An exact 1-D limit challenge - 5

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Vladimir Bondarenko

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May 25, 2005, 7:00:36 AM5/25/05
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Hello,

Is there a person who can calculate using a CAS (yes, please
using any CAS, avoiding fountain-pens/pencils/paper/simply
showing the answer) the exact value of the following limit
where polylog stands for the usual polylogarithm function

limit(polylog(z,z)/z, z= infinity);

?


Best wishes,

Vladimir Bondarenko

GEMM architect
Co-founder, CEO, Mathematical Director
Cyber Tester, LLC
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Vladimir Bondarenko

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May 26, 2005, 10:52:09 AM5/26/05
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Maybe Mr Igor Khavkine could have some time to try the
limit? Just a thought. If you are reading this, I was
delighted with your solutions (especially with 'cheating')

;)

http://groups-beta.google.com/group/sci.math.symbolic/msg/966aab1a20e25964?hl=en


But maybe you are too busy now...

Igor Khavkine

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May 26, 2005, 2:27:22 PM5/26/05
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On 2005-05-25, Vladimir Bondarenko <v...@cybertester.com> wrote:
> Hello,
>
> Is there a person who can calculate using a CAS (yes, please
> using any CAS, avoiding fountain-pens/pencils/paper/simply
> showing the answer) the exact value of the following limit
> where polylog stands for the usual polylogarithm function
>
> limit(polylog(z,z)/z, z= infinity);
>
> ?

I've been staying away from these since for this kind of calculation I'd
prefer not to use a CAS unless necessary. But if you don't mind
cheating, there it is, a pen and paper calculation only:

polylog(n,z) = sum_{k=1 to oo} z^k/k^n

polylog(z,z) = sum_{k=1 to oo} z^k/k^z
= sum_{k=1 to oo} z^k exp(-z ln k)

Ordinarily, the series is only good for |z| < 1, but in the above form,
I think it converges uniformly for all Re(z) > 0.

Also, note that z^(k+1) exp(-z ln(k+1)) / z^k exp(-z ln k)
= z exp(-z ln(1+1/k)). In other words, the k+1'th term is o(k'th term)
as z -> oo. This gives the asymptotic expansion

polylog(z,z) = z + o(z)

polylog(z,z)/z = 1 + o(1) -> 1 as z -> oo.

Maple says:
> f := z -> polylog(z,z)/z;
> evalf(f(10)); evalf(f(100)); evalf(f(1000));
1.012002397 - 0.001575172199 I
-17
1.000000000 + 0.2011537183 10 I
-19
1.000000000 + 0.5109280258 10 I

Igor

A N Niel

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May 26, 2005, 3:26:44 PM5/26/05
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In article <165a3$4296150a$cf702d35$14...@PRIMUS.CA>, Igor Khavkine
<igo...@gmail.com> wrote:

> On 2005-05-25, Vladimir Bondarenko <v...@cybertester.com> wrote:
> > Hello,
> >
> > Is there a person who can calculate using a CAS (yes, please
> > using any CAS, avoiding fountain-pens/pencils/paper/simply
> > showing the answer) the exact value of the following limit
> > where polylog stands for the usual polylogarithm function
> >
> > limit(polylog(z,z)/z, z= infinity);
> >
> > ?
>
> I've been staying away from these since for this kind of calculation I'd
> prefer not to use a CAS unless necessary. But if you don't mind
> cheating, there it is, a pen and paper calculation only:
>
> polylog(n,z) = sum_{k=1 to oo} z^k/k^n
>
> polylog(z,z) = sum_{k=1 to oo} z^k/k^z
> = sum_{k=1 to oo} z^k exp(-z ln k)
>
> Ordinarily, the series is only good for |z| < 1, but in the above form,
> I think it converges uniformly for all Re(z) > 0.

What? You claim it converges for z=1 ? Or z=2 ?

Igor Khavkine

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May 26, 2005, 7:01:32 PM5/26/05
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Damn, you're right! I really should be more careful with sweeping
statements like that. Anyway, I think the above series is still good
asymptotically as z -> oo. And that's good enough to get the limit.

Igor

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