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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 9,169 of 10,432    |
|    clicliclic@freenet.de to antispam@math.uni.wroc.pl    |
|    Re: Announce: FriCAS 1.3.0 has been rele    |
|    22 Sep 16 19:59:38    |
      antispam@math.uni.wroc.pl schrieb:       >       > clicliclic@freenet.de wrote:       > >       > > I suppose "more algebraic functions" include those cube-root integrands       > > whose Trager resultants are polynomials in t^3. When some change in the       > > integrand coefficients makes such integrals non-elementary, does FriCAS       > > 1.3.0 certify them as non-elementary by returning them unevaluated, or       > > does it merely exit with an error message like "implementation       > > incomplete (trace 0)"?       >       > Trager resultant is of form t^3 - a (that us residues are third roots       > ) and integral is nonelementary current implementation will detect it.       > But as you showed adding extra part can change Trager resultant...       >              ... which doesn't sound promising. But I see that I should have been       specific. So, how does FriCAS 1.3.0 react to:               INT(1/(x^3 - 3*x^2 + 7*x - 4)^(1/3), x) = ?               INT(1/(x*(3*x^2 - 6*x + 5)^(1/3)), x) = ?               INT((1 - x^3)^(1/3)/(2 + x), x) = ?               INT((2 + x)/((1 + x + x^2)*(2 + x^3)^(1/3)), x) = ?              All four are 'detuned' versions of pseudo-elliptics considered earlier.                     > > Analogy suggests that Trager's resultants for classical Goursat pseudo-       > > elliptics involving square roots of quartics or cubics are limited to       > > polynomials in t^2. Perhaps the problem integrand (x - SQRT(a^2+1) - a)       > > / ((x + SQRT(a^2+1) - a)*SQRT((x - a)*(x^2 + 1))) could thus also be       > > handled in finite time by hard-coding of the splitting field.       >       > The problem is quite different here.       >              Taking Goursat's elementary viewpoint, integrability detection and       antiderivative computation are closely related for the square-root and       cube-root problems. That's why I would be surprised if the Trager       resultant of this problem integrand were not just a polynomial in t^2.              Martin.              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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