#314 Poverseries error for 1/(1-t)^2

closed
nobody
5
2012-12-04
2003-05-16
Anonymous
No

It seems powerseries(1(a-b*t)^n, t, 0) produces an
error for n=2, but is correct for n=1,3,4:

(C1) POWERSERIES(1/(a-b*t)^2, t, 0);

           INF
           ====
           \       I1  - I1 - 2       I1

(D1) > a b (I1 + 1) t
/
====
I1 = 0

[I believe that powers of a and b above have opposite
signs].

Discussion

  • Nobody/Anonymous

    Logged In: NO

    (C1) verbose:true;
    (D1) TRUE
    (C2) POWERSERIES(1/(a-b*t)^2, t, 0);

    In the first simplification we have returned:
    1
    --------------------
    2 2 2
    b t - 2 a b t + a

    trying to do a rational function expansion of
    1
    --------------------
    2 2 2
    b t - 2 a b t + a
    Using a special rule for expressions of form
    M - N
    (A + C VAR )

    Here we have
    [N = 2, A = - a, C = b, M = 1]

    INF

    \ I1 - I1 - 2 I1
    (D2) > a b (I1 + 1) t
    /
    ====
    I1 = 0
    (C3)

     
  • Martin Rubey

    Martin Rubey - 2003-05-19

    Logged In: YES
    user_id=651552

    This is fixed with the fix for [727542] powerseries wrong/fix

    Martin

    (please, somebody merge it into cvs, it is IMPORTANT!)

     
  • Raymond Toy

    Raymond Toy - 2003-05-28
    • status: open --> closed
     
  • Raymond Toy

    Raymond Toy - 2003-05-28

    Logged In: YES
    user_id=28849

    Bug marked as duplicate. I'll apply your patch for bug 727542.

     

Log in to post a comment.