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An exact asymptotics challenge - 1

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Vladimir Bondarenko

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May 31, 2005, 2:37:18 PM5/31/05
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Hello,

Is there a person who can calculate the exact value of the
following asymptotics using a computer algebra system (CAS),
where hypergeom stands for the hypergeometric function,

asympt(hypergeom([1/z, z],[1+1/z], -1), z, 2);

? (I promise you quite an elegant answer :)


Best wishes,

Vladimir Bondarenko

GEMM architect
Co-founder, CEO, Mathematical Director
Cyber Tester, LLC
13 Dekabristov Str, Simferopol
Crimea 95000, Ukraine

tel: +38-(0652)-447325
tel: +38-(0652)-230243
tel: +38-(0652)-523144
fax: +38-(0652)-510700

http://www.cybertester.com/
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http://www.CAS-testing.org/

C W

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May 31, 2005, 4:38:42 PM5/31/05
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Vladimir Bondarenko wrote:
>
> Hello,
>
> Is there a person who can calculate the exact value of the
> following asymptotics using a computer algebra system (CAS),
> where hypergeom stands for the hypergeometric function,
>
> asympt(hypergeom([1/z, z],[1+1/z], -1), z, 2);

Do these actually originate from problems or do You simply obscure mathematical
formulae ?

G. A. Edgar

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Jun 1, 2005, 10:31:50 AM6/1/05
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In article <1117564638.4...@o13g2000cwo.googlegroups.com>,
Vladimir Bondarenko <v...@cybertester.com> wrote:

> Hello,
>
> Is there a person who can calculate the exact value of the
> following asymptotics using a computer algebra system (CAS),
> where hypergeom stands for the hypergeometric function,
>
> asympt(hypergeom([1/z, z],[1+1/z], -1), z, 2);
>
> ? (I promise you quite an elegant answer :)
>
>

The limit is 1, to get two more terms would need more effort.

In any case, the "hypergeom" is just making it harder. Try instead
the integral:
(1/z)*int(t^(1/z-1)*(1+t)^(-z), t=0..1)

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

Vladimir Bondarenko

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Jun 3, 2005, 6:56:35 AM6/3/05
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C W wrote:

http://groups-beta.google.com/group/sci.math.symbolic/msg/19024b824543e750?hl=en

> Vladimir Bondarenko wrote:
>> asympt(hypergeom([1/z, z],[1+1/z], -1), z, 2);

CW> Do these actually originate from problems or do You simply
CW> obscure mathematical formulae ?


Great question.

Actually, the challenges are a tiny fraction of what we hope
could become the definitive, world's largest benchmark test
for new powerful computer algebra systems to be created within
the next 20-30 years. Maybe, say, Mathematica will evolve into
such a system, who knows. Possibly, AXIOM; there are a huge
stock of bugs and too few developers there, but there are also
some ideas on account of this headache, and we are going to
come back to AXIOM soon.

In our opinion, by several reasons, such a benchmark should
consist of *two* components, a static and a variable ones, -
and in a year or so I am planning to tell publicly much more
about this stuff.

About the origin of the challenges, they are mostly either
the tasks I can solve from a uniform viewpoint and are hard
enough, and/or the tasks I needed to find the exact answer
to thousands cases to be placed as the EXPECTED entries into
the Maple Bugs Encyclopaedia when Maple fails but no other
system produces a correct answer. Plus, just for fun, a small
layer of something I found elsewhere.

Best wishes,

Vladimir Bondarenko

Vladimir Bondarenko

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Jun 4, 2005, 5:48:16 PM6/4/05
to
"G. A. Edgar" <e...@math.ohio-state.edu.invalid>
writes on Wed, Jun 1 2005 10:31 am

http://groups-beta.google.com/group/sci.math.symbolic/msg/039f4d5101ad3a95?hl=en


GA> The limit is 1

Why sure. But how to get this using, say, Maple?


GA> to get two more terms would need more effort.

Absolutely right... but the final answer is nice.


GA> In any case, the "hypergeom" is just making it harder.

Actually, there was no intention to make the things harder.


GA> Try instead the integral: (1/z)*int(t^(1/z-1)*(1+t)^(-z)­,
t=0..1)

Yes, I have tried, no luck yet :)

Maple 9.5.2> asympt(1/z*int(t^(1/z-1)*(1+t)^(-z),t=0..1), z, 1);
Maple 9.5.2> asympt(1/z*int(t^(1/z-1)*(1+t)^(-z),t=0..1), z, 2);

Error, (in asympt) unable to compute series
Error, (in asympt) unable to compute series

I predict that, with Maple 10, the outcome is the same.


With the deepest respect to your computational flair,

Vladimir Bondarenko

Man+machine review of Maple crisis: 1993-2004
http://maple.bug-list.org/MapleCrisis-Review-01.pdf

Vladimir Bondarenko

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Jun 5, 2005, 6:44:19 PM6/5/05
to
A bit of hint...


asympt(hypergeom([1/z, z],[1+1/z], -1), z, 2) = 1 - gamma/z - ?


where gamma = 0.5772... is the Euler-Mascheroni constant

http://mathworld.wolfram.com/Euler-MascheroniConstant.html


So, what is hidden under '?' and how to get it?

;-)


Best wishes,

Vladimir Bondarenko

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Jun 14, 2005, 11:37:43 AM6/14/05
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Actually,

asympt(hypergeom([1/z, z],[1+1/z], -1), z, 2) = 1 - (gamma + ln(z))/z

For example,

Limit[Hypergeometric2F1[1/n, n, 1 + 1/n, -1] -
(1 - EulerGamma/n - Log[n]/n), n -> 10000.]

4.92037 10^(-7)

So, is there a person who can get to this answer using any
combination of commands in computer algebra systems, or if
it is too much ;) then the paper-and-pencil method ?


Best wishes,

Vladimir Bondarenko

Man+machine review of Maple crisis: 1993-2004

http://maple.bug-list.org/MapleCrisis-Review-01.pdf [16 Mb]

VM and GEMM architect


Co-founder, CEO, Mathematical Director
Cyber Tester, LLC
13 Dekabristov Str, Simferopol
Crimea 95000, Ukraine

http://www.cybertester.com/
http://maple.bug-list.org/
http://www.CAS-testing.org/

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