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According to this paper the following finite difference approximation is third-order accurate:

$$\frac{d\rho_j}{dx}\approx\frac{2-\eta}{3}\frac{\rho_{j+1/2}-\rho_{j-1/2}}{\Delta x}+\frac{1+\eta}{3}\frac{\rho_{j-1/2}-\rho_{j-3/2}}{\Delta x}\text{, with }\eta=\frac{u\Delta t}{\Delta x}.$$

But I don't see how this is third-order accurate. If you Taylor expand $\rho_j$ you get: $$\frac{\rho_{j+1/2}-\rho_{j-1/2}}{\Delta x}=\frac{d\rho_i}{dx}+\frac{\Delta x^2}{24}\frac{d^3\rho_j}{dx^3}+O(\Delta x^3),$$ $$\frac{\rho_{j-1/2}-\rho_{j-3/2}}{\Delta x}=\frac{d\rho_i}{dx}-\Delta x\frac{d^2\rho_j}{dx^2}+\frac{13\Delta x^2}{24}\frac{d^3\rho_j}{dx^3}+O(\Delta x^3).$$ Therefore, at best this scheme is first-order accurate? For reference see page 6 of the paper that is linked. Or this image here: enter image description here

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I think you are meant to examine the order of the whole scheme for the equation, rather than just this particular approximation of $\rho_x$. Substitute the whole expression back into the definition of the scheme (top of page 6, with $u$ missing): $$ \frac{\rho^{n+1}_{j+\frac12}-\rho^n_{j+\frac12}}{\delta t} + u \frac{\bar\rho_{j+1}-\bar\rho_j}{\delta x} = 0 $$ and evaluate the residual for a true solution $\rho$ of the equation $\rho_t + u \rho_x = 0$.

For the 1st order scheme I get 1st order (code): $$ \tfrac12 h u(\eta-1)\rho_{xx}, $$ second order: $$ -\tfrac16 h^2u(\eta^2-1)\rho_{xxx}, $$ third order: $$ \tfrac1{24}h^3u(\eta^3-2\eta^2-\eta+2)\rho_{xxxx}, $$ so the orders come out as the paper says.

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