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As you say, this is essentially truncating Taylor's series after the linear term(s). By Lagrange's form of the remainder in one variable (see Wikipedia or your favorite calculus textbook): $$f(x) = f(x_0) + f'(x_0) (x - x_0) + \frac{f''(\xi)}{2!}(x - x_0)^2$$ where $x_0 \le \xi \le x$. The approximation you cite is valid as long as the second-order term ...
This is not quite an answer, but perhaps complements Yiorgos' approach. The idea is to view $x$ as a function of $A$ and compute the derivative. The implicit function theorem provides a convenient way of computing the derivative. Let $\phi(y,A) = Ay-b$, we have ${\partial \phi(y,A) \over \partial y} = A$, and ${\partial \phi(y,A) \over \partial A}(\Delta) ... 1 The relation $$(A+\delta A)(x+\delta x)=b,$$ implies that $$Ax+\delta A (x+\delta x)+A\delta x=b,$$ and since$Ax=b$, the above becomes $$\delta A (x+\delta x)+A\delta x=0,$$ and hence $$\delta x= -A^{-1}\delta A (x+\delta x),$$ which implies that $$\|\delta x\|\le \|A^{-1}\|\|\delta A\|\|x+\delta x\|=\kappa(A)\cdot\frac{\|\delta ... 0 If you replace A by A+δ e_1e_1^T, then by the formulas of Sherman, Woodbury and Morrison the determinant obeys \det(I+uv^T)=1+v^Tu, i.e.,$$\det(A+δ e_1e_1^T)=\det(A)(1+δe_1^TA^{-1}e_1)$$and the inverse follows from$(I+uv^T)(I+cuv^T)=I+(1+c+cv^Tu)uv^T$for$c=-(1+v^Tu)^{-1}$, that is for the perturbed$A\begin{align} (A+δ\, e_1e_1^T)^{-1} ... 0 Clearly the main task here is to reach a precise mathematical question. Let us assume that the setting is that one starts from some vectorU_1$in$\mathbb R^3$, that one first transforms$U_1$into an intermediate vector$U_2$in$\mathbb R^3$using$T_1$and some noise, thus$U_2=T_1U_1+W_1$where$W_1$has mean zero and covariance matrix$\Sigma_1$, and ... 1 More generally, if$x$is twice continuously differentiable, there exists a$\xi\in [0,1]$for which$\frac{x(t+h)-x(t)}{h}=\frac{h}{2!}x''(t+\xi h)$, so that you need to estimate the second derivative:$v_1 = 10 \pm \frac{0.2}{2} v''_{max}$1 Look at the simplest functions$x(t)$and compute the exact expressions$v(h,t) = \frac{x(t+h)-x(t)}{h}$. For$x(t) = at$you have$v(h,t) = \frac{x(t+h)-x(t)}{h} = a$and therefore the error term$O(h)$is zero. For$x(t) = at^2$you have$v(h,t) = 2 at+ah$and$O(h)=ah.\;$Thus the error is constant in time, it only depends on$a,h.$For$x(t) = at^3$... 0 Yes, it should be. I think that "proof" on Wikipedia is bad and isn't sufficient. It is right that the global error of Verlet is$\mathcal{O}(\Delta t^2)\$ but it is no the way to show this. You can find some information in Vesely F. J., Computational Physics: An Introduction (2nd edition ed.). 2001, page 105. Or you can take another approach. By using some ...