Hi,
Thank u everybody for your help.
i'm going to use Maxima and Maple to compute my expression.
Raouf
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On Sat, Aug 30, 2008 at 11:15 AM, Robert Dodier <[EMAIL PROTECTED]> wrote:
> Dunno if it matters but maybe you can handle this directly in Maxima.
>
> foo : sum (1/(k + m)^3, k, 1, inf);
>
> load (simplify_sum);
> simplify_sum (foo);
> => -psi[2](m+1)/2
>
> ev (%, m=2);
> => zeta(3)-9/8
>
> ev (
Raouf wrote:
> I am a newbie in sage and i want to compute an infinite sum with
> parameter m, like sum(1/(k+m)^3) k=1 to infinity.
Dunno if it matters but maybe you can handle this directly in Maxima.
foo : sum (1/(k + m)^3, k, 1, inf);
load (simplify_sum);
simplify_sum (foo);
=> -psi[2](m+1
On Sat, Aug 30, 2008 at 4:34 PM, Ondrej Certik <[EMAIL PROTECTED]> wrote:
> On Sat, Aug 30, 2008 at 6:06 AM, tkeller <[EMAIL PROTECTED]> wrote:
>>
>> I asked this question myself a few months ago, and the easiest 2
>> solutions seem to be utilizing sympy or maxima.
>>
>> Via sympy it is:
>> import
On Sat, Aug 30, 2008 at 6:06 AM, tkeller <[EMAIL PROTECTED]> wrote:
>
> I asked this question myself a few months ago, and the easiest 2
> solutions seem to be utilizing sympy or maxima.
>
> Via sympy it is:
> import sympy
> sympy.var('x')
> print sympy.sum(2**(-x), (x, 1, oo))
>
> I'm taking this
Hi
Thank u Thomas for these idea.
In fact i divide the expression that i want to estimate into 2 part,
the first( where we found the infinite sum with parameters) i can
calculate it with Maple but i don't know how to use Maple into the
Sage's notebook.
And the second part of expression i must do i
I asked this question myself a few months ago, and the easiest 2
solutions seem to be utilizing sympy or maxima.
Via sympy it is:
import sympy
sympy.var('x')
print sympy.sum(2**(-x), (x, 1, oo))
I'm taking this from a question I posed on the sympy message list:
http://groups.google.com/group/sym