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2.33.5  Padé expansion: pade

pade takes 4 arguments

pade returns a rationnal fraction P/Q such that degree(P)<p and P/Q=f (mod xn+1 ) or P/Q=f (mod N ). In the first case, it means that P/Q and f have the same Taylor expansion at 0 up to order n.
Input :

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

Or :

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

Output :

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

To verify input :

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

Output :

1+x+1/2*x^2+1/6*x^3+1/24*x^4+1/120*x^5+x^6*order_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*x^3-1)/(-x^11+x^10-x^9+x^8-x^7+x^6-x^5+x^4- x^3-x^2+x-1)

To verify, input :

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

Output :

1+x-x^12-x^13+2x^15+x^16*order_size(x)

then input :

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

Output :

1+x-x^12-x^13+x^15+x^16*order_size(x)

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


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