Proximity-concentration hypothesis: homogeneous firms and symmetric countries

Hi everyone I am running into a stone wall and don’t seem to be able to solve this, can someone show me the steps how I can insert 1 into 2 and get 3.

1. $ B=\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})} $
2. $ π = φ^{(σ−1)}*B+φ^{(σ−1)}B-f_E-2f_D $
3. $ \frac{2f_D}{f_E} = τ^{(1-σ)} -1 $

I am trying to understand how the symmetric model of the proximity-concentration is derived. Would love to show you guys my steps but to be honest I haven’t gotten much further than

$ π = φ^{(σ−1)}\frac{f_E+f_D}{φ^{(σ−1)}(1+τ^{(1-σ)})}+φ^{(σ−1)}\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}-f_E-2f_D $ and then setting $ π = 0 $ . The paper sets $ π<0 $ but for the purpose of question and my issues setting $ π = 0 $ difficult enough.

Appreciate any and all help.

$ π = φ^{(σ−1)}\frac{f_E+f_D}{φ^{(σ−1)}(1+τ^{(1-σ)})}+φ^{(σ−1)}\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}-f_E-2f_D $

$ π = 2\frac{f_E+f_D}{(1+τ^{(1-σ)})}-f_E-2f_D $

$ (1+τ^{(1-σ)})π = 2(f_E+f_D)-(1+τ^{(1-σ)})f_E-2(1+τ^{(1-σ)})f_D $

$ (1+τ^{(1-σ)})π = 2f_E+2f_D-f_E-τ^{(1-σ)}f_E-2f_D-2τ^{(1-σ)}f_D $

$ (1+τ^{(1-σ)})π = f_E-τ^{(1-σ)}f_E-2τ^{(1-σ)}f_D $

$ (1+τ^{(1-σ)})π = f_E(1-τ^{(1-σ)})-2τ^{(1-σ)}f_D $

use $ \pi = 0 $

$ 0 = f_E(1-τ^{(1-σ)})-2τ^{(1-σ)}f_D $

$ 2τ^{(1-σ)}f_D = f_E(1-τ^{(1-σ)}) $

$ \frac{2f_D}{f_E} = \frac{(1-τ^{(1-σ)}) }{τ^{(1-σ)}} = \frac{1}{τ^{(1-σ)}} - 1 = τ^{(σ-1)} -1 $

in last identity I use $ \frac{1}{τ^{(1-σ)}} = τ^{(σ-1)} $ but you write $ τ^{(1-σ)} $ in expression (3) so this is not quite what you are looking for but maybe you have sign error in this exponent in (3)?

引用自：https://economics.stackexchange.com/questions/43733