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If I define the approximation of the second derivative as $$\delta^2_xV_{i}=\dfrac{D^+_xV_{i}-D^-_xV_{i}}{(x_{i+1}-x_{i-1})/2}$$ where $$D^+_xV_{i}=\dfrac{V_{i+1}-V_i}{x_{i+1}-x_i}, D^-_xV_{i}=\dfrac{V_{i}-V_{i-1}}{x_i-x_{i-1}},$$ If I divide the interval $\Omega=(0,1)$ into 3 subintervals: $\Omega_l=(0,\sigma), \Omega_c=(\sigma,1-\sigma), \Omega_r=(1-\sigma,1),$ where $\sigma=\min\{1/4,2\sqrt{\epsilon}\ln N_x\}$. Then for both $\Omega_l$ and $\Omega_r$ I divide each into $N_x/4$ elements and $\Omega_c$ into $N_x/2$ elements so that $\Omega=(0,1)$ is divided into $N_x$ elements.

How can I show using Taylor expansion or otherwise that $$|\dfrac{d^2}{dx^2}V(x_i)-\delta^2_xV(x_i)|\leq \begin{cases} \dfrac{1}{3}(x_{i+1}-x_{i-1})||\dfrac{d^3V}{dx^3}||&\text{if } x_i=\sigma \text{ or }x_i=1-\sigma\\ \dfrac{1}{12}(x_{i}-x_{i-1})^2||\dfrac{d^4V}{dx^4}||&\text{otherwise.}\end{cases}$$ I can calculate this if the mesh is uniform but here since the mesh is nonuniform I am stuck.

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Apply Taylor series to \begin{align} \frac{\frac{f(x+ah)-f(x)}{ah}-\frac{f(x)-f(x-bh)}{bh}}{(ah+bh)/2} &=\frac{bf(x+ah)-(a+b)f(x)+af(x-bh)}{ab(a+b)h^2/2} \\[1em] &= \frac{\sum_{k=0}^\infty \frac1{k!}(ba^k+a(-b)^k)f^{(k)}(x)h^k-(a+b)f(x)}{ab(a+b)h^2/2} \\[1em] &= \sum_{k=2}^\infty \frac2{k!}\,\frac{a^{k-1}-(-b)^{k-1}}{a-(-b)}\,f^{(k)}(x)h^{k-2}-(a+b) \\[1em] &= f''(x)+\frac13(a-b)f'''(x)h+\frac1{12}(a^2-ab+b^2)f^{(4)}(x)h^2+O(h^3) \end{align} For $a=b$, the linear error term cancels, as required by the symmetry of the situation. If not, then the linear term dominates the error.

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