$$\forall m \land \forall \vec r_{(t1)}(r_{1x},r_{1y},r_{1z}) \land \forall \vec r_{(t2)}(r_{2x},r_{2y},r_{2z}) \land \forall \vec r_{(t3)}(r_{3x},r_{3y},r_{3z}) $$ $$\vec v_{(t1)}=\frac {\vec r_{(t2)} - \vec r_{(t1)}} {\delta t} \tag {1.a} $$ $$\vec v_{(t2)}=\frac {\vec r_{(t3)} - \vec r_{(t2)}} {\delta t} \tag {1.b} $$ $$\vec a_{(t1)}=\frac {\vec v_{(t2)} - \vec v_{(t1)}} {\delta t} \tag {1.c} $$ $$ \vec r_{(t1)} \times m \vec v_{(t1)}=m(r_{1y} v_{1z} -r_{1z} v_{1y},r_{1z} v_{1x} -r_{1x} v_{1z},r_{1x} v_{1y} -r_{1y} v_{1x})= \vec L_{(t1)} \tag {2.a} $$ $$ \vec r_{(t2)} \times m \vec v_{(t2)}=m(r_{2y} v_{2z} -r_{2z} v_{2y},r_{2z} v_{2x} -r_{2x} v_{2z},r_{2x} v_{2y} -r_{2y} v_{2x})= \vec L_{(t2)} \tag {2.b} $$ $$\frac {\vec L_{(t2)}-\vec L_{(t1)}} {\delta t} =(\frac {\delta l_{x}}{\delta t}, \frac {\delta l_{y}}{\delta t},\frac {\delta l_{z}}{\delta t} ) \tag 3$$ $$\frac {\delta l_{x}}{\delta t}=m(r_{2y}v_{2z}-r_{1y}v_{1z}-r_{2z}v_{2y}+r_{1z}v_{1y}) \tag {3.a}$$ $$\frac {\delta l_{y}}{\delta t}=m(r_{2z}v_{2x}-r_{1z}v_{1x}-r_{2x}v_{2z}+r_{1x}v_{1z}) \tag {3.b}$$ $$\frac {\delta l_{z}}{\delta t}=m(r_{2x}v_{2y}-r_{1x}v_{1y}-r_{2y}v_{2x}+r_{1y}v_{1x}) \tag {3.c}$$
$$\vec r_{(t1)}\times m \vec a_{(t1)}=m(r_{1y}a_{1z} - r_{1z}a_{1y}, r_{1z}a_{1x}-r_{1x}a_{1z},r_{1x}a_{1y} -r_{1y}a_{1x}) \tag 4$$
$$r_{1y}a_{1z} - r_{1z}a_{1y}=r_{1y}v_{2z}-r_{1y}v_{1z} -r_{1z}v_{2y}+r_{1z}v_{1y}\tag {4.a}$$ $$r_{1z}a_{1x} - r_{1x}a_{1z}=r_{1z}v_{2x}-r_{1z}v_{1x} -r_{1x}v_{2z}+r_{1x}v_{1z}\tag {4.b}$$ $$r_{1x}a_{1y} - r_{1y}a_{1x}=r_{1x}v_{2y}-r_{1x}v_{1y} -r_{1y}v_{2x}+r_{1y}v_{1x}\tag {4.c}$$
$$(4) \neq (3)$$
Correct pattern $$ \frac {\delta \vec L}{\delta t} = m( \frac {\delta \vec r_{(t1)}} {\delta t} \times \frac {\delta \vec r_{(t2)}} {\delta t} + \vec r_{(t1)} \times \frac {\delta^2 \vec r_{(t1)}} {\delta^2 t}) \tag 5$$
I proved it here but this result $(4) \neq (3)$ is contrary to bookish knowledge and (5) changes the understanding of point rotation, since this formula allows for the existence of non-zero internal moments of force, lawful conservation of angular momentum.
So the question is: I am wrong in the calculations or maybe the present knowledge is wrong?
As I said somewhere, I prefer the symmetric discretizations that have automatically a higher order of accuracy. So with some infinitesimal $δt$ set $t_k=t_0+\frac12k\,δt$. Then $a_0=\frac{v_1-v_{-1}}{δt}+O(δt^2)$, $v_1=\frac{r_2-r_0}{δt}+O(δt^2)$, $v_{-1}$ similarly, and \begin{align} \frac{r_1×v_1-r_{-1}×v_{-1}}{δt} &=\frac{(r_1-r_{-1})×(v_1+v_{-1})}{2δt}+\frac{(r_1+r_{-1})×(v_1-v_{-1})}{2δt} \\ &=\frac{(v_0+O(δt)^2)×(v_0+O(δt)^2)}{2δt}+\frac{(r_0+O(δt)^2)×(a_0+O(δt)^2)}{2} \\ &=r_0×a_0+O(δt^2) \end{align} Now remember that as an infinitesimal $δt$ has a decimal representation with a mind-boggling number of zeros after the decimal dot. This transfers to the remainder $O(δt^2)$. So in all integrations over finite time spans this remainder contributes at most $O(δt)$, which has no appreciable influence.