MaplePrimes - answers and comments on Question, computing a residue

series and term at 1/x

Axel Vogt — Sun, 09 Dec 2012 20:17:55 Z

Using 'series' gives a false result, 'MultiSeries:-series' however works.

f:= x -> 1/x^(1/3)/(x^2+2*x*cos(phi)+1);

series(f(x), x=-exp(I*phi), 2):  simplify(%); # gives residue=0

MultiSeries:-series(f(x), x=-exp(I*phi), 2); # works

For phi = Pi/2 that results in 1/4*I*(3^(1/2)+I)/(x+I), giving
-.250000000000000+.433012701892220*I

Cross check z:= t -> -I+exp(I*t*Pi*2); is a curve around -I
with correct orientation, chosen for phi = Pi/2 and

'Int(f(z(t))*D(z)(t), t=0..1)/(2*Pi*I)';
eval(%, phi=Pi/2);
evalf[10](%);

-.2500000000+.4330127017*I

'confirms' that

unprotect hack

Alejandro Jakubi — Mon, 10 Dec 2012 07:48:18 Z

Yes, this unprotect hack seems to work fine. And the result can be put as shown in the link above, with p=1/3:

unprotect(`series`):
series:=MultiSeries:-series:
residue(1/x^(1/3)/(x^2+2*x*cos(phi)+1), x=-exp(I*phi));
                                   3 exp(phi I)
         - ------------------------------------------------------------
                        1/3              2
           (-exp(phi I))    (7 exp(phi I)  - 8 cos(phi) exp(phi I) + 1)

dividend(%,numer(%),[u->u,simplify@(u->convert(u,exp))]);
                                     1/2 I
                           -------------------------
                                        1/3
                           (-exp(phi I))    sin(phi)

Where I am using the procedure dividend.

series

acer — Mon, 10 Dec 2012 00:47:56 Z

Can one unprotect `series`, assign it to MultiSeries:-series, and then call `residue`?

acer