Deriving the translog production function
The idea is indeed to Taylor expand the production function. To justify it, you can start with the constant elasticity of substitution function, which in the two-factor case can be written as

$$
Y = A[\alpha K^\gamma + (1 - \alpha)L^\gamma]^{1/\gamma} \tag{1}
$$

in this case $X_1 = K$, $X_2 = L$. Now we expand $\ln Y$ around $\gamma = 0$ (recall the CES approximates to a cobb-douglas production function when $\gamma \approx0).$ 


  $\gamma^0$ term


$$
\lim_{\gamma \to 0} \ln Y = \ln (A K^\alpha L^{1 - \alpha}) \tag{2}
$$


  $\gamma^1$ term


\begin{eqnarray}
\lim_{\gamma \to 0} \frac{\partial \ln Y}{\partial \gamma} &amp;=&amp; \lim_{\gamma \to 0}\frac{\alpha  K^{\gamma } \ln (L)+(1-\alpha ) l^{\gamma } \log (L)}{\gamma 
   \left(\alpha  K^{\gamma }+(1-\alpha ) L^{\gamma }\right)}-\frac{\ln \left(\alpha 
   K^{\gamma }+(1-\alpha ) L^{\gamma }\right)}{\gamma ^2}\\
&amp;=&amp; \frac{1}{2} (1 - \alpha) \alpha  (\ln (K)-\ln (L))^2 \tag{3}
\end{eqnarray}

Up to first order we have then

\begin{eqnarray}
\ln Y &amp;\approx&amp; \color{blue}{(\ln Y)_{\gamma = 0}} + \color{red}{\left(\frac{\partial \ln Y}{\partial \gamma}\right)_{\gamma = 0} \gamma} \\
&amp;\stackrel{(2),(3)}{=}&amp; \color{blue}{\ln A + \alpha \ln K + (1 - \alpha) \ln L} + \color{red}{\frac{1}{2}\alpha(1 - \alpha)\gamma [\ln K - \ln L]^2} \\
&amp;=&amp; \ln A + \alpha \ln X_1 + (1 - \alpha) \ln X_2 + \frac{\alpha \gamma (1 -\alpha)}{2}\left[\ln^2 X_1 -2\ln X_1\ln X_2 + \ln^2 X_2 \right] \\
&amp;=&amp; \alpha_0 + \sum_{i = 1}^2 \alpha_i \ln X_i + \frac{1}{2}\sum_{i, j = 1}^2 \beta_{ij}\ln X_i \ln X_j \tag{3}
\end{eqnarray}

This is naturally extended to $n &gt; 2$ as 

$$
\ln Y = \alpha_0 + \sum_{i = 1}^n \alpha_i \ln X_i + \frac{1}{2}\sum_{i, j = 1}^n \beta_{ij}\ln X_i \ln X_j
$$