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Re: Re: Solve weirdness again

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  • Subject: [mg35072] Re: [mg35043] Re: Solve weirdness again
  • From: Andrzej Kozlowski <andrzej at>
  • Date: Mon, 24 Jun 2002 03:20:54 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

Well actually Mathematica can solve this, provided one eliminates 
mistakes in the input and uses rational numbers rather than floating 
point ones.


 From In[3]:=
Solve::ifun: Inverse functions are being used by Solve, so some 
solutions may \
not be found.


It is a little strange why this does not work without Rationalize since, 
I think, Solve normally applies Rationalize anyway, at least in the 
multivariate case. What is even stranger that NSolve also requires 
Rationalize in this case. Looks looks to me like there is a minor bug 


On Friday, June 21, 2002, at 12:54  PM, David W. Cantrell wrote:

> "Joshua A. Solomon" <J.A.Solomon at> wrote:
>> There was a transcription error in my original message.
>> Allow me to try again.
>> Why doesn't
>> Solve[{k*.01^p==10,k*.1^p==1},{k,p}]
>> produce
>> {{k->10.,p->-1.}}
>> ?
> There appears to be another error remaining. If your equations are now
> correct, then I presume you meant to ask why {{k->0.1,p->-1.0}} is not
> produced.
> I would say that Mathematica _should_ be able to solve the system. The
> equations do not really "involve variables in an essentially 
> nonalgebraic
> way", as Mathematica uses that phrase.
> Here's one way to help Mathematica:
> In[1]:=
> Eliminate[{k*a^p==b,k*c^p==d},k]
> Out[1]=
> a^p*d == b*c^p
> In[2]:=
> Solve[%,p]
> Solve::ifun: Inverse functions are being used by Solve, so some 
> solutions
> may not be found.
> Out[2]=
> {{p -> -(Log[b/d]/(-Log[a] + Log[c]))}}
> Then to get k, just substitute the value obtained for p in either of the
> original equations.
> It's unfortunate that Mathematica can't do this on its own under just 
> the
> Solve command. It really should be able to do so, it seems to me.
> Regards,
>   David
> --
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Andrzej Kozlowski
Toyama International University

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