We've done a little work on being able to handle stuff like this, but
it's still rudimentary (and it basically uses list comprehensions
along the lines of the other answer. We'd welcome a ticket about this
specific use case.
Also, think of the Maxima solve capability as using dummy variables;
tha
wow, thank you. I would not have thought of using list comprehensions here.
Is there a way to do
[s for s in sol if ]
thanks
Robin
On Fri, Dec 9, 2011 at 11:48 AM, achrzesz wrote:
>
>
> On Dec 8, 9:36 pm, robin hankin wrote:
>> hello
>>
>> I have been playing with assumptions(). I want t
On Dec 8, 9:36 pm, robin hankin wrote:
> hello
>
> I have been playing with assumptions(). I want to assume a>b
> but solve() gives me a solution which is not consistent with this:
>
> sage: var('a b')
> (a, b)
> sage: assume(a>b)
> sage: assumptions()
> [a > b]
> sage: solve([a+b==2,a-b==0],a,
In the last post one can replace 0.99 by 1
but I wanted to exclude the following situation:
sage: n=var('n')
sage: x=var('x')
sage: x=(1/2)^(1/(n+1))
sage: limit(x^(n+1)/(1-x),n=+oo)
+Infinity # OK
On 16 Kwi, 15:36, achrzesz wrote:
> I must correct myself
> W... alpha:
> Assuming[x>-0.99,Assumin
I must correct myself
W... alpha:
Assuming[x>-0.99,Assuming[x<0.99,Limit[x^(n+1)/(1-x),n->+Infinity]]
gives correct answer 0
On 16 Kwi, 09:27, achrzesz wrote:
> In W... alpha
> Assuming[x>-0.99,x<0.99];Limit[x^(n+1)/(1-x),n->+Infinity]
> remains unevaluated, so Maxima, Sage are nol alone
>
> On 1
In W... alpha
Assuming[x>-0.99,x<0.99];Limit[x^(n+1)/(1-x),n->+Infinity]
remains unevaluated, so Maxima, Sage are nol alone
On 16 Kwi, 08:18, achrzesz wrote:
> One can discuss if in limits of f(x,n) as n-->oo
> x may depend on n or not but in the following version:
>
> sage: assume(x>-0.99,x<0.99
One can discuss if in limits of f(x,n) as n-->oo
x may depend on n or not but in the following version:
sage: assume(x>-0.99,x<0.99)
sage: n=var('n')
sage: sage: limit(x^(n+1)/(1-x), n=infinity)
-limit(x^(n + 1), n, +Infinity)/(x - 1)
sage: assume(x>0)
sage: sage: limit(x^(n+1)/(1-x), n=infinit
(Respectfully) Then why does splitting the range of values for x into
positive, zero and negative ranges work?
Joal Heagney
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Without additional assumption that x is constant
the limit is not zero (take for example x=(1/2)^(1/(n+1))
(W... alpha:
Assuming[x=const,x<1,x>0];Limit[x^(n+1)/(1-x),n->+Infinity] 0 OK,
Assuming[x<1,x>0];Limit[x^(n+1)/(1-x),n->+Infinity] unevaluated OK)
On 15 Kwi, 06:00, Dan Drake wrote:
> Why d
