Deriving the translog production function

Ive been having difficulty deriving the translog production function defined as:

$$\ln y=\alpha_0+\sum_{i=1}^n\alpha_i \ln x_i+\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n\ \beta_{ij}\ln x_i\ln x_j$$

I know we start with a log-log production function. $$\ln y=\alpha_0+\sum_{i=1}^n\alpha_i\ln x_i$$

the next step from what i recall is to take the taylor series of this function around the point $$x_i=0$$. the reason why this is an issue is because $$\ln(0)$$ is undefined.

How exactly is this function derived?

The translog is a second order Taylor development of $$\log y = f(\log x)$$ in $$\log x$$ around an arbitrary point $$x_0 \ne 0$$:

$$\log y = f(\log x_0) + \frac{\partial f}{\partial \log x^T}(\log x_0)\cdot(\log x-\log x_0) + \frac{1}{2}(\log x-\log x_0)^T\frac{\partial^2 f}{\partial \log x\log x^T}(\log x_0)\cdot(\log x-\log x_0)$$ $$=\alpha_0+\alpha_x^T \log x+\frac{1}{2}\log x^TB\log x$$ where this last reparameterization is obtained when defining: $$\alpha_0=f(\log x_0) - \frac{\partial f}{\partial \log x^T}(\log x_0)\cdot\log x_0 + \frac{1}{2}\log x_0^T\frac{\partial^2 f}{\partial \log x\log x^T}(\log x_0)\cdot\log x_0$$ $$\alpha_x=\frac{\partial f}{\partial \log x}(\log x_0) - \frac{\partial^2 f}{\partial \log x\log x^T}(\log x_0)\log x_0$$ $$B=\frac{\partial^2 f}{\partial \log x\log x^T}(\log x_0).$$

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

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

Up to first order we have then

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

This is naturally extended to $$n > 2$$ as

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