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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

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/

http://maple.bug-list.org/

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

May 26, 2005, 10:52:09 AM5/26/05

to

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')

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...

May 26, 2005, 2:27:22 PM5/26/05

to

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);

>

> ?

> 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

May 26, 2005, 3:26:44 PM5/26/05

to

In article <165a3$4296150a$cf702d35$14...@PRIMUS.CA>, Igor Khavkine

<igo...@gmail.com> wrote:

<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 ?

May 26, 2005, 7:01:32 PM5/26/05

to

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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