How can the differential of an arbitrary function $f(t,S_t)$, through a second-order Taylor Expansion, become:
$df = \frac{∂f}{∂t}dt+ \frac{∂f}{∂S_t}dS_t + \frac{1}{2}\frac{∂²f}{∂S^2_t}(dSt_t)^2$
What are the steps to arrive at this equation? Is the function expanded first and differentiated next? Or are "f,x, and a" of an ordinary TE replaced before expansion?
First we need a preamble on multivariate calculus. The following formula is due to differential of a bivariate function:$$df={\partial f\over\partial x}dx+{\partial f\over\partial y}dy$$(this can be generalized to multivariate calculus with number of variables more that 2 but we're not about to look for it) and if we substitute $x=t$ and $y=S_t$ we obtain$${df\over dt}={\partial f\over\partial t}+{\partial f\over\partial S_t}{dS_t\over dt}$$also$${d^2f\over dt^2}={d\over dt}{df\over dt}={\partial^2 f\over \partial t^2}+{\partial^2 f\over \partial t\partial S_t}{dS_t\over dt}+{\partial f\over \partial S_t}{d^2 S_t\over d t^2}$$Now we are able to write the 2nd-order Taylor series around $0$:$$f(t,S_t)=f(0,S_0)+\left\{{\partial f\over\partial t}+{\partial f\over\partial S_t}{dS_t\over dt}\right\}_{t=0}\cdot t+{1\over 2}\left\{{\partial^2 f\over \partial t^2}+{\partial^2 f\over \partial t\partial S_t}{dS_t\over dt}+{\partial f\over \partial S_t}{d^2 S_t\over d t^2}\right\}_{t=0}\cdot t^2+O(t^3)$$where $O(.)$ denotes Big-O Notation.