how do we rewrite the sum of two function to multiplication of them,and write the quadratic equation formula without seeing the quadratic equation

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I don't understand why can we do this in this paper

$\gamma ^{x_1}_1(\alpha,\rho)=\frac{(1-\rho)\alpha P_s g_1}{(1-\rho)N_0+\mu N_0}$

$\gamma ^{x_2}_1(\alpha,\rho)=\frac{(1-\rho)(1-\alpha)P_s g_1}{(1-\rho)\alpha P_sg_1+(1-\rho)N_0+\mu N_0}$

$\gamma ^{MRC}_2(\alpha,\rho)=\frac{(1-\alpha)P_sg_2}{\alpha P_s g_2+N_0+\mu N_0}+\frac{\rho \eta P_s g_1 g_3}{N_0+\mu N_0}$

$C_1(\alpha,\rho)=\frac{1}{2}log_2(1+\gamma ^{x_1}_1(\alpha,\rho))$ $C_2(\alpha,\rho)=\frac{1}{2}log_2(1+min\{\gamma ^{x_2}_1(\alpha,\rho),\gamma ^{MRC}_2(\alpha,\rho)\})$

and the author said if the objective function is

$\max\limits _{\alpha,\rho} w_1C_1(\alpha,\rho)+w_2C_2(\alpha,\rho)$

s.t. $0\le \alpha \le 1,0 \le \rho \le 1$

We can rewrite it to

$\max\limits _{\alpha,\rho} (1+\gamma ^{x_1}_1(\alpha,\rho))(1+\gamma ^{MRC}_2(\alpha,\rho))^{w_2/w_1}$

s.t. $\gamma ^{x_2}_1(\alpha,\rho) \ge \gamma ^{MRC}_2(\alpha,\rho),0\le \alpha \le 1,0 \le \rho \le 1 $

As a further step.we cam rewrite the optimal problem again to

$\max\limits _{\alpha,\rho} (1+\gamma ^{x_1}_1(\alpha,\rho))(1+\gamma ^{MRC}_2(\alpha,\rho))^{w_2/w_1}$

s.t. $0\le \alpha \le 1,0 \le \rho \le \tilde \rho(\alpha) $

and $\tilde \rho(\alpha)=\frac{b-\sqrt{b^2-4ac}}{2a},\bar \gamma =P_s /N_0,\alpha=\frac{\eta \bar \gamma g_1 g_3(\alpha \bar \gamma g_1+1) }{1+\mu}$

$\bar \gamma =P_s /N_0,$

$b=\frac{\eta \bar \gamma g_1 g_3(\alpha \bar \gamma g_1+\mu +1)}{1+\mu}-\frac{(1-\alpha) \bar \gamma g_2(\alpha \bar \gamma g_1+1))}{\alpha \bar \gamma g_2+\mu +1} + (1-\alpha)\bar \gamma g_1$

$c=(1-\alpha)\bar \gamma g_1 -\frac{(1-\alpha) \bar \gamma g_2(\alpha \bar \gamma g_1+\mu+1))}{\alpha \bar \gamma g_2+\mu +1}$

Does anyone know how can we rewrite the first objected function to the third objected function?

Because in the first objected function,

we do plus $C_1$ and $C_2$,

but begin in the second objected function,we multiple them together

In the third optimize problem, $\frac{b-\sqrt{b^2-4ac}}{2a}$ just appear suddenly in the constraint,i don't know why and how will it appear in the constraint.Does anyone know about them?

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Without going through all the Algebra which looks like quite a lot of algebra it looks like he took the function and absorbed one of the constants into it so it looked something like $$f(\alpha,\rho):= C_1(\alpha,\rho)+\frac{w_1}{w_2}C_2(\alpha,\rho)$$ and since $C_1$ and $C_2$ are logarithms you can write them as a product. He notes that the logarithm is monotonic which I assume is his justification for dropping the logarithm afterward. The going from sum to product is definitely by messing with the constants then putting the constants as exponents and merging as a product. The $$\frac{b-\sqrt{b^2-4ac}}{2a}$$ is the quadratic formula which probably comes about somewhere during the algebra. I would recommend in order to understand it try to reverse engineer it and multiply out all of the items and try and simplify the algebra to get the claim. Often in papers authors compress multiple pages of algebra into a simple line for the sake of brevity.