn+1 )
(meaning that P/Q and f have the same
Taylor expansion at 0 up to order n) or
P/Q=exprf (mod R ), respecively.
Examples.
Input:
pade(exp(x),x,5,3)
or:
pade(exp(x),x,x^6,3)
Output:
−3 x2−24 x−60
x3−9 x2+36 x−60
To verify: Input:
taylor((3*x^2+24*x+60)/(-x^3+9*x^2-36*x+60))
Output:
1+x+
x2
2
+
x3
6
+
x4
24
+
x5
120
+x6order_size
⎛
⎝
x
⎞
⎠
which is the 5th-order series expansion of exp(x) at x=0.
Input:
pade((x^15+x+1)/(x^12+1),x,12,3)
or:
pade((x^15+x+1)/(x^12+1),x,x^13,3)
Output:
x+1
Input:
pade((x^15+x+1)/(x^12+1),x,14,4)
or:
pade((x^15+x+1)/(x^12+1),x,x^15,4)
Output:
2 x3+1
x11−x10+x9−x8+x7−x6+x5−x4+x3+x2−x+1
To verify: Input:
series(ans(),x=0,15)
Output:
1+x−x12−x13+2 x15+x16order_size
⎛
⎝
x
⎞
⎠
Then: Input:
series((x^15+x+1)/(x^12+1),x=0,15)
Output:
1+x−x12−x13+x15+x16order_size
⎛
⎝
x
⎞
⎠
These two expressions have the same 14th-order series expansion at x=0.