Re: Some bugs in Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg59205] Re: [mg59188] Some bugs in Mathematica
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Tue, 2 Aug 2005 00:42:27 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
$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 at hotmail.com" <akhmel at 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}}
> ]
>
> (*******************************************************************
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