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Re: The integration test suites for Sage.
Posted:
Sep 6, 2013 2:51 PM


>> Albert used Maxima 5.28 whereas I used Sage 5.10. I do not know which >> Maxima version Sage 5.10 uses. They might be different.
> The Maxima integrator would be undergoing noticeable development then. A > pleasant surprise.
Well, I don't know. I just switched from Sage 5.10 to 5.11 and there are differences with regard to the Charlwood problems! Problem 8 for example now has a monster solution; so long that I did not care to check if it is right or wrong.
> Charlwood_problem(43) > integrand : tan(x)/sqrt(tan(x)^4 + 1) > antideriv : 1/4*sqrt(2)*arctanh(1/2*(tan(x)^21)*sqrt(2)/sqrt(tan(x)^4+1)) > maxima : 1/4*sqrt(2)*arcsinh(2*sin(x)^2  1)
> After sign inversion the Maxima result appears to be correct on > the real axis.
Yes. And what about
diff(1/4*sqrt(2)*arctanh(1/2*(tan(x)^21)*sqrt(2)/sqrt(tan(x)^4+1)),x) = tan(x)/sqrt((tan(x))^4+1) versus diff(1/4*sqrt(2)*arcsinh(cos(2*x)),x) = sin(2*x)/sqrt(cos(4*x)+3)
tan(x)/sqrt((tan(x))^4+1) = sin(2*x)/sqrt(cos(4*x)+3) on the real axis?
> But then Maxima doesn't claim to deliver antiderivatives for the > entire complex plane, or does it?
What are rules of the game anyway: Does the 'Charlwood test' require antiderivatives for the entire complex plane or only for the real line? Charlwood writes: "We consider integrals of real elementary functions of a single real variable in the examples that follow."
>> Charlwood_problem(49) >> integrand : arcsin(x/sqrt(x^2 + 1)) >> antideriv : x*arcsin(x/sqrt(x^2 + 1)) + arctan(sqrt(2*x^2 + 1)) >> maxima : x*arcsin(x/sqrt(x^2 + 1))  1/2*(2*I*x^2 + I)/ sqrt(2*x^2  1)  1/2*I*sqrt(2*x^2  1)  1/2*I*log(sqrt(2*x^2  1)  1) + 1/2*I*log(sqrt(2*x^2  1) + 1)
> The Sage/Maxima result is more than just deficient: it is incorrect for > 1/SQRT(2) < x < 1/SQRT(2) on the real axis.
Ok.
Peter



