Open benruijl opened 4 years ago
Don’t forget that the expansion is relative to the leading term. You can try rat(ep^4+ep^5+ep^6+ep^7+ep^8,1) and see what happens.
Jos
On 8 Jun 2020, at 14:41, Ben Ruijl notifications@github.com wrote:
When all terms in the numerator depend on the expansion variable and have power > 1, polyratfun(expand,....) does not truncate the numerator.
S eps; CF rat; Polyratfun rat(expand,eps,3);
L F = rat(eps^4,1); Print +s; .end Gives rat(eps^4,1) instead of rat(1,1).
Similary, it will not expand rat(eps^4 + eps^5,1); and rat(eps^2 + eps^5,1); It will expand rat(eps + eps^5,1); and rat(1 + eps^5,1);.
This is in contradiction with the manual, where it says: "eventually all terms with powers of x that are greater than ’power’ will be discarded".
— You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub https://github.com/vermaseren/form/issues/348, or unsubscribe https://github.com/notifications/unsubscribe-auth/ABJPCEQ6KIYPYWVZKKYADHDRVTMA5ANCNFSM4NYL3XWQ.
I figured something like this is happening, but then we should write that in the manual more clearly.
I interpret "eventually all terms with powers of x that are greater than ’power’ will be discarded" as a statement about the absolute power of x.
When all terms in the numerator depend on the expansion variable and have power > 1,
polyratfun(expand,....)
does not truncate the numerator.Gives
rat(eps^4,1)
instead ofrat(1,1)
.Similary, it will not expand
rat(eps^4 + eps^5,1);
andrat(eps^2 + eps^5,1);
It will expandrat(eps + eps^5,1);
andrat(1 + eps^5,1);
.This is in contradiction with the manual, where it says: "eventually all terms with powers of x that are greater than ’power’ will be discarded".