#288 another taylor bug

open
nobody
5
2012-12-18
2003-04-03
No

(C1) display2d:false;

(D1) FALSE
(C2) psi(yb,n):=yb^n/(n*(1/2-(yb+1)/((yb+1)^2+1))^(n/2));

(D2) PSI(yb,n):=yb^n/(n*(1/2-(yb+1)/((yb+1)^2+1))^(n/2))
(C3) taylor(psi(yb,2),yb,0,1);

(D3) +0
(C4) taylor(ratsimp(psi(yb,2)),yb,0,1);

(D4) 2+2*yb

(which is correct)

Martin

Discussion

  • Martin Rubey

    Martin Rubey - 2003-04-03

    Logged In: YES
    user_id=651552

    In fact, that was only half the story: Taylor always gives zero when the order of
    expansion is one less than the second argument to psi. I tried to set taylordepth to a
    higher value, but that doesn't change anything...

    (C3) taylor(psi(yb,5),yb,0,4);

    (D3) +0
    (C4) taylor(psi(yb,5),yb,0,5);

    (D4) 32/5+16*yb+20*yb^2+14*yb^3+23*yb^4/4+43*yb^5/40

    unfortunately, the ratsimp trick doesn't work here:

    (C5) taylor(ratsimp(psi(yb,5)),yb,0,5);

    TAYLOR encountered an unfamiliar singularity in:
    ABS(yb)
    -- an error. Quitting. To debug this try DEBUGMODE(TRUE);)

    Martin

     
  • Robert Dodier

    Robert Dodier - 2006-04-09
    • labels: 460522 --> Lisp Core - Taylor
     
  • Robert Dodier

    Robert Dodier - 2012-12-18

    On looking at this again, it seems that the problem is that psi(yb, n) is undefined at zero. If psi is redefined (e.g. via ratsimp) in such a way as to make it defined at zero, the problem goes away. Another way to avoid the undefined point is to expand around a point "a" and then take the limit as a goes to zero. So if I were going to try to fix it, I would look first at how taylor handles errors when evaluating at the given point.

    Another problem observed: with the stated definition of psi,

    taylor (ratsimp (psi (yb, 5)), yb, 0, 1);
    

    causes a stack overflow (Maxima 5.29+commits, ECL 12.2.1).

     
  • Rupert Swarbrick

    The stack overflow is avoided with the patch I posted last night to bug 2520 (although this just causes taylor to error out). Probably an improvement, but still not great.

     

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