• To: mathgroup at smc.vnet.net
• From: "Roger L. Bagula" <rlbtftn at netscape.net>
• Date: Fri, 11 Jun 2004 23:59:14 -0400 (EDT)
• References: <ca3gvi\$rto\$1@smc.vnet.net> <200406100643.CAA29482@smc.vnet.net> <40C89DA6.1010403@wolfram.com> <cabqd9\$orm\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```I tried your approach. Works about the same as mine.
what works is setting the solution  equal to y and solving for t
The soltion looks like a square root function or the other way a
parabolic function.
Anyway a function like:
s[t_]=y=a+t^s
for s in the range of 1/2 or 1/3 about.

Roger L. Bagula wrote:
> Dear Daniel Lichtblau,
> As a signal goes it doesn't look good as an answer, but thanks for your
>
>
>>The InverseFunction wrapper means, not surprisingly, that it is an
>>inverse function. Not a reciprocal (which would be written in
>
>
> 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.
>
>
> 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
>>
>>
>>
>>>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
>>
>>
>
>

```

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