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Submitted by : (unknown) at: 2007-11-17T22:09:23-08:00 (9 years ago)
Name :
Axiom Version :
Category : Severity : Status :
Optional subject :  
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axiom
D(product((1-q^(n-i))/(1-q^(m-i)),i=0..m-1),q)

\label{eq1}{\left({\prod_{
\displaystyle
{i = 0}}^{
\displaystyle
{m - 1}}{{{q^{\left(n - i \right)}}- 1}\over{{q^{\left(m - i \right)}}- 1}}}\right)}\ {\left({\sum_{
\displaystyle
{i = 0}}^{
\displaystyle
{m - 1}}{{{{\left(- m + i \right)}\ {q^{\left(m - i - 1 \right)}}\ {q^{\left(n - i \right)}}}+{{\left({{\left(n - i \right)}\ {q^{\left(m - i \right)}}}- n + i \right)}\ {q^{\left(n - i - 1 \right)}}}+{{\left(m - i \right)}\ {q^{\left(m - i - 1 \right)}}}}\over{{{\left({q^{\left(m - i \right)}}- 1 \right)}\ {q^{\left(n - i \right)}}}-{q^{\left(m - i \right)}}+ 1}}}\right)}(1)
Type: Expression(Integer)
axiom
f:=operator 'f;
axiom
D(product(f(i,q),i=0..m),q)

\label{eq2}{\left(\prod_{
\displaystyle
{i = 0}}^{
\displaystyle
m}{f \left({i , \: q}\right)}\right)}\ {\left(\sum_{
\displaystyle
{i = 0}}^{
\displaystyle
m}{{{f_{, 2}}\left({i , \: q}\right)}\over{f \left({i , \: q}\right)}}\right)}(2)
Type: Expression(Integer)

I'll try to correct this tomorrow...

Martin

Fortunately, a fix is quite easy, since we know how to differentiate products according to Leibniz rule: Here is a patch to combfunc.spad.pamphlet that also fixes some leftover problems with differentiating sums without bounds and displaying sums and products with and without bounds:

combfunc.spad.pamphlet.patch

property change --kratt6, Tue, 04 Oct 2005 07:38:49 -0500 reply
Status: open => fix proposed

test April release --Bill Page, Wed, 21 Jun 2006 06:23:50 -0500 reply
axiom
)version
Value = "FriCAS 2010-12-08 compiled at Tuesday April 5, 2011 at 13:07:45 " D(product((1-q^(n-i))/(1-q^(m-i)),i=0..m-1),q)

\label{eq3}{\left({\prod_{
\displaystyle
{i = 0}}^{
\displaystyle
{m - 1}}{{{q^{\left(n - i \right)}}- 1}\over{{q^{\left(m - i \right)}}- 1}}}\right)}\ {\left({\sum_{
\displaystyle
{i = 0}}^{
\displaystyle
{m - 1}}{{{{\left(- m + i \right)}\ {q^{\left(m - i - 1 \right)}}\ {q^{\left(n - i \right)}}}+{{\left({{\left(n - i \right)}\ {q^{\left(m - i \right)}}}- n + i \right)}\ {q^{\left(n - i - 1 \right)}}}+{{\left(m - i \right)}\ {q^{\left(m - i - 1 \right)}}}}\over{{{\left({q^{\left(m - i \right)}}- 1 \right)}\ {q^{\left(n - i \right)}}}-{q^{\left(m - i \right)}}+ 1}}}\right)}(3)
Type: Expression(Integer)
axiom
f:=operator 'f;
axiom
D(product(f(i,q),i=0..m),q)

\label{eq4}{\left(\prod_{
\displaystyle
{i = 0}}^{
\displaystyle
m}{f \left({i , \: q}\right)}\right)}\ {\left(\sum_{
\displaystyle
{i = 0}}^{
\displaystyle
m}{{{f_{, 2}}\left({i , \: q}\right)}\over{f \left({i , \: q}\right)}}\right)}(4)
Type: Expression(Integer)

fixed in Patch 46 or so --kratt6, Fri, 27 Oct 2006 03:12:40 -0500 reply
Status: fix proposed => closed




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