Prove the implicit method has $\mathcal{O}(\Delta{t}+(\Delta{x})^2)$ as truncation error for the heat equation (finite difference). 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?
 A: 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.
