I have a relatively simple expression here that is not simplifying:
$$ \frac{2 s_0 \left(\sqrt{\gamma ^5 s_0}+\sqrt{\gamma ^9 s_0}\right)+\sqrt{\gamma ^3 s_0}+2 \sqrt{\gamma ^7 s_0}+\sqrt{\gamma ^{11} s_0}+\sqrt{\gamma ^7 s_0^5}}{\gamma \left(\gamma ^2+\gamma s_0+1\right){}^2} $$
$Assumptions = {(s0 | γ) ∈ Reals, γ > 0, s0 > 0};
(Sqrt[s0 γ^3] + 2 Sqrt[s0 γ^7] + Sqrt[s0^5 γ^7] +
Sqrt[s0 γ^11] + 2 s0 (Sqrt[s0 γ^5] + Sqrt[s0 γ^9]))/(γ (1 +
s0 γ + γ^2)^2) // Simplify
(Sqrt[s0 γ^3] + 2 Sqrt[s0 γ^7] + Sqrt[s0^5 γ^7] + Sqrt[s0 γ^11] +
2 s0 (Sqrt[s0 γ^5] + Sqrt[s0 γ^9]))/(γ (1 +
s0 γ + γ^2)^2) == Sqrt[s0 γ] // Simplify
The output is:
(Sqrt[s0 γ^3] + 2 Sqrt[s0 γ^7] + Sqrt[s0^5 γ^7] + Sqrt[s0 γ^11] +
2 s0 (Sqrt[s0 γ^5] + Sqrt[s0 γ^9]))/(γ (1 + s0 γ + γ^2)^2)
True
Why is Mathematica not simplifying to this much simpler form $\sqrt{s_0 \gamma}$? I think my assumptions should be enough. I can do the simplification by hand
Assuming[{γ>0,s0>0},(Sqrt[s0 γ^3]+2 Sqrt[s0 γ^7]+Sqrt[s0^5 γ^7]+Sqrt[s0 γ^11]+2 s0 (Sqrt[s0 γ^5]+Sqrt[s0 γ^9]))/(γ (1+s0 γ+γ^2)^2)//Refine//Simplify]
$\endgroup$Sqrt
:Sqrt[z^2]
is not automatically converted toz
. (And then they recommend the usage ofPowerExpand
for positive, realz
.) $\endgroup$