Re: Overly complicated reductions?
- To: mathgroup at smc.vnet.net
- Subject: [mg77674] Re: Overly complicated reductions?
- From: dh <dh at metrohm.ch>
- Date: Thu, 14 Jun 2007 06:20:01 -0400 (EDT)
- References: <f4om08$7bj$1@smc.vnet.net>
Hi David,
Reduce gives the correct answer. If you do not belive it, simply replace
x in Exp[-2x] by the solution and you will get y.
Obviously for y=0 there is no solution (or infinity if you like)
2Pi n comes in because Exp[2PI n]==1
If you only want a real solution, replace y==.. by {y==..,Element[x,Reals]}
hope this helps, Daniel
David Rees wrote:
> Consider f(x)=e^(-2x)
>
> I wanted to retreive the inverse function f^-1(x), Mathematica to the
> rescue:
> \!\(Reduce[y == E\^\(\(-2\) x\), x]\)
>
> \!\(C[1] \[Element] Integers && y != 0 &&
> x == 1\/2\ \((2\ \[ImaginaryI]\ \[Pi]\ C[1] + Log[1\/y])\)\)
>
> This can't be right, I can rearrange it to just Ln(x)/-2 on paper. What did
> I do wrong?
>
> Thanks
>
>
>