I have the heat equation: $\frac{\partial u}{\partial t} - \nu \frac{\partial^2 u}{\partial x^2} = f$ for $(t,x)\in (0,+\infty) \times (0,1)$. For the implicit method (finite difference), the truncation error is defined like: \begin{equation*} \varepsilon(u;t,x)=\frac{u(t+\Delta t,x)-u(t,x)}{\Delta t} - \nu \frac{u(t+\Delta t,x-\Delta x)-2u(t+\Delta t,x)+u(t+\Delta t,x+\Delta x)}{(\Delta x)^2}-f(t+\Delta t,x). \end{equation*} I have applied Taylor's formula: \begin{equation*} \dfrac{u(t + \Delta t, x) - u(t,x)}{\Delta t} = \dfrac{\partial u}{\partial t} (t,x) + \dfrac{\Delta t}{2!}\dfrac{\partial^2u}{\partial t^2}(t,x) + \dfrac{(\Delta t)^2}{3!}\dfrac{\partial^3u}{\partial t^3}(\xi_x, x) \end{equation*} with $\xi_x \in (t,t+\Delta t)$
Now, I don´t know how to apply Taylor's formula to the sencond fraction. Concretly in $u(t+\Delta t,x-\Delta x)$ and $u(t+\Delta t,x+\Delta x)$. In the last term I have use: \begin{equation*} f(t+\Delta t,x) = f(t,x)+\dfrac{\partial f}{\partial t}(t,x) + \dfrac{\partial^2 f}{\partial t^2}(\mu_x,x) \end{equation*} with $\mu_x \in (t,t+\Delta t)$. Any helps to continue?
If you do the expansion in $(t+Δt,x)$, then there are only 3 terms to expand parallel to the time and space axis, instead of 4 with 2 in diagonal directions. The difference of the error to an expansion in $(t,x)$ is all in higher order error terms, the principal behavior is not influenced.
This means that you only have to insert the expansions in $x$ direction. As the terms occur symmetrically in $Δx$, the sum will only contain the even terms in the Taylor expansion, so $$ u(t+Δt,x+Δx)+u(t+Δt,x-Δx)=2u(t+Δt,x)+u_{xx}(t+Δt,x)Δx^2+\frac1{12}\partial_x^4u(t+Δt,x)Δx^4+O(Δx^6) $$ and the expansion solely in $t$ direction $$ u(t,x)=u(t+Δt,x)- u_t(t+Δt,x)Δt+\frac12u_{tt}(t+Δt,x)Δt^2+O(Δt^3)... $$ so that \begin{align} &\frac{u(t+Δt,x)-u(t,x)}{Δt} \\[.5em]&\qquad=u_t(t+Δt,x)-\frac12u_{tt}(t+Δt,x)Δt+O(Δt^2) \\[1em] &\frac{u(t+Δt,x+Δx)-2u(t+Δt,x)+u(t+Δt,x-Δx)}{Δx^2} \\[.5em]&\qquad=u_{xx}(t+Δt,x)+\frac1{12}∂_x^4u(t+Δt,x)Δx^2+O(Δx^4) \\[1em] ε(u;t,x)&=[u_t(t+Δt,x)-νu_{xx}(t+Δt,x)-f(t+Δt,x)] \\[.5em]&\qquad-\frac12u_{tt}(t+Δt,x)Δt+\frac1{12}∂_x^4u(t+Δt,x)Δx^2+... \\[.5em]&=O(Δt+Δx^2) \end{align} as claimed.