f



Re: Re: Some bugs in Mathematica

Looking at the sum in more detail:

s1[n_] :=
    Sum[x^k*(Gamma[n-k-1/2]*Gamma[k+1/2])/
          (Gamma[n-k-1]*Gamma[k+1]),{k,0,n-1}];

Calculating each term separately

TableForm[t=Table[(Gamma[n-k-1/2]*Gamma[k+1/2])/(Gamma[n-k-1]
*Gamma[k+1]),
      {n,5},{k,0,n-1}],
  TableHeadings->{Automatic,Table[i,{i,0,4}]}]

Summing each row of the table

Tr/@t

{0, Pi/2, Pi, (3*Pi)/2, 2*Pi}

For example, for n=2, the term for k=0 is

((Gamma[n-k-1/2]*Gamma[k+1/2])/(Gamma[n-k-1]*Gamma[k+1])/.
{n->2,k->0}) ==
  Gamma[3/2]*Gamma[1/2]/(Gamma[1]*Gamma[1])==
  (1/2)*Gamma[1/2]*Gamma[1/2]==Pi/2

True

and the term for k=1 is

((Gamma[n-k-1/2]*Gamma[k+1/2])/(Gamma[n-k-1]*Gamma[k+1])/.
{n->2,k->1})==
  Gamma[1/2]*Gamma[3/2]/(Gamma[0]*Gamma[2])==0

True

Alternatively, generalizing the sum

s1[n_,x_]=
  Sum[x^k*(Gamma[n-k-1/2]*Gamma[k+1/2])/
        (Gamma[n-k-1]*Gamma[k+1]),{k,0,n-1}]

(Sqrt[Pi]*Gamma[n - 1/2]*Hypergeometric2F1[1/2, 2 - n, 3/2 - n, x])/Gamma
[n - 1]

Table[s1[n,x],{n,5}]

{0, Pi/2, (1/4)*Pi*(x + 3), (3/16)*Pi*(x^2 + 2*x + 5), 
  (1/32)*Pi*(5*x^3 + 9*x^2 + 15*x + 35)}

%/.x->1

{0, Pi/2, Pi, (3*Pi)/2, 2*Pi}

And@@Table[(s1[n]==s1[n,x]==(n-1)*Pi/2)/.x->1,{n,25}]

True


Bob Hanlon

> 
> From: "Alex Khmelnitsky" <akhmel@hotmail.com>
> Date: 2005/08/01 Mon AM 10:07:41 EDT
> To: "Bob Hanlon" <hanlonr@cox.net>
> Subject: Re:  Some bugs in Mathematica
> 
> 1) You are wrong. My sum is equal to 1 for any n and the respectable 
program
> like Mathematica claims to be should know it.
> 
> 2) The integral is elementary, a doesn't have to be equal to b to give a
> logarithm.
> 
> 3) I had enough patience, so you better trust me.
> 
> Alex
> 
> ----- Original Message ----- 
> From: "Bob Hanlon" <hanlonr@cox.net>
> To: <akhmel@hotmail.com>; <mathgroup@smc.vnet.net>
> Cc: <hanlonr@cox.net>
> Sent: Monday, August 01, 2005 7:08 AM
> Subject: Re:  Some bugs in Mathematica
> 
> 
> > $Version
> >
> > 5.2 for Mac OS X (June 20, 2005)
> >
> > Question 1:
> >
> > s1[n_]:=Sum[(Gamma[n-k-1/2]*Gamma[k+1/2])/
> >        (Gamma[n-k-1]*Gamma[k+1]),
> >       {k,0,n-1}];
> >
> > For arbitrary n, this sum is unevaluated
> >
> > s1[n]
> >
> > Sum[(Gamma[k + 1/2]*Gamma[-k + n - 1/2])/
> >    (Gamma[k + 1]*Gamma[-k + n - 1]),
> >   {k, 0, n - 1}]
> >
> > For specific nonnegative integer values of n
> >
> > Table[s1[n],{n,5}]
> >
> > {0, Pi/2, Pi, (3*Pi)/2, 2*Pi}
> >
> > For nonnegative integer values of n, this sum appears to be (n-1)*Pi/2
> >
> > And@@Table[s1[n]==(n-1)*Pi/2,{n,25}]
> >
> > True
> >
> > Question 2:
> >
> > int1=Integrate[1/(r*Sqrt[r^2-a^2]*
> >          Sqrt[r^2-b^2]),r]
> >
> > -((Sqrt[1 - a^2/r^2]*Sqrt[1 - b^2/r^2]*
> >    AppellF1[1, 1/2, 1/2, 2, a^2/r^2,
> >     b^2/r^2])/(2*Sqrt[r^2 - a^2]*
> >    Sqrt[r^2 - b^2]))
> >
> > int2=Simplify[int1,Element[r, Reals]]
> >
> > -(AppellF1[1, 1/2, 1/2, 2, a^2/r^2,
> >    b^2/r^2]/(2*r^2))
> >
> > These simplify to a log if b equals a
> >
> > Simplify[int1/.b->a]
> >
> > Log[1 - a^2/r^2]/(2*a^2)
> >
> > int2 /. b->a
> >
> > Log[1 - a^2/r^2]/(2*a^2)
> >
> > Question 3:
> >
> > Your last integral did not return in the amount of time I was willing to
> > wait.
> >
> >
> > Bob Hanlon
> >
> > On 8/1/05 1:05 AM, "akhmel@hotmail.com" <akhmel@hotmail.com> 
wrote:
> >
> > > Dear Mr. Lichtblau,
> > >
> > > I attach hereby Mathematica file with 3 examples, which you might be
> > > interested to look at.
> > >
> > > 1) Example of summation which gives an obviously wrong result, 0
> > > instead of
> > > 1.
> > > 2) Example of indefinate integration, which gives appel function
> > > instead of
> > > elementary logarithm.
> > > 3) Definite integration which doesn't give any result at all, though
> > > again,
> > > it is nothing but elementary function.
> > >
> > > Your comments will be appreciated.
> > >
> > > Thanks,
> > >
> > > Alex
> > >
> > >
> > > Notebook[{
> > >
> > > Cell[CellGroupData[{
> > > Cell[BoxData[
> > >     \(Sum[\(Gamma[n - k - 1/2] Gamma[k + 1/2]\)\/\(Gamma[n - k - 
1]
> > > Gamma[k + \
> > > 1]\), {k, 0, n - 1}]\)], "Input"],
> > >
> > > Cell[BoxData[
> > >     \(0\)], "Output"]
> > > }, Open  ]],
> > >
> > > Cell[CellGroupData[{
> > >
> > > Cell[BoxData[
> > >     \(Integrate[1\/\(r \(\@\( r\^2 - a\^2\)\) \@\(r\^2 - b\^2\)\),
> > >       r]\)], "Input"],
> > >
> > > Cell[BoxData[
> > >     \(\(-\(\(\@\(1 - a\^2\/r\^2\)\ \@\(1 - b\^2\/r\^2\)\ AppellF1[1,
> > > 1\/2,
> > >               1\/2, 2, a\^2\/r\^2,
> > >               b\^2\/r\^2]\)\/\(2\ \@\(\(-a\^2\) + r\^2\)\ \@\(\(-b\^2\)
> > > + \
> > > r\^2\)\)\)\)\)], "Output"]
> > > }, Open  ]],
> > >
> > > Cell[BoxData[
> > >     \(\(\(-\(\(\@\(1 - a\^2\/r\^2\)\ \@\(1 - b\^2\/r\^2\)\ AppellF1[1,
> > > 1\/2,
> > >               1\/2, 2, a\^2\/r\^2,
> > >               b\^2\/r\^2]\)\/\(2\ \@\(\(-a\^2\) + r\^2\)\ \@\(\(-b\^2\)
> > > + \
> > > r\^2\)\)\)\)\(\[IndentingNewLine]\)
> > >     \)\)], "Input"],
> > >
> > > Cell[CellGroupData[{
> > >
> > > Cell[BoxData[
> > >     \(Integrate[
> > >       1\/\(\(r\^3\) \(\@\(r\^2 - a\^2\)\) \@\(r\^2 - b\^2\)\), {r, x,
> > >         a}]\)], "Input"],
> > >
> > > Cell[BoxData[
> > >     \(Integrate::"gener" \(\(:\)\(\ \)\)
> > >       "Unable to check convergence."\)], "Message"],
> > >
> > > Cell[BoxData[
> > >     \(\[Integral]\_x\%a\(
> > >             1\/\(r\^3\ \@\(\(-a\^2\) + r\^2\)\ \@\(\(-b\^2\) +
> > > r\^2\)\)\) \
> > > \[DifferentialD]r\)], "Output"]
> > > }, Open  ]]
> > > },
> > > FrontEndVersion->"4.2 for Microsoft Windows",
> > > ScreenRectangle->{{0, 720}, {0, 407}},
> > > WindowSize->{496, 249},
> > > WindowMargins->{{0, Automatic}, {Automatic, 0}}
> > > ]
> > >
> > > (*******************************************************************
> > > Cached data follows.  If you edit this Notebook file directly, not
> > > using Mathematica, you must remove the line containing CacheID at
> > > the top of  the file.  The cache data will then be recreated when
> > > you save this file from within Mathematica.
> > > *******************************************************************)
> > >
> > > (*CellTagsOutline
> > > CellTagsIndex->{}
> > > *)
> > >
> > > (*CellTagsIndex
> > > CellTagsIndex->{}
> > > *)
> > >
> > > (*NotebookFileOutline
> > > Notebook[{
> > >
> > > Cell[CellGroupData[{
> > > Cell[1776, 53, 127, 2, 44, "Input"],
> > > Cell[1906, 57, 35, 1, 29, "Output"]
> > > }, Open  ]],
> > >
> > > Cell[CellGroupData[{
> > > Cell[1978, 63, 103, 2, 48, "Input"],
> > > Cell[2084, 67, 228, 4, 71, "Output"]
> > > }, Open  ]],
> > > Cell[2327, 74, 259, 5, 98, "Input"],
> > >
> > > Cell[CellGroupData[{
> > > Cell[2611, 83, 126, 3, 48, "Input"],
> > > Cell[2740, 88, 107, 2, 24, "Message"],
> > > Cell[2850, 92, 148, 3, 46, "Output"]
> > > }, Open  ]]
> > > }
> > > ]
> > > *)
> > >
> >
> >
> >
> 

0
hanlonr (2279)
8/2/2005 7:15:04 AM
comp.soft-sys.math.mathematica 28821 articles. 0 followers. Follow

4 Replies
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[PageSpeed] 14

You changed my example by adding factor x^k. Mathematica does work in
this case. It does not work when this factor is absent. So, your
example is useless. There is a bug and they need to correct it.

Alex


Bob Hanlon wrote:
> Looking at the sum in more detail:
>
> s1[n_] :=
>     Sum[x^k*(Gamma[n-k-1/2]*Gamma[k+1/2])/
>           (Gamma[n-k-1]*Gamma[k+1]),{k,0,n-1}];
>
> Calculating each term separately
>
> TableForm[t=Table[(Gamma[n-k-1/2]*Gamma[k+1/2])/(Gamma[n-k-1]
> *Gamma[k+1]),
>       {n,5},{k,0,n-1}],
>   TableHeadings->{Automatic,Table[i,{i,0,4}]}]
>
> Summing each row of the table
>
> Tr/@t
>
> {0, Pi/2, Pi, (3*Pi)/2, 2*Pi}
>
> For example, for n=2, the term for k=0 is
>
> ((Gamma[n-k-1/2]*Gamma[k+1/2])/(Gamma[n-k-1]*Gamma[k+1])/.
> {n->2,k->0}) ==
>   Gamma[3/2]*Gamma[1/2]/(Gamma[1]*Gamma[1])==
>   (1/2)*Gamma[1/2]*Gamma[1/2]==Pi/2
>
> True
>
> and the term for k=1 is
>
> ((Gamma[n-k-1/2]*Gamma[k+1/2])/(Gamma[n-k-1]*Gamma[k+1])/.
> {n->2,k->1})==
>   Gamma[1/2]*Gamma[3/2]/(Gamma[0]*Gamma[2])==0
>
> True
>
> Alternatively, generalizing the sum
>
> s1[n_,x_]=
>   Sum[x^k*(Gamma[n-k-1/2]*Gamma[k+1/2])/
>         (Gamma[n-k-1]*Gamma[k+1]),{k,0,n-1}]
>
> (Sqrt[Pi]*Gamma[n - 1/2]*Hypergeometric2F1[1/2, 2 - n, 3/2 - n, x])/Gamma
> [n - 1]
>
> Table[s1[n,x],{n,5}]
>
> {0, Pi/2, (1/4)*Pi*(x + 3), (3/16)*Pi*(x^2 + 2*x + 5),
>   (1/32)*Pi*(5*x^3 + 9*x^2 + 15*x + 35)}
>
> %/.x->1
>
> {0, Pi/2, Pi, (3*Pi)/2, 2*Pi}
>
> And@@Table[(s1[n]==s1[n,x]==(n-1)*Pi/2)/.x->1,{n,25}]
>
> True
>
>
> Bob Hanlon
>
> >
> > From: "Alex Khmelnitsky" <akhmel@hotmail.com>
> > Date: 2005/08/01 Mon AM 10:07:41 EDT
> > To: "Bob Hanlon" <hanlonr@cox.net>
> > Subject: Re:  Some bugs in Mathematica
> >
> > 1) You are wrong. My sum is equal to 1 for any n and the respectable
> program
> > like Mathematica claims to be should know it.
> >
> > 2) The integral is elementary, a doesn't have to be equal to b to give a
> > logarithm.
> >
> > 3) I had enough patience, so you better trust me.
> >
> > Alex
> >
> > ----- Original Message -----
> > From: "Bob Hanlon" <hanlonr@cox.net>
> > To: <akhmel@hotmail.com>; <mathgroup@smc.vnet.net>
> > Cc: <hanlonr@cox.net>
> > Sent: Monday, August 01, 2005 7:08 AM
> > Subject: Re:  Some bugs in Mathematica
> >
> >
> > > $Version
> > >
> > > 5.2 for Mac OS X (June 20, 2005)
> > >
> > > Question 1:
> > >
> > > s1[n_]:=Sum[(Gamma[n-k-1/2]*Gamma[k+1/2])/
> > >        (Gamma[n-k-1]*Gamma[k+1]),
> > >       {k,0,n-1}];
> > >
> > > For arbitrary n, this sum is unevaluated
> > >
> > > s1[n]
> > >
> > > Sum[(Gamma[k + 1/2]*Gamma[-k + n - 1/2])/
> > >    (Gamma[k + 1]*Gamma[-k + n - 1]),
> > >   {k, 0, n - 1}]
> > >
> > > For specific nonnegative integer values of n
> > >
> > > Table[s1[n],{n,5}]
> > >
> > > {0, Pi/2, Pi, (3*Pi)/2, 2*Pi}
> > >
> > > For nonnegative integer values of n, this sum appears to be (n-1)*Pi/2
> > >
> > > And@@Table[s1[n]==(n-1)*Pi/2,{n,25}]
> > >
> > > True
> > >
> > > Question 2:
> > >
> > > int1=Integrate[1/(r*Sqrt[r^2-a^2]*
> > >          Sqrt[r^2-b^2]),r]
> > >
> > > -((Sqrt[1 - a^2/r^2]*Sqrt[1 - b^2/r^2]*
> > >    AppellF1[1, 1/2, 1/2, 2, a^2/r^2,
> > >     b^2/r^2])/(2*Sqrt[r^2 - a^2]*
> > >    Sqrt[r^2 - b^2]))
> > >
> > > int2=Simplify[int1,Element[r, Reals]]
> > >
> > > -(AppellF1[1, 1/2, 1/2, 2, a^2/r^2,
> > >    b^2/r^2]/(2*r^2))
> > >
> > > These simplify to a log if b equals a
> > >
> > > Simplify[int1/.b->a]
> > >
> > > Log[1 - a^2/r^2]/(2*a^2)
> > >
> > > int2 /. b->a
> > >
> > > Log[1 - a^2/r^2]/(2*a^2)
> > >
> > > Question 3:
> > >
> > > Your last integral did not return in the amount of time I was willing to
> > > wait.
> > >
> > >
> > > Bob Hanlon
> > >
> > > On 8/1/05 1:05 AM, "akhmel@hotmail.com" <akhmel@hotmail.com>
> wrote:
> > >
> > > > Dear Mr. Lichtblau,
> > > >
> > > > I attach hereby Mathematica file with 3 examples, which you might be
> > > > interested to look at.
> > > >
> > > > 1) Example of summation which gives an obviously wrong result, 0
> > > > instead of
> > > > 1.
> > > > 2) Example of indefinate integration, which gives appel function
> > > > instead of
> > > > elementary logarithm.
> > > > 3) Definite integration which doesn't give any result at all, though
> > > > again,
> > > > it is nothing but elementary function.
> > > >
> > > > Your comments will be appreciated.
> > > >
> > > > Thanks,
> > > >
> > > > Alex
> > > >
> > > >
> > > > Notebook[{
> > > >
> > > > Cell[CellGroupData[{
> > > > Cell[BoxData[
> > > >     \(Sum[\(Gamma[n - k - 1/2] Gamma[k + 1/2]\)\/\(Gamma[n - k -
> 1]
> > > > Gamma[k + \
> > > > 1]\), {k, 0, n - 1}]\)], "Input"],
> > > >
> > > > Cell[BoxData[
> > > >     \(0\)], "Output"]
> > > > }, Open  ]],
> > > >
> > > > Cell[CellGroupData[{
> > > >
> > > > Cell[BoxData[
> > > >     \(Integrate[1\/\(r \(\@\( r\^2 - a\^2\)\) \@\(r\^2 - b\^2\)\),
> > > >       r]\)], "Input"],
> > > >
> > > > Cell[BoxData[
> > > >     \(\(-\(\(\@\(1 - a\^2\/r\^2\)\ \@\(1 - b\^2\/r\^2\)\ AppellF1[1,
> > > > 1\/2,
> > > >               1\/2, 2, a\^2\/r\^2,
> > > >               b\^2\/r\^2]\)\/\(2\ \@\(\(-a\^2\) + r\^2\)\ \@\(\(-b\^2\)
> > > > + \
> > > > r\^2\)\)\)\)\)], "Output"]
> > > > }, Open  ]],
> > > >
> > > > Cell[BoxData[
> > > >     \(\(\(-\(\(\@\(1 - a\^2\/r\^2\)\ \@\(1 - b\^2\/r\^2\)\ AppellF1[1,
> > > > 1\/2,
> > > >               1\/2, 2, a\^2\/r\^2,
> > > >               b\^2\/r\^2]\)\/\(2\ \@\(\(-a\^2\) + r\^2\)\ \@\(\(-b\^2\)
> > > > + \
> > > > r\^2\)\)\)\)\(\[IndentingNewLine]\)
> > > >     \)\)], "Input"],
> > > >
> > > > Cell[CellGroupData[{
> > > >
> > > > Cell[BoxData[
> > > >     \(Integrate[
> > > >       1\/\(\(r\^3\) \(\@\(r\^2 - a\^2\)\) \@\(r\^2 - b\^2\)\), {r, x,
> > > >         a}]\)], "Input"],
> > > >
> > > > Cell[BoxData[
> > > >     \(Integrate::"gener" \(\(:\)\(\ \)\)
> > > >       "Unable to check convergence."\)], "Message"],
> > > >
> > > > Cell[BoxData[
> > > >     \(\[Integral]\_x\%a\(
> > > >             1\/\(r\^3\ \@\(\(-a\^2\) + r\^2\)\ \@\(\(-b\^2\) +
> > > > r\^2\)\)\) \
> > > > \[DifferentialD]r\)], "Output"]
> > > > }, Open  ]]
> > > > },
> > > > FrontEndVersion->"4.2 for Microsoft Windows",
> > > > ScreenRectangle->{{0, 720}, {0, 407}},
> > > > WindowSize->{496, 249},
> > > > WindowMargins->{{0, Automatic}, {Automatic, 0}}
> > > > ]
> > > >
> > > > (*******************************************************************
> > > > Cached data follows.  If you edit this Notebook file directly, not
> > > > using Mathematica, you must remove the line containing CacheID at
> > > > the top of  the file.  The cache data will then be recreated when
> > > > you save this file from within Mathematica.
> > > > *******************************************************************)
> > > >
> > > > (*CellTagsOutline
> > > > CellTagsIndex->{}
> > > > *)
> > > >
> > > > (*CellTagsIndex
> > > > CellTagsIndex->{}
> > > > *)
> > > >
> > > > (*NotebookFileOutline
> > > > Notebook[{
> > > >
> > > > Cell[CellGroupData[{
> > > > Cell[1776, 53, 127, 2, 44, "Input"],
> > > > Cell[1906, 57, 35, 1, 29, "Output"]
> > > > }, Open  ]],
> > > >
> > > > Cell[CellGroupData[{
> > > > Cell[1978, 63, 103, 2, 48, "Input"],
> > > > Cell[2084, 67, 228, 4, 71, "Output"]
> > > > }, Open  ]],
> > > > Cell[2327, 74, 259, 5, 98, "Input"],
> > > >
> > > > Cell[CellGroupData[{
> > > > Cell[2611, 83, 126, 3, 48, "Input"],
> > > > Cell[2740, 88, 107, 2, 24, "Message"],
> > > > Cell[2850, 92, 148, 3, 46, "Output"]
> > > > }, Open  ]]
> > > > }
> > > > ]
> > > > *)
> > > >
> > >
> > >
> > >
> >

0
akhmel (29)
8/3/2005 5:48:42 AM
On 3 Aug 2005, at 07:20, akhmel@hotmail.com wrote:

>>> From: "Alex Khmelnitsky" <akhmel@hotmail.com>
>>>
>>> Date: 2005/08/01 Mon AM 10:07:41 EDT
>>> To: "Bob Hanlon" <hanlonr@cox.net>
>>> Subject:  Re:  Some bugs in Mathematica
>>>
>>> 1) You are wrong. My sum is equal to 1 for any n and the respectable
>>>
>> program
>>
>>> like Mathematica claims to be should know it.

1. This is obviously false. Take n=1. Than your sum is

Sum[(Gamma[1/2 - k ]*Gamma[k + 1/2])/
    (Gamma[ - k ]*Gamma[k + 1]), {k, 0, 0}]

which is simply Gamma[1/2]*Gamma[1/2]/(Gamma[0]*Gamma[1])

The issue is what is Gamma[0]? The Gamma function has a simple pole  
at 0, so Mathematica correctly gives


Gamma[0]


ComplexInfinity

That forces the answer to be 0, thus contradicting your claim that  
the sum is always 1.

2. This group is not a place for reporting Mathematica bugs. If you  
wan to report a bug you shoudl report it directly to WRI's technical  
support.
This is a Mathematica discussion group. No employees of Wolfram are  
obliged to read any posting to this group or reply to them. Those who  
choose to do so, do so voluntarily because they want to help  
Mathematica users. Referring to them by name in your postings  is not  
going to make them reply  and I would guess that if you make yourself  
sound as if you were demanding rather than asking for an answer they  
are more likely to ignore you. Which they are fully entitled to do.

Andrzej Kozlowski





0
akoz (2415)
8/4/2005 6:25:03 AM
Just in case you decided to object that n was supposed to be >1, a  
trivial computation by hand shows that for n=2 the answer is

(Gamma[3/2]*Gamma[1/2])/(Gamma[1]*Gamma[1]) + (Gamma[1/2]*Gamma[3/2])/ 
(Gamma[0]*Gamma[2])

The first term is Pi/2 and the second is 0 so the answer comes to Pi/ 
2 and not 1.

What made you claim it is 1?

Andrzej Kozlowski


On 3 Aug 2005, at 23:09, Andrzej Kozlowski wrote:

>
> On 3 Aug 2005, at 07:20, akhmel@hotmail.com wrote:
>
>
>>>> From: "Alex Khmelnitsky" <akhmel@hotmail.com>
>>>>
>>>>
>>>> Date: 2005/08/01 Mon AM 10:07:41 EDT
>>>> To: "Bob Hanlon" <hanlonr@cox.net>
>>>> Subject:  Re:  Some bugs in Mathematica
>>>>
>>>> 1) You are wrong. My sum is equal to 1 for any n and the  
>>>> respectable
>>>>
>>>>
>>> program
>>>
>>>
>>>> like Mathematica claims to be should know it.
>>>>
>
> 1. This is obviously false. Take n=1. Than your sum is
>
> Sum[(Gamma[1/2 - k ]*Gamma[k + 1/2])/
>    (Gamma[ - k ]*Gamma[k + 1]), {k, 0, 0}]
>
> which is simply Gamma[1/2]*Gamma[1/2]/(Gamma[0]*Gamma[1])
>
> The issue is what is Gamma[0]? The Gamma function has a simple pole  
> at 0, so Mathematica correctly gives
>
>
> Gamma[0]
>
>
> ComplexInfinity
>
> That forces the answer to be 0, thus contradicting your claim that  
> the sum is always 1.
>
> 2. This group is not a place for reporting Mathematica bugs. If you  
> wan to report a bug you shoudl report it directly to WRI's  
> technical support.
> This is a Mathematica discussion group. No employees of Wolfram are  
> obliged to read any posting to this group or reply to them. Those  
> who choose to do so, do so voluntarily because they want to help  
> Mathematica users. Referring to them by name in your postings  is  
> not going to make them reply  and I would guess that if you make  
> yourself sound as if you were demanding rather than asking for an  
> answer they are more likely to ignore you. Which they are fully  
> entitled to do.
>
> Andrzej Kozlowski
>
>
>
>
>
>

0
akoz (2415)
8/4/2005 6:25:48 AM
Andrzej Kozlowski wrote:
> 1. This is obviously false. Take n=1. Than your sum is
>
> Sum[(Gamma[1/2 - k ]*Gamma[k + 1/2])/
>     (Gamma[ - k ]*Gamma[k + 1]), {k, 0, 0}]
>
> which is simply Gamma[1/2]*Gamma[1/2]/(Gamma[0]*Gamma[1])
>
> The issue is what is Gamma[0]? The Gamma function has a simple pole
> at 0, so Mathematica correctly gives
>
>
> Gamma[0]
>
>
> ComplexInfinity
>
> That forces the answer to be 0, thus contradicting your claim that
> the sum is always 1.
>
> 2. This group is not a place for reporting Mathematica bugs. If you
> wan to report a bug you shoudl report it directly to WRI's technical
> support.
> This is a Mathematica discussion group. No employees of Wolfram are
> obliged to read any posting to this group or reply to them. Those who
> choose to do so, do so voluntarily because they want to help
> Mathematica users. Referring to them by name in your postings  is not
> going to make them reply  and I would guess that if you make yourself
> sound as if you were demanding rather than asking for an answer they
> are more likely to ignore you. Which they are fully entitled to do.
>
> Andrzej Kozlowski

You are right in one aspect; I made two misprints when typing formula.
Instead of Gamma[n-k-1], it should be \Pi Gamma[n-k]. I apologize for
the
confusion. Nevertheless, my claim still stands: the corrected sum is
equal
to 1, while Mathematica gives 0.

I do not accept your remark that your newsgroup is not the right place
to
expose Mathematica bugs. It looks like you're being paid by Mathematica
and
are trying to shut up any criticism. It just doesn't look good.
Mathematica
definitely needs improvement and their staff should be grateful to
people
like myself, who points out their mistakes.

Thanks,

Alex

0
akhmel (29)
8/5/2005 6:06:59 AM
Reply:

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