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Re: can't Solve[1 - x == x^r, x]?

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I don't think that this can be solved for x[r]; however, you can also use ContourPlot to plot it.

ContourPlot[1 - x == x^r, {r, 0, 1}, {x, 0, 1/2},
 AspectRatio -> 1/GoldenRatio,
 FrameLabel -> {r, x}]


Bob Hanlon

---- Scott Centoni <scentoni@hotmail.com> wrote: 

=============
I want to solve the equation

1 - x == x^r (1)

for x[r] and for r[x], particularly in the region 0<=r<=1 and 0<=x<=1/2 where both are real. Solving for r works well:

> In[1]:= Solve[1 - x == x^r, r]
>
> During evaluation of In[1]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
>
> Out[1]= {{r -> Log[1 - x]/Log[x]}}

However solving for x does not:

> In[2]:= Solve[1 - x == x^r, x]
>
> During evaluation of In[2]:= Solve::tdep: The equations appear to involve the variables to be solved for in an essentially non-algebraic way. >>
>
> Out[2]= Solve[1 - x == x^r, x]

I've tried Reduce and InverseFunction, but they just left the expression unevaluated. For plotting purposes, I used

> Nx[r_] := x /. FindRoot[1 - x == x^r, {x, .13}]

> Plot[{.5 r^.5, .5 r, Nx[r]}, {r, 0, 1}]

The plot is fairly close to .5 r^.5 over this interval, if that helps. Is there really no way for Mathematica to express this function other than as a numerical expression like this? Not even a big, messy formula involving PolyLog and hypergeometric functions?

Thanks,
Scott Centoni


--

Bob Hanlon

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Re: can't Solve[1 - x == x^r, x]? Bob Hanlon <hanlonr@cox.net> - 2011-07-01 00:43 +0000

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