#1059 infinite sums with simpsum : true

closed
nobody
Lisp Core (472)
5
2007-09-26
2007-01-11
Barton Willis
No

Should be inf, not inf - 1:

(%i2) sum(exp(k),k,minf,inf),simpsum;
(%o2) inf-1

I think 'und' is a better value for this sum:

(%i3) sum(k,k,minf,inf),simpsum;
(%o3) 0

Should be 'und', not 'undefined':

(%i4) sum((-1)^k,k,0,inf),simpsum;
(%o4) undefined

Again, should be 'und'

(%i5) sum((-1)^k,k,minf,inf),simpsum;
(%o5) undefined-1

At least limit(und-1) --> und, but limit(undefined-1) -> undefined-1.

Discussion

  • Robert Dodier
    Robert Dodier
    2007-01-13

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    Maybe 1-argument limit should be called to clean up results of infinite summations. Although limit has its own problems so I am hesitant about that ...

     
  • Robert Dodier
    Robert Dodier
    2007-01-13

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    Compare the following

    (%i1) integrate(x, x, 0, inf);
    Integral is divergent
    -- an error. Quitting. To debug this try debugmode(true);
    (%i2) sum(k, k, 0, inf), simpsum;
    (%o2) inf

    Since maxima is not very good with infinities (that is inf-inf = 0 and inf/inf=1), sum should signal an error instead of returning inf for divergent sums.

    Andrej

     
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    Fixed in cvs:

    (%i2) sum(exp(k),k,minf,inf),simpsum;
    (%o2) inf
    (%i3) sum(k,k,minf,inf),simpsum;
    (%o3) und
    (%i4) sum((-1)^k,k,0,inf),simpsum;
    (%o4) und
    (%i5) sum((-1)^k,k,minf,inf),simpsum;
    (%o5) und

    Andrej