On 2017-01-22, Dima Pasechnik wrote:
> (%i1) limit((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x),x,inf);
> (%o1) 2
>
> Although calling this version of Maxima as
> sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity)
>
> gives -Infinity, still, as
I think Dima very very confusingly actually agreed that this is a bug in
maxima that is in the newest version.
I think your options are:
- use the sympy option, or
- fix the bug in maxima, or
- change what sage uses by default, or
- wait
On Sun, Jan 22, 2017 at 12:49 PM Peleg Michaeli w
So... I couldn't really follow: should I open a new ticket, or should it be
solved by integrating the new maxima?
On Sunday, 22 January 2017 19:37:50 UTC+2, Dima Pasechnik wrote:
>
>
>
> On Sunday, January 22, 2017 at 4:13:53 PM UTC, William wrote:
>>
>> On Sun, Jan 22, 2017 at 7:14 AM, Peleg Mic
On Sunday, January 22, 2017 at 4:13:53 PM UTC, William wrote:
>
> On Sun, Jan 22, 2017 at 7:14 AM, Peleg Michaeli > wrote:
> > Hi,
> >
> > sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity)
> > -Infinity
> >
> > This is a wrong answer. It should be 2.
> >
> > Replacing
On Sun, Jan 22, 2017 at 7:14 AM, Peleg Michaeli wrote:
> Hi,
>
> sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity)
> -Infinity
>
> This is a wrong answer. It should be 2.
>
> Replacing 3/2 in the power by 1, 2, or 3 (at least) gives correct answers
> (2, inf, inf). Replacing i
Hi,
sage: ((2^(2*x+1)+(2^x*x^100)^(3/2))/(4^x-100*2^x)).limit(x=infinity)
-Infinity
This is a wrong answer. It should be 2.
Replacing 3/2 in the power by 1, 2, or 3 (at least) gives correct answers
(2, inf, inf). Replacing it by 5/2 given a wrong answer again.
Is this related to a known bug?
