On Mon, 10 Nov 2008 17:36:46 -0800 (PST)
cesarnda [EMAIL PROTECTED] wrote:
Actually this sum can't be done by Maxima, but Derive can do it (even
an old version of derive). do you have an idea of how this problem is
planning to be solved?
As Robert Dodier pointed out, Maxima can actually do
On Thu, Nov 13, 2008 at 10:31 AM, Burcin Erocal [EMAIL PROTECTED] wrote:
On Mon, 10 Nov 2008 17:36:46 -0800 (PST)
cesarnda [EMAIL PROTECTED] wrote:
Actually this sum can't be done by Maxima, but Derive can do it (even
an old version of derive). do you have an idea of how this problem is
On Thu, 13 Nov 2008 04:09:56 -0600
Jason Grout [EMAIL PROTECTED] wrote:
Burcin Erocal wrote:
Returning to the question of how Sage plans to handle this, the
short answer is I am working on it. :)
Yeah!
We will have a completely new implementation of summation,
On Thu, 13 Nov 2008 12:32:54 +0100
Ondrej Certik [EMAIL PROTECTED] wrote:
On Thu, Nov 13, 2008 at 10:31 AM, Burcin Erocal [EMAIL PROTECTED]
wrote:
snip
We will have a completely new implementation of summation,
independent of the one in Maxima. At the moment I have an
implementation
Burcin Erocal wrote:
On Thu, 13 Nov 2008 04:09:56 -0600
Jason Grout [EMAIL PROTECTED] wrote:
Burcin Erocal wrote:
Returning to the question of how Sage plans to handle this, the
short answer is I am working on it. :)
Yeah!
We will have a completely new implementation of summation,
On Tue, Nov 11, 2008 at 3:15 AM, cesarnda [EMAIL PROTECTED] wrote:
that is the output I was expecting, but it is not the input I gave.
Obviously,
1/x - 1/(x+1) = 1/(x*(x+1))
but, if the right hand side can be done why the left hand side can't?
This is the bug I was talking about...
It
On Nov 10, 7:15 pm, cesarnda [EMAIL PROTECTED] wrote:
that is the output I was expecting, but it is not the input I gave.
Obviously,
1/x - 1/(x+1) = 1/(x*(x+1))
but, if the right hand side can be done why the left hand side can't?
This is the bug I was talking about...
Thanks for pointing
On Mon, Nov 10, 2008 at 5:29 PM, cesarnda [EMAIL PROTECTED] wrote:
how could I compute this:
sum_{ x = 1}^{\infty} 1/x - 1/(x+1)
or
sum(1/x-1/(x+1),x,1, infinity)
directly in Sage, without calling maxima or sympy?
Unfortunately, this isn't implemented yet. See:
Actually this sum can't be done by Maxima, but Derive can do it (even
an old version of derive). do you have an idea of how this problem is
planning to be solved?
On 10 nov, 19:30, William Stein [EMAIL PROTECTED] wrote:
On Mon, Nov 10, 2008 at 5:29 PM, cesarnda [EMAIL PROTECTED] wrote:
how
On Mon, Nov 10, 2008 at 5:36 PM, cesarnda [EMAIL PROTECTED] wrote:
Actually this sum can't be done by Maxima, but Derive can do it (even
an old version of derive). do you have an idea of how this problem is
planning to be solved?
Is this the answer you were expecting?
(%i6)
that is the output I was expecting, but it is not the input I gave.
Obviously,
1/x - 1/(x+1) = 1/(x*(x+1))
but, if the right hand side can be done why the left hand side can't?
This is the bug I was talking about...
On 10 nov, 19:51, Mike Hansen [EMAIL PROTECTED] wrote:
On Mon, Nov 10, 2008 at
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