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MathGroup Archive 2004

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Re: Re: LogIntegral^(-1)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48724] Re: [mg48682] Re: LogIntegral^(-1)
  • From: "Roger L. Bagula" <rlbtftn at netscape.net>
  • Date: Fri, 11 Jun 2004 03:53:14 -0400 (EDT)
  • References: <ca3gvi$rto$1@smc.vnet.net> <200406100643.CAA29482@smc.vnet.net> <40C89DA6.1010403@wolfram.com>
  • Sender: owner-wri-mathgroup at wolfram.com

Dear Daniel Lichtblau,
As a signal goes it doesn't look good as an answer, but thanks for your 
reply.

> The InverseFunction wrapper means, not surprisingly, that it is an 
> inverse function. Not a reciprocal (which would be written in 
> OutputForm as LogIntegral[...]^(-1) rather than LogIntegral^(-1)[...]).

This is not a real clear distinction
 and I've never seen it in any of
 my Mathematica books or software, so thans for giving me that!
I frankly would rather it was ^(-1) instead of an inverse function.

It's not the kind of result I had hoped for.

But I'm glad I did get a reply.

Daniel Lichtblau wrote:

> Roger L. Bagula wrote:
>
>> I really need to know if this is:
>> Li(x)^(-1)
>> or
>> ArcLi(x)
>> since it makes a great difference and Mathematica notation on this 
>> seems unclear.
>
>
> It's very clear.
>
> n[t] = s[t]/(Exp[D[s[t],t]/w]-1);
> n1[t_] = FullSimplify[D[n[t],t]];
>
> In[10]:= InputForm[soln = DSolve[n1[t]==0, s[t], t]]
>
> Solve::ifun: Inverse functions are being used by Solve, so some 
> solutions may
>      not be found; use Reduce for complete solution information.
>
> InverseFunction::ifun:
>    Inverse functions are being used. Values may be lost for multivalued
>     inverses.
>
> Out[10]//InputForm=
> {{s[t] -> (-1 + InverseFunction[LogIntegral, 1, 1][
>   E^(C[1]/w)*w*(t + C[2])])/E^(C[1]/w)}}
>
> The InverseFunction wrapper means, not surprisingly, that it is an 
> inverse function. Not a reciprocal (which would be written in 
> OutputForm as LogIntegral[...]^(-1) rather than LogIntegral^(-1)[...]).
>
>
>> One approximation ( Euler's I think) of the distribution of primes is
>> Pi(n)=Li(n)--> n/log(n): asymptotic
>
>
> This is a bit misleading. Pi(n) is approximated by Li(n) but they are 
> not equal.
>
>
>> [...]
>
>
>
> Daniel Lichtblau
> Wolfram Research
>
>

-- 
Respectfully, Roger L. Bagula
tftn at earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
URL :  http://home.earthlink.net/~tftn
URL :  http://victorian.fortunecity.com/carmelita/435/ 



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