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Submitted by : (unknown) at: 2007-11-17T22:22:01-08:00 (9 years ago)
Name :
Axiom Version :
Category : Severity : Status :
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When I type:

fricas
ex1:=(x-x/(x+exp(x)/(x*log(x+x))))*exp(2)

\label{eq1}{{{\left({{x}^{3}}-{{x}^{2}}\right)}\ {{e}^{2}}\ {\log \left({2 \  x}\right)}}+{x \ {{e}^{2}}\ {{e}^{x}}}}\over{{{{x}^{2}}\ {\log \left({2 \  x}\right)}}+{{e}^{x}}}(1)
Type: Expression(Integer)
fricas
ex2:=D(ex1, x)

\label{eq2}{\left(
\begin{array}{@{}l}
\displaystyle
{{{x}^{4}}\ {{e}^{2}}\ {{\log \left({2 \  x}\right)}^{2}}}+{{\left({3 \ {{x}^{2}}}-{2 \  x}\right)}\ {{e}^{2}}\ {{e}^{x}}\ {\log \left({2 \  x}\right)}}+ 
\
\
\displaystyle
{{{e}^{2}}\ {{{e}^{x}}^{2}}}-{x \ {{e}^{2}}\ {{e}^{x}}}
(2)
Type: Expression(Integer)
fricas
ex3:=simplify(ex2)

\label{eq3}{\left(
\begin{array}{@{}l}
\displaystyle
{{{x}^{4}}\ {{e}^{2}}\ {{\log \left({2 \  x}\right)}^{2}}}+{{\left({3 \ {{x}^{2}}}-{2 \  x}\right)}\ {{e}^{x + 2}}\ {\log \left({2 \  x}\right)}}+ 
\
\
\displaystyle
{{e}^{{2 \  x}+ 2}}-{x \ {{e}^{x + 2}}}
(3)
Type: Expression(Integer)
fricas
integrate(ex3, x)

\label{eq4}{{{{x}^{3}}\ {{{e}^{2}}^{2}}\ {\log \left({2 \  x}\right)}}+{{\left(x + 1 \right)}\ {{e}^{2}}\ {{e}^{x + 2}}}}\over{{{{x}^{2}}\ {{e}^{2}}\ {\log \left({2 \  x}\right)}}+{{e}^{x + 2}}}(4)
Type: Union(Expression(Integer),...)

then Axiom 3.9 returns unevaluated integral.

Note, that if I remove the exp(2) factor, then Axiom is able to compute the integral.

fricas
ex4:=(x-x/(x+exp(x)/(x*log(x+x))))

\label{eq5}{{{\left({{x}^{3}}-{{x}^{2}}\right)}\ {\log \left({2 \  x}\right)}}+{x \ {{e}^{x}}}}\over{{{{x}^{2}}\ {\log \left({2 \  x}\right)}}+{{e}^{x}}}(5)
Type: Expression(Integer)
fricas
ex5:=D(ex4, x)

\label{eq6}{{{{x}^{4}}\ {{\log \left({2 \  x}\right)}^{2}}}+{{\left({3 \ {{x}^{2}}}-{2 \  x}\right)}\ {{e}^{x}}\ {\log \left({2 \  x}\right)}}+{{{e}^{x}}^{2}}-{x \ {{e}^{x}}}}\over{{{{x}^{4}}\ {{\log \left({2 \  x}\right)}^{2}}}+{2 \ {{x}^{2}}\ {{e}^{x}}\ {\log \left({2 \  x}\right)}}+{{{e}^{x}}^{2}}}(6)
Type: Expression(Integer)
fricas
ex6:=simplify(ex5)

\label{eq7}{{{{x}^{4}}\ {{\log \left({2 \  x}\right)}^{2}}}+{{\left({3 \ {{x}^{2}}}-{2 \  x}\right)}\ {{e}^{x}}\ {\log \left({2 \  x}\right)}}+{{e}^{2 \  x}}-{x \ {{e}^{x}}}}\over{{{{x}^{4}}\ {{\log \left({2 \  x}\right)}^{2}}}+{2 \ {{x}^{2}}\ {{e}^{x}}\ {\log \left({2 \  x}\right)}}+{{e}^{2 \  x}}}(7)
Type: Expression(Integer)
fricas
integrate(ex6, x)

\label{eq8}{{{{x}^{3}}\ {\log \left({2 \  x}\right)}}+{{\left(x + 1 \right)}\ {{e}^{x}}}}\over{{{{x}^{2}}\ {\log \left({2 \  x}\right)}}+{{e}^{x}}}(8)
Type: Union(Expression(Integer),...)

We claim that Axiom is good at integration, therefore critical --kratt6, Wed, 10 May 2006 01:28:56 -0500 reply
Severity: normal => critical

We claim that Axiom is good at integration, therefore critical --kratt6, Wed, 10 May 2006 01:29:14 -0500 reply
Category: Axiom Mathematics => Axiom Library

Related example --unknown, Wed, 10 May 2006 06:22:25 -0500 reply
If I try:
fricas
integrate(simplify(((x^4*log(2*x)^2+(3*x^2-2*x)*exp(x)*log(2*x)+exp(2*x)-x*exp(x))/(x^4*log(2*x)^2+2*x^2*exp(x)*log(2*x)+exp(2*x))-1)*exp(2)), x)

\label{eq9}{{{e}^{2}}\ {{e}^{x + 2}}}\over{{{{x}^{2}}\ {{e}^{2}}\ {\log \left({2 \  x}\right)}}+{{e}^{x + 2}}}(9)
Type: Union(Expression(Integer),...)

I see that Axiom failed to recognize that e^4 - (e^2)^2 is zero. AFAICS similar problem appears in original example: when I trace rischNormalize I see that both e^2 and e^4 appear in "normalized" function. But this may cause trouble later, since Axiom assumes that different kernels are algebraically independent. I do not know if rischNormalize is supposed to simplify constants (maybe this shoud be done by a separtate pass) but failing to recognize "transcendental" zero may give wrong results and otherwise Risch algorithm may declare an integrable function as non integrable.

fixed in FriCAS revision 475 --kratt6, Fri, 10 Apr 2009 03:20:47 -0700 reply
Status: open => fixed somewhere




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