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I'd like to prove that the order of the two-steps Backward Differentiation Formula is $O(k^2)$. For this, I write $$y'_i\approx \dfrac{y_{i}-{\tfrac {4}{3}}y_{i-1}+{\tfrac {1}{3}}y_{i-2}}{2k/3}=\dfrac{1}{k}\left(\dfrac{3}{2}-2E^{-1}+\dfrac{1}{2}E^{-2}\right)=\dfrac{1}{k}\left(\nabla+\dfrac{\nabla^2}{2}\right).$$

So, the approximation is $$D= \dfrac{1}{k}\left(\nabla+\dfrac{\nabla^2}{2}+\dfrac{\nabla^3}{3}+\dfrac{\nabla^4}{4}+\cdots\right)\approx \dfrac{1}{k}\left(\nabla+\dfrac{\nabla^2}{2}\right)$$ and the error $$ \dfrac{1}{k}\left(\dfrac{\nabla^3}{3}+\dfrac{\nabla^4}{4}+\cdots\right)$$

How could I proceed?

Thank you in advance!

Quiet_waters
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    The difference operator is usually denoted with the \Delta $\Delta$ symbol, not the gradient \nabla $\nabla$. /// You could be more explicit in writing that here $\Delta=1-E^{-1}$, and $E=e^{kD}$, so that $kD=-\ln(1-\Delta)=$ logarithm MacLarin-Taylor expansion. With all that, your approach is correct. You could, alternatively, also insert the exponential or Taylor shift series for $E^{-1}=e^{-kD}$ directly. – Lutz Lehmann Aug 16 '22 at 07:11
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    @LutzLehmann Thank you! $\nabla$ is the backward difference, I forgot to say this. – Quiet_waters Aug 16 '22 at 20:34

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You need to notice that the higher powers of the difference operator come with a like power of $k$, $$ \Delta^m=(1-e^{-kD})^m=(kD-\tfrac12(kD)^2+...)^m =(kD)^m+\tfrac{m}2(kD)^{m+1}+... \\~\text{or}~\\ [\Delta^my]_i= k^my^{(m)}(x_i)+O(k^{m+1}). $$ This then implies that the remainder terms start with $O(k^2)$.

Lutz Lehmann
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