Hello, I found out that trying to calculate log(a*x+b*x*x*log(x)).series(x==0,3) does not terminate. This is because in the function the argument has a pole even if a log(a*x) is subtracted. The attached patch should solve this. With this patch the result of the calculation is (log(x*a))+(a^(-1)*b*log(x))*x+(-1/2*a^(-2)*b^2*log(x)^2)*x^2+Order(x^3) as expected. Have a nice weekend, Chris Dams
Hi. On Fri, Apr 16, 2004 at 06:03:50PM +0000, Chris Dams wrote:
Hello,
I found out that trying to calculate
log(a*x+b*x*x*log(x)).series(x==0,3)
does not terminate. This is because in the function the argument has a pole even if a log(a*x) is subtracted. The attached patch should solve this. With this patch the result of the calculation is
(log(x*a))+(a^(-1)*b*log(x))*x+(-1/2*a^(-2)*b^2*log(x)^2)*x^2+Order(x^3)
as expected.
Your patch has been applied. Thanks! Bye, and nice weekend too, Jens
participants (2)
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Chris Dams
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Jens Vollinga