Three derivative matching theorems will be proven, one theorem per derivative. The first two theorems have been proven in [10, 11] for the pure state. In here they are proven for the Poisson state. The third theorem is entirely new.

The structure of the proofs is somewhat different than in [10, 11] since in here the focus is on factorial moments. It seems that the equation system for factorial moments is more compact than for other types of moments. As an artifact of that, the theorems do not contain the error terms *ε* that were used in [10, 11]. The theorems proven in here are more generic since they hold even for multi particle reactions, not just binary reactions. Again, as stated previously, all components of $\overrightarrow{\mu}$ are taken strictly larger than zero. If one of the components is zero the XARNES ansatz does not work.

**Theorem 1**. If the base and the exact factorial moments match at

*t = t*_{0}, and the particle number distribution function is the Poisson distribution, i.e., if

${\rho}_{\widehat{m}}\left({t}_{0}\right)={v}_{\widehat{m}}\left({t}_{0}\right)={\overrightarrow{\mu}}^{\widehat{m}};\widehat{m}\in {\mathbf{\Omega}}_{\xi}$

(59)

then the first order derivatives in time also match for the base factorial moments:

$\frac{d}{dt}{\rho}_{\widehat{m}}\left(t\right){|}_{t={t}_{0}}=\frac{d}{dt}{\nu}_{\widehat{m}}\left(t\right){|}_{t={t}_{0}}$

(60)

for all $\widehat{m}\in {\mathbf{\Omega}}_{\xi}$.

*Proof*. The theorem can be easily proven by considering Eq. (58) with *h =* 1. By assumptions of the theorem one has that ${p}_{\widehat{m}}^{\left[0\right]}-{\psi}_{\widehat{m}}^{\left[0\right]}=0$ which eliminates the sum over $\widehat{m}\in {\mathbf{\Omega}}_{\xi}$ in (58). From Lemma 2 it follows that ${\rho}_{\stackrel{\u0304}{m}}^{\left[0\right]}-{\psi}_{\stackrel{\u0304}{m}}^{\left[0\right]}=0$ which eliminates the sum over $\stackrel{\u0304}{m}\in {\stackrel{\u0304}{\mathbf{\Omega}}}_{\xi}$. This finally proves the theorem.

**Theorem 2**. If the base and the exact factorial moments match at

*t = t*_{0}, and the particle number distribution function is the Poisson distribution, i.e., if

${\rho}_{\widehat{m}}\left({t}_{0}\right)={\nu}_{\widehat{m}}\left({t}_{0}\right)={\overrightarrow{\mu}}^{\widehat{m}};\widehat{m}\in {\mathbf{\Omega}}_{\xi}$

(61)

and if Ω

_{
R
} ⊂ Ω

*
ξ
*, then the second order derivatives in time also match:

$\frac{{d}^{2}}{d{t}^{2}}{\rho}_{\widehat{m}}\left(t\right){|}_{t={t}_{0}}=\frac{{d}^{2}}{d{t}^{2}}{v}_{\widehat{m}}\left(t\right){|}_{t={t}_{0}};\widehat{m}\in {\mathbf{\Omega}}_{\xi}$

(62)

*Proof*. The proof of this theorem is somewhat lengthier. To prove the theorem one needs to consider Eq. (58) with *h* = 2. If the assumptions of the theorem are valid, by theorem 1, ${p}_{\widehat{m}}^{\left[1\right]}-{\psi}_{\widehat{m}}^{\left[1\right]}=0$ which eliminates the sum over $\widehat{m}\in {\mathbf{\Omega}}_{\xi}$ in (58). Thus what is left to show is that a difference ${\rho}_{\stackrel{\u0304}{m}}^{\left[1\right]}-{\psi}_{\stackrel{\u0304}{m}}^{\left[1\right]}$ with $\stackrel{\u0304}{m}\in r{\mathbf{\Omega}}_{\xi}$ vanishes.

A straight forward application of the time derivative on the moment closure function leads to the following expression,

${\psi}_{\stackrel{\u0304}{m}}^{\left[1\right]}=\sum _{\widehat{m}\in {\mathbf{\Omega}}_{\xi}}\frac{\partial \psi \stackrel{\u0304}{m}}{\partial \psi \widehat{m}}{v}_{\widehat{m}}^{\left[1\right]}=\sum _{\widehat{m}\in {\mathbf{\Omega}}_{\xi}}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}{\overrightarrow{\mu}}^{\stackrel{\u0304}{m}-\widehat{m}}{v}_{\widehat{m}}^{\left[1\right]}$

(63)

Also, the use of the exact, and the XARNES equation systems to evaluate

$\phantom{\rule{0.3em}{0ex}}{\rho}_{\stackrel{\u0304}{m}}^{\left[1\right]}$ and

${\psi}_{\stackrel{\u0304}{m}}^{\left[1\right]}$ gives

$\begin{array}{c}{\rho}_{\stackrel{\u0304}{m}}^{\left[1\right]}-{\psi}_{\stackrel{\u0304}{m}}^{\left[1\right]}=\sum _{\overrightarrow{c}\in {\mathbf{\Omega}}_{R}}\left[\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill \overrightarrow{c}\hfill \end{array}\right)-\sum _{\phantom{\rule{0.3em}{0ex}}\widehat{m}\in {\mathbf{\Omega}}_{\xi}}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}\left(\begin{array}{c}\hfill \widehat{m}\hfill \\ \hfill \overrightarrow{c}\hfill \end{array}\right)\right]\times \\ \sum _{\alpha =1}^{R}{\lambda}_{\alpha}{\mathbf{\Gamma}}_{\alpha}\left(\overrightarrow{c}\right){\overrightarrow{\mu}}^{\stackrel{\u0304}{m}+{\overrightarrow{u}}_{\alpha}-\overrightarrow{c}}\end{array}$

(64)

This difference is zero provided

$\sum _{\widehat{m}\in {\mathbf{\Omega}}_{\xi}}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}\left(\begin{array}{c}\hfill \widehat{m}\hfill \\ \hfill \overrightarrow{c}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill \overrightarrow{c}\hfill \end{array}\right)$

(65)

for every $\overrightarrow{c}\in {\mathbf{\Omega}}_{R}$. Please note that this condition is almost identical to the equation that characterize the *γ* coefficients of the XARNES ansatz. In one replaces $\overrightarrow{c}$ with $\widehat{m}$, and Ω _{
R
} with Ω _{
ξ
} in the equation above, then the equation obtained in such a way would be identical to Eq. (25) or (44). Thus if Ω _{
R
} ⊂ Ω _{
ξ
} then the equation above is contained in the condition that defines the *γ* coefficients, and the equation is automatically valid. This proves the theorem.

The third order derivatives will be investigated in the same vein as the first and the second order derivatives. The result will be formulated in a precise mathematical theorem. However, before stating the next theorem, it is useful to generalize the space of contraction vectors as follows.

**Definition 12**. Vector space of sums of contraction vectors

${\overrightarrow{c}}_{1}+{\overrightarrow{c}}_{2}+\cdots +{\overrightarrow{c}}_{h}$ where each of the vectors in the sum is from Ω

*
R
*, will be denoted as

${\mathbf{\Omega}}_{h\otimes R}=\left\{{\overrightarrow{c}}_{1}+{\overrightarrow{c}}_{2}+\cdots +{\overrightarrow{c}}_{h}|{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\cdots \phantom{\rule{0.3em}{0ex}},{\overrightarrow{c}}_{h}\in {\mathbf{\Omega}}_{R}\right\}$

(66)

and *h* is an integer and obeys *h* ≥ 1.

**Theorem 3**. If the base and the exact factorial moments match at

*t* =

*t*_{0} and the particle number distribution function is the Poisson distribution, i.e., if

${\rho}_{\widehat{m}}\left({t}_{0}\right)={\nu}_{\widehat{m}}\left({t}_{0}\right)={\overrightarrow{\mu}}^{\widehat{m}};\widehat{m}\in {\mathbf{\Omega}}_{\xi}$

(67)

and if

*Ω*_{2⊗R}⊂

*Ω*_{
ξ
} , then the third order derivatives in time also match

$\frac{{d}^{3}}{d{t}^{3}}{\rho}_{\widehat{m}}\left(t\right){|}_{t={t}_{0}}=\frac{{d}^{3}}{d{t}^{3}}{\nu}_{\widehat{m}}\left(t\right){|}_{t={t}_{0}};\widehat{m}\in {\mathbf{\Omega}}_{\xi}$

(68)

*Proof*. The theorem can be proven by considering Eq. (58) with *h* = 3. By theorem 2 one has that ${\rho}_{\widehat{m}}^{\left[2\right]}-{\psi}_{\widehat{m}}^{\left[2\right]}=0$. What is left to show is that all differences ${\rho}_{\stackrel{\u0304}{m}-{\psi}_{\stackrel{\u0304}{m}}^{\left[2\right]}}^{\left[2\right]}$ with $\stackrel{\u0304}{m}\in {\stackrel{\u0304}{\mathbf{\Omega}}}_{\xi}$ vanish as well. Unfortunately this is a highly nontrivial task.

By the use of the standard calculus one can show that

${\psi}_{\stackrel{\u0304}{m}}^{\left[2\right]}={\psi}_{\stackrel{\u0304}{m},\u2020}^{\left[2\right]}+{\psi}_{\stackrel{\u0304}{m},\u2021}^{\left[2\right]}$

(69)

where

$\begin{array}{c}{\psi}_{\stackrel{\u0304}{m},\u2020}^{\left[2\right]}=\sum _{{\widehat{m}}_{1},{\widehat{m}}_{2}\in {\mathbf{\Omega}}_{\xi}}\left({\gamma}_{{\widehat{m}}_{1}}^{\stackrel{\u0304}{m}\gamma}{\gamma}_{{\widehat{m}}_{2}}^{\stackrel{\u0304}{m}\gamma}-{\gamma}_{{\widehat{m}}_{1}}^{\stackrel{\u0304}{m}\gamma}{\delta}_{{\widehat{m}}_{1},{\widehat{m}}_{2}}\right)\times \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{\mu}^{\stackrel{\u0304}{m}-{\widehat{m}}_{1}-{\widehat{m}}_{2}}{v}_{{\widehat{m}}_{1}}^{\left[1\right]}{v}_{{\widehat{m}}_{2}}^{\left[1\right]}\end{array}$

(70)

and

${\psi}_{\stackrel{\u0304}{m},\u2021}^{\left[2\right]}=\sum _{\widehat{m}\in {\mathbf{\Omega}}_{\xi}}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}{\overrightarrow{\mu}}^{\stackrel{\u0304}{m}-\widehat{m}}{\nu}_{\widehat{m}}^{\left[2\right]}$

(71)

By using the XARNES equations, and identity (44) for

*γ* coefficients, the

${\psi}_{\stackrel{\u0304}{m},\u2020}^{\left[2\right]}$ can be expressed as

$\begin{array}{c}{\psi}_{\stackrel{\u0304}{m},\u2020}^{\left[2\right]}=,\phantom{\rule{2.77695pt}{0ex}}\sum _{{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\in {\mathbf{\Omega}}_{R}}\left[\left(\begin{array}{c}\stackrel{\u0304}{m}\hfill \\ {\overrightarrow{c}}_{1}\hfill \end{array}\right)\left(\begin{array}{c}\stackrel{\u0304}{m}\hfill \\ {\overrightarrow{c}}_{2}\hfill \end{array}\right)-\sum _{{}_{\widehat{m}\in \mathbf{\Omega}\xi}}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}\left(\begin{array}{c}\widehat{m}\hfill \\ \hfill {\overrightarrow{c}}_{1}\hfill \end{array}\right)\left(\begin{array}{c}\widehat{m}\hfill \\ {\overrightarrow{c}}_{2}\hfill \end{array}\right)\right]\times \hfill \\ \phantom{\rule{0.5em}{0ex}}\sum _{\alpha ,\beta}{\lambda}_{\alpha}{\lambda}_{\beta}{\mathbf{\Gamma}}_{\alpha}\left({\overrightarrow{c}}_{1}\right){\mathbf{\Gamma}}_{\beta}\left({\overrightarrow{c}}_{2}\right){\overrightarrow{\mu}}^{\stackrel{\u0304}{m}+\overrightarrow{u}\alpha +{\overrightarrow{u}}_{\beta}-{\overrightarrow{c}}_{1}-{\overrightarrow{c}}_{2}}\hfill \end{array}$

(72)

which vanishes provided

$\sum _{\widehat{m}\in {\mathbf{\Omega}}_{\xi}}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}\left(\begin{array}{c}\hfill \widehat{m}\hfill \\ \hfill {\overrightarrow{c}}_{1}\hfill \end{array}\right)\left(\begin{array}{c}\hfill \widehat{m}\hfill \\ \hfill {\overrightarrow{c}}_{2}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {\overrightarrow{c}}_{1}\hfill \end{array}\right)\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {\overrightarrow{c}}_{2}\hfill \end{array}\right)$

(73)

for any ${\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\in {\mathbf{\Omega}}_{R}$. This is indeed true by corollary 3 under the assumptions of the theorem.

What is left to show is that

${\rho}_{\stackrel{\u0304}{m}}^{\left[2\right]}-{\psi}_{\stackrel{\u0304}{m},\u2021}^{\left[2\right]}=0$. A strategy for proving this is as follows. If one could show that the following identity holds for the exact moments

${\Delta}_{\stackrel{\u0304}{m}}\equiv {\rho}_{\stackrel{\u0304}{m}}^{\left[2\right]}-\sum _{\widehat{m}\in \mathbf{\Omega}\xi}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}{\overrightarrow{\mu}}^{\stackrel{\u0304}{m}-\widehat{m}}{\rho}_{\widehat{m}}^{\left[2\right]}=0$

(74)

then it is clear that

${\rho}_{\stackrel{\u0304}{m}}^{\left[2\right]}-{\psi}_{\stackrel{\u0304}{m},\u2021}^{\left[2\right]}=0$ would be true since one could write

${\rho}_{\stackrel{\u0304}{m}}^{\left[2\right]}-{\psi}_{\stackrel{\u0304}{m},\u2021}^{\left[2\right]}=\sum _{\widehat{m}\in \mathbf{\Omega}\xi}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}{\overrightarrow{\mu}}^{\stackrel{\u0304}{m}-\widehat{m}}\left({\rho}_{\widehat{m}}^{\left[2\right]}-{\psi}_{\widehat{m}}^{\left[2\right]}\right)$

(75)

and this expression would vanish by Theorem 2.

Somewhat naive recursive application of the exact equations of motion results in

$\begin{array}{c}{\rho}_{\stackrel{\u0304}{m}}^{\left[2\right]}=\sum _{\alpha ,\beta}{\lambda}_{\alpha}{\lambda}_{\beta}\sum _{{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\in {\mathbf{\Omega}}_{R}}\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {\overrightarrow{c}}_{1}\hfill \end{array}\right)\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}+{\overrightarrow{u}}_{\alpha}-{\overrightarrow{c}}_{1}\hfill \\ \hfill {\overrightarrow{c}}_{2}\hfill \end{array}\right)\times \\ {\mathbf{\Gamma}}_{\alpha}\left({\overrightarrow{c}}_{1}\right){\mathbf{\Gamma}}_{\beta}\left({\overrightarrow{c}}_{2}\right){\overrightarrow{\mu}}^{\stackrel{\u0304}{m}+{\overrightarrow{u}}_{\alpha}-{\overrightarrow{u}}_{\beta}-{\overrightarrow{c}}_{1}-{\overrightarrow{c}}_{2}}\phantom{\rule{0.3em}{0ex}}\end{array}$

(76)

By using a tedious manipulation of the binomial coefficients the expression above can be converted into a more useful form

$\begin{array}{c}{\rho}_{\stackrel{\u0304}{m}}^{\left[2\right]}=\sum _{\alpha ,\beta}{\lambda}_{\alpha}{\lambda}_{\beta}\sum _{{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2},\overrightarrow{d}\in {\mathbf{\Omega}}_{R}}\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {c}_{1}\hfill \end{array}\right)\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}-{\overrightarrow{c}}_{1}\hfill \\ \hfill {c}_{2}\hfill \end{array}\right)\times \\ \left(\frac{{\overrightarrow{u}}_{\alpha}}{d}\right){\mathbf{\Gamma}}_{\alpha}\left({\overrightarrow{c}}_{1}\right){\mathbf{\Gamma}}_{\beta}\left({\overrightarrow{c}}_{2}+\overrightarrow{d}\right){\overrightarrow{\mu}}^{\stackrel{\u0304}{m}+{\overrightarrow{u}}_{\alpha}+{\overrightarrow{u}}_{\beta}-{\overrightarrow{c}}_{1}-{\overrightarrow{c}}_{2}-\overrightarrow{d}}\end{array}$

(77)

In fact, it is easier to start from (77) and obtain (76). First, the manipulation requires that the sum of ${\overrightarrow{c}}_{2}$ and $\overrightarrow{d}$ is changed, into the sum over ${\overrightarrow{c}}_{2}={\overrightarrow{c}}_{2}+\overrightarrow{d}$ and ${\overrightarrow{c}}_{3}={\overrightarrow{c}}_{2}$After that the Vandermonde identity needs to be used which consumes the sum over ${\overrightarrow{c}}_{{3}^{\prime}}$, resulting finally in (77).

By using the fact that

$\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {\overrightarrow{c}}_{1}\hfill \end{array}\right)\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}-{\overrightarrow{c}}_{1}\hfill \\ \hfill {\overrightarrow{c}}_{2}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {\overrightarrow{c}}_{1}+{\overrightarrow{c}}_{2}\hfill \end{array}\right)\left(\begin{array}{c}\hfill {\overrightarrow{c}}_{1}+{\overrightarrow{c}}_{2}\hfill \\ \hfill {\overrightarrow{c}}_{2}\hfill \end{array}\right)$

(78)

Eq (77) can be written in the most useful form as

$\begin{array}{c}\phantom{\rule{1em}{0ex}}{\rho}_{\stackrel{\u0304}{m}}^{\left[2\right]}=\sum _{{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\in {\mathbf{\Omega}}_{R}}\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {\overrightarrow{c}}_{1}+{\overrightarrow{c}}_{2}\hfill \end{array}\right)\times \\ \sum _{\alpha ,\beta}{\mathbf{\Lambda}}_{\alpha}{,}_{\beta}\left(\overrightarrow{\mu},{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\right){\overrightarrow{\mu}}^{\stackrel{\u0304}{m}+{\overrightarrow{u}}_{\alpha}+{\overrightarrow{u}}_{\beta}-{\overrightarrow{c}}_{1}-{\overrightarrow{c}}_{2}}\end{array}$

(79)

The exact form of a coefficient ${\mathbf{\Lambda}}_{\alpha ,\beta}\left(\overrightarrow{\mu},{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\right)$ can be found if needed and, in fact, it is a series in $\overrightarrow{\mu}$. However, the exact form of these coefficients is not relevant for the discussion that follows.

Let us use (79) in (74). This gives

$\begin{array}{c}{\mathbf{\Delta}}_{\stackrel{\u0304}{m}}=\sum _{{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\in {\mathbf{\Omega}}_{R}}\left[\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {\overrightarrow{c}}_{1}+{\overrightarrow{c}}_{2}\hfill \end{array}\right)-\sum _{{\widehat{m}}_{1}\in {\mathbf{\Omega}}_{\xi}}{\gamma}_{{\widehat{m}}_{1}}^{\stackrel{\u0304}{m}}\left(\begin{array}{c}\hfill {\widehat{m}}_{1}\hfill \\ \hfill {\overrightarrow{c}}_{1}+{\overrightarrow{c}}_{2}\hfill \end{array}\right)\right]\times \\ \sum _{\alpha ,\beta}{\mathbf{\Lambda}}_{\alpha ,\beta}\left(\overrightarrow{\mu},{\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}\right){\overrightarrow{\mu}}^{\stackrel{\u0304}{m}+{\overrightarrow{u}}_{\alpha}+{\overrightarrow{u}}_{\beta}-{\overrightarrow{c}}_{1}-{\overrightarrow{c}}_{2}}\end{array}$

(80)

and

${\mathbf{\Delta}}_{\stackrel{\u0304}{m}}$ is zero if the following condition is met,

$\sum _{\widehat{m}\in {\mathbf{\Omega}}_{\xi}}{\gamma}_{\widehat{m}}^{\stackrel{\u0304}{m}}\left(\begin{array}{c}\hfill \widehat{m}\hfill \\ \hfill {\overrightarrow{c}}_{1,2}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \stackrel{\u0304}{m}\hfill \\ \hfill {\overrightarrow{c}}_{1,2}\hfill \end{array}\right)$

(81)

for every ${\overrightarrow{c}}_{1,2}\in {\mathbf{\Omega}}_{2\otimes R}$. Eq. (81) is satisfied by the assumption of the theorem which states that *Ω*_{2⊗R}⊂ *Ω*_{
ξ
} . In such a case equations in (81) are a subset of the equations satisfied by the *γ* coefficients which are given in (44) and are automatically valid.

The three theorems proven so far are suggestive of the fact that one might try to prove the following conjecture.

**Conjecture 1. If** the base and the exact factorial moments match at

*t* =

*t*_{0}, and the particle number distribution function is the Poisson distribution, i.e., if

${\rho}_{\widehat{m}}\left({t}_{0}\right)={\nu}_{\widehat{m}}\left({t}_{0}\right)={\overrightarrow{\mu}}^{\widehat{m}};\widehat{m}\in {\mathbf{\Omega}}_{\xi}$

(82)

and if

*Ω*_{h⊗R}⊂

*Ω*_{
ξ
} , where

*h* is an arbitrary integer such that

*h* ≥ 1, then the time derivatives with orders

*D* = 0, 1, 2,

**...**,

*h* + 1 will also match

$\frac{{d}^{D}}{d{t}^{D}}{\rho}_{\widehat{m}}\left(t\right){|}_{t={t}_{0}}=\frac{{d}^{D}}{d{t}^{D}}{v}_{\widehat{m}}\left(t\right){|}_{t={t}_{0}};\widehat{m}\in {\mathbf{\Omega}}_{\xi}$

(83)

*Eventual. Inductive* proof for general *h* could be used. However, the problems is that one would need to inspect a difference ${\rho}_{\stackrel{\u0304}{m}}^{\left[h\right]}-{\psi}_{\stackrel{\u0304}{m}}^{\left[h\right]}$ for arbitrary *h* and show that it vanishes under some conditions. Presumably, the main condition, apart from the standard requirements, e.g. such as having the Poisson initial condition, would be that *Ω*_{h⊗R}⊂ *Ω*_{
ξ
} . For example, one can easily see that an expressions such as the one shown in (53) will appear if one tries to calculate higher time derivatives of ${\psi}_{\stackrel{\u0304}{m}}$. However, as demonstrated for the *h* = 2 (*D* = 3) case the computation of ${\psi}_{\stackrel{\u0304}{m}}^{\left[h\right]}$ for higher values of *h* is a rather cumbersome and technical procedure. Generalization to higher orders is apparently very hard but not impossible.

There are couple of reasons why such a conjecture might be valid. First, the structure of the proofs of theorems 1-3 (the cases *h* = 0, 1, 2) suggests such a possibility. Second, there is numerical evidence from a previous study [17] that increase in *ξ* improves the accuracy of the XARNES method. In the context of the theorems discussed in here, increase in *ξ* implies that the initial *Ω*_{
ξ
} set becomes larger. This in turncan make the condition *Ω*_{h⊗R}⊂ *Ω*_{
ξ
} valid for a larger value of *h*. Finally as a result of that a larger number of derivatives would match which would explain the observed accuracy improvements in the studied benchmark cases. Finally, third, this conjecture has be checked by Mathematica for the *T* = 1 case and two binary reactions 2*X*_{1} → 0 and 2*X*_{1} → *X*_{1}, both with *h* = 0, 1, 2, 3, 4, 5 and *ξ* = 2, 4, 6, 8, 10 respectively, and one multi particle reaction 3*X*_{1} → 2*X*_{1} with *h* = 0, 1, 2, 3 and *ξ* = 3, 6, 9.