I have several equations of this form, and I'm trying to find $\alpha, \rho \in (0, 1)$ so that the equation is true for certain values of the parameters, for example $R_1 = 1.05, R_2 = 1.10, \theta = 0.90$.

\begin{equation} \theta \left(R_{1} \alpha - \alpha + 1\right) \left(R_{1}^{\frac{\rho}{\rho - 1}} \alpha - \alpha + 1\right)^{- \frac{1}{\rho} \left(\rho - 1\right)} = \left(R_{2} \alpha - \alpha + 1\right) \left(R_{2}^{\frac{\rho}{\rho - 1}} \alpha - \alpha + 1\right)^{- \frac{1}{\rho} \left(\rho - 1\right)} \end{equation}

Using a numerical solver (brute force with a fine grid over $\alpha$ and $\rho$) and a somewhat-unhelpful Sympy plotenter image description here

I don't think there are any solutions to the problem in the unit interval, but in general, are there algebraic techniques that can help me prove that? My primary algebraic attempt was taking logs of both sides to simplify the equation a bit, but that didn't lead me anywhere.

  • $\begingroup$ Try looking up bijection. $\endgroup$ – Simply Beautiful Art Nov 28 '16 at 1:09
  • $\begingroup$ @SimpleArt Any other details on what you mean? I mean I know about bijective functions, but are you saying I need to write the equation as a bijective function $f: \mathbb{R}^2 \to \mathbb{R}$ and set $f = 0$ to show that it doesn't have a solution? "Bijection" could apply in a lot of different ways in this context and I can break equations like this into myriad subfunctions. $\endgroup$ – Michael A Nov 28 '16 at 1:26
  • $\begingroup$ @Moo One of our work servers has Mathematica, so I can look into that when I'm back in the office later this week. $\endgroup$ – Michael A Nov 28 '16 at 1:33
  • $\begingroup$ @Moo Thanks. I guess I was hoping for algebraic instead of computational techniques (hence the title) because I've already experimented with numerical methods and I was looking for something more rigorous. $\endgroup$ – Michael A Nov 28 '16 at 2:09

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