Padé expansion

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≡ expr· f(mod R ), respecively.

pade(exp(x),x,5,3)

or:

pade(exp(x),x,x^6,3)

−3 x²−24 x−60

x³−9 x²+36 x−60

To verify:

taylor((3*x^2+24*x+60)/(-x^3+9*x^2-36*x+60))

1+x+

x²

x³

x⁴

x⁵

120

+x⁶ order_size

⎛
⎝

⎞
⎠

which is the 5th-order series expansion of exp(x) at x=0.

pade((x^15+x+1)/(x^12+1),x,12,3)

or:

pade((x^15+x+1)/(x^12+1),x,x^13,3)

x+1

pade((x^15+x+1)/(x^12+1),x,14,4)

or:

pade((x^15+x+1)/(x^12+1),x,x^15,4)

2 x³+1

x¹¹−x¹⁰+x⁹−x⁸+x⁷−x⁶+x⁵−x⁴+x³+x²−x+1

To verify:

series(ans(),x=0,15)

1+x−x¹²−x¹³+2 x¹⁵+x¹⁶ order_size

⎛
⎝

⎞
⎠

Then:

series((x^15+x+1)/(x^12+1),x=0,15)

1+x−x¹²−x¹³+x¹⁵+x¹⁶ order_size

⎛
⎝

⎞
⎠

These two expressions have the same 14th-order series expansion at x=0.