e^{t\mathrm{ad}_{F}}=\begin{bmatrix}1&4a^{2}t^{2}&2at\\\ 0&1&0\\\
0&4at&1\end{bmatrix}.$ (82)

We now calculate their product:

$\displaystyle
e^{t\cdot\mathrm{ad}_{G}}e^{t\cdot\mathrm{ad}_{F}}=\begin{bmatrix}1&8a^{2}t^{2}&2at\\\
8b^{2}t^{2}&1+8abt^{2}+64a^{2}b^{2}t^{4}&2bt+16a^{2}bt^{3}\\\
-4bt&4at-32a^{2}bt^{3}&1-8abt^{2}\end{bmatrix}.$ (83)

With some help from SymPy [51], we now calculate $P$ and $D$ in the
diagonalization $e^{t\cdot\mathrm{ad}_{G}}e^{t\cdot\mathrm{ad}_{F}}=PDP^{-1}$:

$\displaystyle
P=\left[\begin{matrix}-\frac{1}{2bt}&\frac{-abt-i\sqrt{ab\left(1-abt^{2}\right)}}{2b}&\frac{-abt+i\sqrt{ab\left(1-abt^{2}\right)}}{2b}\\\
-\frac{1}{2at}&\frac{-abt+i\sqrt{ab\left(1-abt^{2}\right)}}{2a}&\frac{-abt-i\sqrt{ab\left(1-abt^{2}\right)}}{2a}\\\
1&1&1\end{matrix}\right].$ (84)

Here $i=\sqrt{-1}$ is the imaginary unit.

We can thereby compute $P^{-1}(G+e^{tad_{G}}F)$ as follows:

$\displaystyle
P^{-1}(G+e^{tad_{G}}F)=\left[\begin{matrix}\frac{2abt}{abt^{2}-1}\\\
\frac{abt\left(3abt-4a^{2}b^{2}t^{3}+i\left(4abt^{2}-1\right)\sqrt{ad\left(1-abt^{2}\right)}\right)}{a^{2}b^{2}t^{3}-abt+i\left(1-abt^{2}\right)\sqrt{ab\left(1-abt^{2}\right)}}\\\
i\frac{abt\left(2abt-2a^{2}b^{2}t^{3}-i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)}{\sqrt{ab\left(1-abt^{2}\right)}\left(1-abt^{2}\right)}\end{matrix}\right].$
(85)

On the other hand,

$\displaystyle D=\textrm{diag}\left(\begin{bmatrix}1\\\
8a^{2}b^{2}t^{4}-8abt^{2}-4t\sqrt{ab\left(abt^{2}-1\right)}\left(2abt^{2}-1\right)+1\\\
8a^{2}b^{2}t^{4}-8abt^{2}+4t\sqrt{ab\left(abt^{2}-1\right)}\left(2abt^{2}-1\right)+1\end{bmatrix}\right).$
(86)

Thus, we use (86) to calculate $\log(D)(D-I)^{-1}$ by applying the series of
matrix logarithm (68):

$\log(D)(D-I)^{-1}=\sum_{j=0}^{\infty}\frac{(I-D)^{j}}{j+1}\\\
=\textrm{diag}\begin{bmatrix}1\\\
\sum_{j=0}^{\infty}{\frac{4^{j}t^{j}}{j+1}(-2a^{2}b^{2}t^{3}+2abt+\sqrt{ab(abt^{2}-1)}(2abt^{2}-1))^{j}}\\\
\sum_{j=0}^{\infty}{\frac{4^{j}t^{j}}{j+1}(-2a^{2}b^{2}t^{3}+2adt-\sqrt{ab(abt^{2}-1)}(2abt^{2}-1))^{j}}\end{bmatrix}.$
(87)

Note that by assumption, $t\leq\eta<1/\sqrt{ab}$, which implies $t^{2}<1/ab$,
or $abt^{2}-1<0$. Thus, (87) can be rewritten in terms of complex numbers:

$\displaystyle\log\left(D\right)\left(D-I\right)^{-1}=\textrm{diag}\left(\begin{bmatrix}1\\\
\sum_{j=0}^{\infty}{\frac{4^{j}t^{j}}{j+1}\left(2abt-2a^{2}b^{2}t^{3}+i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)^{j}}\\\
\sum_{j=0}^{\infty}{\frac{4^{j}t^{j}}{j+1}\left(2abt-2a^{2}b^{2}t^{3}-i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)^{j}}\end{bmatrix}\right)=$
$\displaystyle-\textrm{diag}\left(\begin{bmatrix}-1\\\
\frac{1}{4t\left(2abt-2a^{2}b^{2}t^{3}+i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)}\log\left[1-4t\left(2abt-2a^{2}b^{2}t^{3}+i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)\right]\\\
\frac{1}{4t\left(2abt-2a^{2}b^{2}t^{3}-i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)}\log\left[1-4t\left(2abt-2a^{2}b^{2}t^{3}-i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)\right]\end{bmatrix}\right),$
(88)

using the Taylor series for
$-\log\left(1-x\right)=\sum_{k=1}^{\infty}\frac{x^{k}}{k}=\sum_{j=0}^{\infty}\frac{x^{j+1}}{j+1}=x\sum_{j=0}^{\infty}\frac{x^{j}}{j+1}$.
Note in particular that this means we are using the principal branch of the
logarithm.

####  C.2.2 Simplification of the integrand

We now start substituting the variables we calculated into Theorem 2. We begin
by multiplying (88) and (85) to get:

$\displaystyle\log\left(D\right)\left(D-I\right)^{-1}P^{-1}(G+e^{t\mathrm{ad}_{G}}F)$
$\displaystyle=-\begin{bmatrix}\frac{2abt}{1-abt^{2}}\\\
\frac{ab\log\left[1-4t\left(2abt-2a^{2}b^{2}t^{3}+i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)\right]}{4\left(2abt-2a^{2}b^{2}t^{3}+i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)}\frac{3abt-4a^{2}b^{2}t^{3}+i\left(4abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}}{a^{2}b^{2}t^{3}-abt+i\left(1-abt^{2}\right)\sqrt{ab\left(1-abt^{2}\right)}}\\\
i\frac{ab\log\left[1-4t\left(2abt-2a^{2}b^{2}t^{3}-i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)\right]}{4\sqrt{ab\left(1-abt^{2}\right)}\left(1-abt^{2}\right)}\end{bmatrix}.$
(89)

The second, rightmost factor in the second row of (89) simplifies to

$\displaystyle\frac{-i}{4\left(1-abt^{2}\right)\sqrt{ab(1-abt^{2})}}.$ (90)

Thus, we plug (90) back into (89) and simplify (89) to

$\displaystyle\frac{ab}{4i\left(1-abt^{2}\right)\sqrt{ab(1-abt^{2})}}\left[\begin{matrix}-8it\sqrt{ab(1-abt^{2})}\\\
-\log\left[1-4t\left(2abt\left(1-abt^{2}\right)+i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)\right]\\\
\log\left[1-4t\left(2abt\left(1-abt^{2}\right)-i\left(2abt^{2}-1\right)\sqrt{ab\left(1-abt^{2}\right)}\right)\right]\end{matrix}\right].$
(91)

We now let $g(t)=\sqrt{ab(1-abt^{2})}$. This allows us to rewrite (84) and
(91) in a compact way:

$\displaystyle P$
$\displaystyle=\left[\begin{matrix}-\frac{1}{2bt}&\frac{-abt-
ig\left(t\right)}{2b}&\frac{-abt+ig\left(t\right)}{2b}\\\
-\frac{1}{2at}&\frac{-abt+ig\left(t\right)}{2a}&\frac{-abt-
ig\left(t\right)}{2a}\\\ 1&1&1\end{matrix}\right]$ (92)
$\displaystyle\log\left(D\right)\left(D-I\right)^{-1}P^{-1}(G+e^{tab_{G}}F)$
$\displaystyle=\frac{a^{2}b^{2}}{4ig^{3}(t)}\left[\begin{matrix}-8itg(t)\\\
-\log\left[1-4t\left(2tg^{2}\left(t\right)+i\left(2abt^{2}-1\right)g\left(t\right)\right)\right]\\\
\log\left[1-4t\left(2tg^{2}\left(t\right)-i\left(2abt^{2}-1\right)g\left(t\right)\right)\right]\end{matrix}\right].$
(93)

Using (92) and (93) to assemble the entire integrand of (81), we have

$\displaystyle P\log\left(D\right)\left(D-I\right)^{-1}P^{-1}(G+e^{tad_{G}}F)$
$\displaystyle=\frac{a^{2}b^{2}}{4ig^{3}(t)}\left[\begin{matrix}-\frac{1}{2bt}&\frac{-abt-
ig\left(t\right)}{2b}&\frac{-abt+ig\left(t\right)}{2b}\\\
-\frac{1}{2at}&\frac{-abt+ig\left(t\right)}{2a}&\frac{-abt-
ig\left(t\right)}{2a}\\\
1&1&1\end{matrix}\right]\left[\begin{matrix}-8itg(t)\\\
-\log\left[1-4t\left(2tg^{2}\left(t\right)+i\left(2abt^{2}-1\right)g\left(t\right)\right)\right]\\\
\log\left[1-4t\left(2tg^{2}\left(t\right)-i\left(2abt^{2}-1\right)g\left(t\right)\right)\right]\end{matrix}\right].$
(94)

Let $z\left(t\right):=abt+ig\left(t\right)$ and
$\nu\left(t\right):=\log\left[1-4t\left(2tg^{2}\left(t\right)-i\left(2abt^{2}-1\right)g\left(t\right)\right)\right]$.
Then, (94) implies that

$\displaystyle
P\log\left(D\right)\left(D-I\right)^{-1}P^{-1}(G+e^{tad_{G}}F)=\frac{a^{2}b^{2}}{2g^{3}\left(t\right)}\left[\begin{matrix}\frac{1}{2b}\Im\left\\{z\left(t\right)\overline{\nu\left(t\right)}\right\\}+\frac{2g(t)}{b}\\\
\frac{1}{2a}\Im\left\\{\overline{z\left(t\right)\nu\left(t\right)}\right\\}+\frac{2g(t)}{a}\\\
\Im\left\\{\nu\left(t\right)\right\\}-4tg(t)\end{matrix}\right],$ (95)

where $\Im\left(c\right)$ is the imaginary component of any $c\in\mathbb{C}$.

Noticing the recurrence of the expression in (95), we are motivated to expand
out $\nu(t)$ to

$\displaystyle\nu\left(t\right)$
$\displaystyle=\log\left[1-8abt^{2}(1-abt^{2})+4it(2abt^{2}-1)\sqrt{ab(1-abt^{2})}\right].$
(96)

By routine algebraic manipulations, we can show that the complex number within
the logarithm of (96) has modulus one:

$\displaystyle(1-8abt^{2}(1-abt^{2}))^{2}+(4t(2abt^{2}-1)\sqrt{ab(1-abt^{2})})^{2}=1.$
(97)

Thus, in light of (96) and (97), we can use Euler’s formula, double angle
formulae, and then trigonometry to rewrite $\nu(t)$ as the following:

$\displaystyle\nu\left(t\right)=4i\arcsin(\sqrt{abt^{2}}).$ (98)

Hence, by (95) and (98), we can simplify the integrand of (81) as follows:

$\displaystyle\widetilde{H}_{\eta}\left(p,q\right)$
$\displaystyle=\frac{a^{2}b^{2}}{\eta}\left[\begin{matrix}p^{2}&q^{2}&pq\end{matrix}\right]\int_{0}^{\eta}\frac{1}{2g^{3}\left(t\right)}\left[\begin{matrix}2at\arcsin(\sqrt{abt^{2}})+\frac{1}{b}2g(t)\\\
2bt\arcsin(\sqrt{abt^{2}})+\frac{1}{a}2g(t)\\\
-4\arcsin(\sqrt{abt^{2}})-4tg(t)\\\ \end{matrix}\right]dt.$ (99)

Finally, we split the integrand in (99) into the two terms shown in the vector
and integrate them separately with the help of Wolfram Mathematica [83]. Doing
so and adding together the results appropriately gives us the following:

$\displaystyle\widetilde{H}_{\eta}\left(p,q\right)$
$\displaystyle=\frac{a^{2}b^{2}}{\eta}\left[\begin{matrix}p^{2}&q^{2}&pq\end{matrix}\right]\left[\begin{matrix}\frac{\arcsin(\sqrt{ab\eta^{2}})}{4ab^{2}\sqrt{ab(1-ab\eta^{2})}}\\\
\frac{\arcsin(\sqrt{ab\eta^{2}})}{4a^{2}b\sqrt{ab(1-ab\eta^{2})}}\\\
-\frac{2ab\eta}{4}\frac{\arcsin(\sqrt{ab\eta^{2}})}{a^{2}b^{2}\sqrt{ab(1-ab\eta^{2})}}\\\
\end{matrix}\right]$
$\displaystyle=\frac{1}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}\arcsin(\sqrt{ab\eta^{2}})(ap^{2}+bq^{2}-2abpq\eta).$
(100)

When $ab<0$, we first show the following proper definition of inverse sine
functions. Suppose we are interested in finding $y=\arcsin{ix}$, which we can
write equivalently as follows:

$\displaystyle y=\arcsin{ix}=i\operatorname{arcsinh}(-x).$

Then the modified Hamiltonian takes the following form:

$\displaystyle\frac{\arcsin(\sqrt{ab\eta^{2}})}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}(ap^{2}+bq^{2}-2abpq\eta)=\frac{\operatorname{arcsinh}(\sqrt{-ab\eta^{2}})}{\sqrt{-ab\eta^{2}(1-ab\eta^{2})}}(ap^{2}+bq^{2}-2abpq\eta).$

####  C.2.3 Convergence radius

The argument above is only valid if the conditions of Theorem 6 are actually
satisfied. Thus, we need to ensure when
$\left|\left|e^{t\cdot\mathrm{ad}_{G}}e^{t\cdot\mathrm{ad}_{F}}-I\right|\right|_{\textrm{op}}=\sigma_{\max}\left(D-I\right)<1$,
or equivalently, when

$\displaystyle\left|-8abt^{2}(1-abt^{2})-4it(2abt^{2}-1)\sqrt{ab(1-abt^{2})}\right|<1,$
(101a)
$\displaystyle\left|-8abt^{2}(1-abt^{2})+4it(2abt^{2}-1)\sqrt{ab(1-abt^{2})}\right|<1$
(101b)

for all $t\in\left[0,\eta\right]$, as taken directly from from (86). Noting
the equivalence of forms (96) and (98) for $\nu(t)$, (101) can be rewritten as

$\displaystyle\left|e^{\overline{\nu(t)}}-1\right|=\left|e^{-4i\arcsin
z}-1\right|<1,$ $\displaystyle\left|e^{\nu(t)}-1\right|=\left|e^{4i\arcsin
z}-1\right|<1,$

respectively, where we have let $z:=\sqrt{abt^{2}}$. From here, we split the
work into two cases based on the sign of $ab$.

##### Case 1: $ab\geq 0$.

In this case, $\arcsin z$ is real, and since $\arcsin$ only takes arguments
from $-1$ to $1$, we get a contradiction when $z>1$. Since $e^{-4i\arcsin
z}=\overline{e^{4i\arcsin z}}$, $|e^{-4i\arcsin z}-1|<1$ if and only if
$|e^{4i\arcsin z}|<1$. When $0<4\arcsin{z}<\frac{\pi}{3}$, we have that
$|e^{4i\arcsin z}-1|<1$. Similarly, when $\frac{5\pi}{3}<4\arcsin{z}<2\pi$,
$|e^{4i\arcsin z}-1|<1$ again. Thus, the absolute convergence criteria in
Theorem 6 is satisfied whenever
$z\in\left(0,\sin(\frac{\pi}{12})\right)\cup\left(\sin(\frac{5\pi}{12}),1\right)$.
We now show that this actually implies that the modified Hamiltonian (17) is
absolutely convergent whenever $|z|\in(0,1)$.

For any $\eta\in(0,\sqrt{\frac{1}{ab}})$, we pick
$\eta^{\prime}=\max(\sqrt{\sin{\frac{11\pi}{24}}},\eta)$. By construction
$\eta^{\prime}\in\left(\sin(\frac{5\pi}{12}),1\right)$ and thus the series
(17) converges absolutely for $\eta^{\prime}$. We now compare the terms of the
two series:

$\left|\frac{\eta^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}-1}\left\\{G^{r_{1}}F^{s_{1}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}(p,q)}{n\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}}\right|\\\
\leq\left|\frac{(\eta^{\prime})^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}-1}\left\\{G^{r_{1}}F^{s_{1}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}(p,q)}{n\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}}\right|.$
(102)

By dominated convergence theorem, the modified Hamiltonian (17) converges
absolutely for $\eta$ chosen above.

##### Case 2: $ab<0$.

When $ab<0$, we let $F^{\prime}=|a|p^{2},G^{\prime}=|b|q^{2}$. From Appendix
C.1, we know that $\left\\{G^{r_{1}}F^{s_{1}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}(p,q)$ is always a monomial of $p,q$ and $a,b$,
regardless of the signs of $p$ and $q$. We have the following equality:

$\displaystyle\left|\frac{\eta^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}-1}\left\\{G^{r_{1}}F^{s_{1}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}(p,q)}{n\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}}\right|$
$\displaystyle\qquad\qquad=\frac{\eta^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}-1}\left|\left\\{G^{r_{1}}F^{s_{1}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}(p,q)\right|}{n\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}}$
$\displaystyle\qquad\qquad=\frac{\eta^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}-1}\left|\left\\{G^{\prime
r_{1}}F^{\prime s_{1}}\cdots G^{\prime r_{n}}F^{\prime
s_{n}}\right\\}(p,q)\right|}{n\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}}$
$\displaystyle\qquad\qquad=\left|\frac{\eta^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}-1}\left\\{G^{\prime
r_{1}}F^{\prime s_{1}}\cdots G^{\prime r_{n}}F^{\prime
s_{n}}\right\\}(p,q)}{n\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}}\right|.$

Since the modified Hamiltonian (17) converges absolutely whenever
$\sqrt{|a||b|t^{2}}<1$, it converges absolutely for $F^{\prime}$ and
$G^{\prime}$. Since the absolute value of the terms is the same, the
convergence is also absolute when $ab<0$, but except when
$\sqrt{|ab|t^{2}}<1$.

Figure 2 shows the graph of the function
$\eta\mapsto\frac{\arcsin\left(\sqrt{ab\eta^{2}}\right)}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}$
that appears in the final expression (100) of the modified Hamiltonian. Note
that the Taylor expansion of
$\eta\mapsto\frac{\arcsin\left(\sqrt{ab\eta^{2}}\right)}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}$
is as follows [83]:

$\displaystyle\frac{\arcsin\left(\sqrt{ab\eta^{2}}\right)}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}=1+\frac{2(\eta\sqrt{ab})^{2}}{3}+\frac{8(\eta\sqrt{ab})^{4}}{15}+\frac{16(\eta\sqrt{ab})^{6}}{35}+\frac{128(\eta\sqrt{ab})^{8}}{215}+\cdots,$

which is precisely what the simplified series formula (79) for
$\widetilde{H}_{\eta}$ suggests. One may further check that the same expansion
holds when $ab<0$.

Figure 2: The function
$\eta\mapsto\frac{\arcsin\left(\sqrt{ab\eta^{2}}\right)}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}$
with different choices of $a$ and $b$.

###  C.3 Verification: Hamiltonian flow of modified Hamiltonian

In this subsection, we verify Theorem 2 in the single-variable quadratic case.
To this end, we have the following lemma:

######  Lemma 10.

Define the modified Hamiltonian as

$\tilde{H}_{\eta}(p,q)=\frac{\arcsin(\sqrt{ab\eta^{2}})}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}(ap^{2}+bq^{2}-2abpq\eta).$

If $\left(p(t),q(t)\right)$ follows the Hamiltonian flow of the modified
Hamiltonian:

$\frac{d}{dt}\begin{pmatrix}p(t)\\\
q(t)\end{pmatrix}=\begin{pmatrix}-\frac{\partial}{\partial
q}\widetilde{H}_{\eta}(p(t),q(t))\\\ \frac{\partial}{\partial
p}\widetilde{H}_{\eta}(p(t),q(t))\end{pmatrix},$

and starts at

$\begin{pmatrix}p(0)\\\ q(0)\end{pmatrix}=\begin{pmatrix}p_{0}\\\
q_{0}\end{pmatrix},$

then we have:

$\begin{pmatrix}p(\eta)\\\ q(\eta)\end{pmatrix}=\begin{pmatrix}1&-2b\eta\\\
2a\eta&1-4ab\eta^{2}\end{pmatrix}\begin{pmatrix}p_{0}\\\ q_{0}\end{pmatrix},$

which coincides with the iterates of symplectic Euler (5) when
$H(p,q)=ap^{2}+bq^{2}$.

###### Proof.

For $\eta>0$, let

$C_{\eta}=\frac{\arcsin(\sqrt{ab\eta^{2}})}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}$

so that the modified Hamiltonian above becomes:

$\tilde{H}_{\eta}(p,q)=C_{\eta}(ap^{2}+bq^{2}-2abpq\eta).$

The Hamiltonian flow is:

$\displaystyle\frac{d}{dt}\begin{pmatrix}p(t)\\\
q(t)\end{pmatrix}=\begin{pmatrix}2C_{\eta}b(-q(t)+a\eta p(t))\\\
2C_{\eta}a(p(t)-b\eta q(t))\end{pmatrix}=2C_{\eta}\begin{pmatrix}ab\eta&-b\\\
a&-ab\eta\end{pmatrix}\begin{pmatrix}p(t)\\\ q(t)\end{pmatrix}.$ (103)

Using the matrix exponential, the solution to the Hamiltonian flow (103) at
time $t=\eta$ is:

$\displaystyle\begin{pmatrix}p(\eta)\\\ q(\eta)\end{pmatrix}=\exp\left(2\eta
C_{\eta}\begin{pmatrix}ab\eta&-b\\\
a&-ab\eta\end{pmatrix}\right)\begin{pmatrix}p_{0}\\\ q_{0}\end{pmatrix}.$
(104)

We compute the matrix exponential in (104) as follows:

$\displaystyle\exp\left(2\eta C_{\eta}\begin{pmatrix}ab\eta&-b\\\
a&-ab\eta\end{pmatrix}\right)$ $\displaystyle=\begin{bmatrix}E_{11}&E_{12}\\\
E_{21}&E_{22}\end{bmatrix},$

where the coefficients are as follows:

$\displaystyle E_{11}$
$\displaystyle=\frac{\left(ab\eta+\sqrt{ab(ab\eta^{2}-1)}\right)\exp\left(2\sqrt{ab(ab\eta^{2}-1)}\eta
C_{\eta}\right)}{2\sqrt{ab(ab\eta^{2}-1)}}$
$\displaystyle\quad-\frac{\left(ab\eta-\sqrt{ab(ab\eta^{2}-1)}\right)\exp\left(-2\sqrt{ab(ab\eta^{2}-1)}\eta
C_{\eta}\right)}{2\sqrt{ab(ab\eta^{2}-1)}}$ $\displaystyle E_{12}$
$\displaystyle=\frac{\sqrt{b}\exp\left(-2\sqrt{ab(ab\eta^{2}-1)}\eta
C_{\eta}\right)}{2\sqrt{a(ab\eta^{2}-1)}}-\frac{\sqrt{b}\exp\left(2\sqrt{ab(ab\eta^{2}-1)}\eta
C_{\eta}\right)}{2\sqrt{a(ab\eta^{2}-1)}}$ $\displaystyle E_{21}$
$\displaystyle=\frac{\sqrt{a}\exp\left(2\sqrt{ab(ab\eta^{2}-1)}\eta
C_{\eta}\right)}{2\sqrt{b(ab\eta^{2}-1)}}-\frac{\sqrt{a}\exp\left(-2\sqrt{ab(ab\eta^{2}-1)}\eta
C_{\eta}\right)}{2\sqrt{b(ab\eta^{2}-1)}}$ $\displaystyle E_{22}$
$\displaystyle=\frac{\left(-ab\eta+\sqrt{ab(ab\eta^{2}-1)}\right)\exp\left(2\sqrt{ab(ab\eta^{2}-1)}\eta
C_{\eta}\right)}{2\sqrt{ab(ab\eta^{2}-1)}}$
$\displaystyle\quad-\frac{\left(-ab\eta-\sqrt{ab(ab\eta^{2}-1)}\right)\exp\left(-2\sqrt{ab(ab\eta^{2}-1)}\eta
C_{\eta}\right)}{2\sqrt{ab(ab\eta^{2}-1)}}.$

To simplify the coefficients above, we let
$S=\sqrt{ab\eta^{2}(1-ab\eta^{2})}$, $T=ab\eta^{2}$. We further calculate:

$\displaystyle\exp\left(2\sqrt{ab(ab\eta^{2}-1)}\eta C_{\eta}\right)$
$\displaystyle=\exp\left(2\sqrt{ab(ab\eta^{2}-1)}\eta\frac{\arcsin(\sqrt{ab\eta^{2}})}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}\right)$
$\displaystyle=\exp\left(2i\arcsin\left(\sqrt{ab\eta^{2}}\right)\right)$
$\displaystyle=1-2T+2iS$

Similarly, we have:

$\displaystyle\exp\left(-2\sqrt{ab\left(ab\eta^{2}-1\right)}\eta
C_{\eta}\right)=1-2T-2iS.$

Hence, we could simplify the coefficients in the following way:

$\displaystyle\begin{pmatrix}E_{11}&E_{12}\\\
E_{21}&E_{22}\end{pmatrix}=\begin{pmatrix}1&-2b\eta\\\
2a\eta&1-4ab\eta^{2}\end{pmatrix}.$

The symplectic Euler algorithm, (5), in this case is:

$\displaystyle\begin{pmatrix}p_{\text{new}}\\\ q_{\text{new}}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}p_{0}-\eta(2bq_{0})\\\
q_{0}+\eta(2ap_{\text{new}})\end{pmatrix}=\begin{pmatrix}p_{0}-2b\eta q_{0}\\\
q_{0}+2a\eta\left(p_{0}-2b\eta
q_{0}\right)\end{pmatrix}=\begin{pmatrix}1&-2b\eta\\\
2a\eta&1-4ab\eta^{2}\end{pmatrix}\begin{pmatrix}p_{0}\\\ q_{0}\end{pmatrix}.$

Since the coefficients match, we confirm that the iterations are indeed
interpolated by the Hamiltonian flow generated by the modified Hamiltonian
$\widetilde{H}_{\eta}$. ∎

###  C.4 The multivariate case

In this subsection we derive the closed-form of the modified Hamiltonian in
the multivariate quadratic case by utilizing results from monovariate case.

The general idea of the proof is the following: we first derive connections
between the iterated Poisson brackets in multivariate and univariate cases. We
then reorder the series form (17) of the modified Hamiltonian in the
multivariate case $\widetilde{H}^{F,G}_{\eta}(p,q)$ and relate the reordered
series with the series form of the modified Hamiltonian in the univariate case
to get closed-form expression of $\widetilde{H}^{F,G}_{\eta}(p,q)$.
Furthermore, this allows us to determine the absolute convergence radius of
the reordered form, which then implies equivalence between the original form
and the reordered form.

####  C.4.1 Notations and settings

We assume that $p,q\in\mathbb{R}^{d}$, where $d\geq 2$. We assume that
$B,C\in\mathbb{R}^{d\times d}$ are symmetric matrices such that $BC$ is
positive semi-definite. Let $Q^{-1}\Lambda Q$ be the diagonalization of $BC$.
We define $F(p)=p^{\top}Bp$, $G(q)=q^{\top}Cq$, and $H^{F,G}(p,q)=F(p)+G(q)$.
In using results from last subsection, we assume that $u,v\in\mathbb{R}$, and
define $f(u)=au^{2},g(v)=bv^{2}$, where $ab>0$, and let
$H^{f,g}(u,v)=f(u)+g(v)$.

We want to find the closed-form expression of modified Hamiltonian for the
following symplectic Euler updates:

$\displaystyle p_{k+1}=p_{k}-2\eta Cq_{k},\quad q_{k+1}=q_{k}+2\eta Bp_{k+1}$

Throughout the subsection, we let $\star$ denote the tuple
$(r_{1},s_{1},\dots,r_{n},s_{n})$, i.e.,

$\star:=(r_{1},s_{1},\dots,r_{n},s_{n}).$

We denote the rank (for its definition, check the review on iterated Poisson
bracket in Section 6) of the Poisson bracket corresponding to the tuple
$\star$ with $\mathcal{N}(\star)$, denote
$\mathcal{M}(\star):=\sum_{i=1}^{n}(r_{i}+s_{i}),$ and denote the set of
tuples of rank $N$ with $\mathcal{S}(N)$. Finally, we define the following
function to express the constant coefficient in the $\widetilde{H}_{\eta}$
series:

$\displaystyle\tau(\star)=\tau((r_{1},s_{1},\cdots,r_{n},s_{n}))={\frac{\left(-1\right)^{n-1}}{n}{{\frac{1}{\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}}}}}$

For simplicity of notations, we define

$\displaystyle I_{p,q}^{F,G}$
$\displaystyle:=\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}(p,q)$ $\displaystyle I_{u,v}^{f,g}$
$\displaystyle:=\left\\{g^{r_{1}}f^{s_{1}}g^{r_{2}}f^{s_{2}}\cdots
g^{r_{n}}f^{s_{n}}\right\\}(u,v)$

Results from Appendix C.1 imply that $I_{u,v}^{f,g}(\star)$ must be a monomial
containing $a,b,u,v$. We denote the constant factor within
$I_{u,v}^{f,g}(\star)$ with $J(\star)$, denote the power of $a$ within
$I_{u,v}^{f,g}(\star)$ with $L_{a}(\star)$ and the power of $b$ within
$I_{u,v}^{f,g}(\star)$ with $L_{b}(\star)$. The following proposition is a
direct consequence of our work above:

######  Proposition 1.

$I_{u,v}^{f,g}(\star)$ must be one of the following forms:

  * •

$J(\star)a^{L_{a}(\star)}b^{L_{b}(\star)}u^{2}$,

  * •

$J(\star)a^{L_{a}(\star)}b^{L_{b}(\star)}v^{2}$, or

  * •

$J(\star)a^{L_{a}(\star)}b^{L_{b}(\star)}uv$.

We define the following subsets of possible tuples:

$\displaystyle\mathcal{S}_{pp}(n):=\mathcal{S}_{uu}(n)$
$\displaystyle:=\\{\star\in\mathcal{S}(n)\mid I_{u,v}^{f,g}(\star)\text{
contains }uu\\}$ $\displaystyle\mathcal{S}_{qq}(n):=\mathcal{S}_{uu}(n)$
$\displaystyle:=\\{\star\in\mathcal{S}(n)\mid I_{u,v}^{f,g}(\star)\text{
contains }uu\\}$ $\displaystyle\mathcal{S}_{pq}(n):=\mathcal{S}_{uv}(n)$
$\displaystyle:=\\{\star\in\mathcal{S}(n)\mid I_{u,v}^{f,g}(\star)\text{
contains }uv\\}$

Notice that
$\star\in\mathcal{S}_{uu}(n)\cup\mathcal{S}_{uu}(n)\cup\mathcal{S}_{uv}(n)$
implies that $I_{u,v}^{f,g}(\star)\neq 0$.

Finally, we define the following functions to connect multivariate and
univariate quadratic cases.

$\displaystyle
T^{\text{pure}}_{\eta}(\lambda):=\begin{cases}\frac{\arcsin(\sqrt{\lambda\eta^{2}})}{\sqrt{\lambda\eta^{2}(1-\lambda\eta^{2})}}&\lambda>0,\\\
\frac{\operatorname{arcsinh}(\sqrt{-\lambda\eta^{2}})}{\sqrt{-\lambda\eta^{2}(1-\lambda\eta^{2})}}&\lambda<0.\end{cases},\quad
T^{\text{cross}}_{\eta}(\lambda):=\begin{cases}\frac{-2\sqrt{\lambda}\arcsin(\sqrt{\lambda\eta^{2}})}{\sqrt{(1-\lambda\eta^{2})}}&\lambda>0,\\\
\frac{2\sqrt{-\lambda}\operatorname{arcsinh}(\sqrt{-\lambda\eta^{2}})}{\sqrt{(1-\lambda\eta^{2})}}&\lambda<0.\end{cases}$

####  C.4.2 Connections between the iterated Poisson brackets in two cases

######  Lemma 11.

If $\star\in\mathcal{S}_{uu}(n)$, then
$L_{a}(\star)=L_{b}(\star)+1=\frac{\mathcal{M}(\star)+1}{2}$. If
$\star\in\mathcal{S}_{uu}(n)$, then
$L_{b}(\star)=L_{a}(\star)+1=\frac{\mathcal{M}(\star)+1}{2}$. If
$\star\in\mathcal{S}_{uv}(n)$, then
$L_{a}(\star)=L_{b}(\star)=\frac{\mathcal{M}(\star)}{2}$.

###### Proof.

This can be verified via an inductive argument based on
$\mathcal{M}(\star)=\\{1,2,3,\cdots\\}$. As the base cases, the proposed lemma
holds for $M=1$, where $f=au^{2}$, $g=bv^{2}$ and $M=2$, where
$\\{f,g\\}=-4abuv=-\\{g,f\\}$. Suppose that the lemma holds for any tuples
$\star$ such that $\mathcal{M}(\star)\leq n$, we verify that it also holds for
any tuples $\star$ such that $\mathcal{M}(\star)\leq n+1$. For the sake of
simplicity we only verify the case where $\\{\cdot,f\\}$ is applied to
$I_{u,v}^{f,g}(\star)$. We denote the new tuple in this case as
$\star^{\prime}$ and remark that
$\mathcal{M}(\star^{\prime})=\mathcal{M}(\star)+1$.

  * •

If
$I_{u,v}^{f,g}(\star)=J(\star)a^{\frac{\mathcal{M}(\star)+1}{2}}b^{\frac{\mathcal{M}(\star)-1}{2}}u^{2}$,
then

$\displaystyle\\{I_{u,v}^{f,g}(\star),f\\}$
$\displaystyle=\\{J(\star)a^{\frac{\mathcal{M}(\star)+1}{2}}b^{\frac{\mathcal{M}(\star)-1}{2}}u^{2},au^{2}\\}$
$\displaystyle=0.$

  * •

If
$I_{u,v}^{f,g}(\star)=J(\star)a^{\frac{\mathcal{M}(\star)-1}{2}}b^{\frac{\mathcal{M}(\star)+1}{2}}v^{2}$,
then

$\displaystyle\\{I_{u,v}^{f,g}(\star),f\\}$
$\displaystyle=\\{J(\star)a^{\frac{\mathcal{M}(\star)-1}{2}}b^{\frac{\mathcal{M}(\star)+1}{2}}v^{2},au^{2}\\}$
$\displaystyle=J(\star^{\prime})a^{\frac{\mathcal{M}(\star)+1}{2}}b^{\frac{\mathcal{M}(\star)+1}{2}}uv$
$\displaystyle=J(\star^{\prime})a^{\frac{\mathcal{M}(\star^{\prime})}{2}}b^{\frac{\mathcal{M}(\star^{\prime})}{2}}uv.$

  * •

If
$I_{u,v}^{f,g}(\star)=J(\star)a^{\frac{\mathcal{M}(\star)}{2}}b^{\frac{\mathcal{M}(\star)}{2}}uv$,
then

$\displaystyle\\{I_{u,v}^{f,g}(\star),f\\}$
$\displaystyle=\\{J(\star)a^{\frac{\mathcal{M}(\star)}{2}}b^{\frac{\mathcal{M}(\star)}{2}}uv,au^{2}\\}$
$\displaystyle=J(\star^{\prime})a^{\frac{\mathcal{M}(\star)+2}{2}}b^{\frac{\mathcal{M}(\star)}{2}}u^{2}$
$\displaystyle=J(\star^{\prime})a^{\frac{\mathcal{M}(\star^{\prime})+1}{2}}b^{\frac{\mathcal{M}(\star^{\prime})}{2}}u^{2}.$

Thus, the induction argument is verified in all cases. ∎

######  Lemma 12.

If $\star\in\mathcal{S}_{uu}(n)$, then
$I_{p,q}^{F,G}(\star)=J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star)-1}{2}}Bp$.
If $\star\in\mathcal{S}_{uu}(n)$, then
$I_{p,q}^{F,G}(\star)=J(\star)q^{\top}C(BC)^{\frac{\mathcal{M}(\star)-1}{2}}q$.
If $\star\in\mathcal{S}_{uv}(n)$, then
$I_{p,q}^{F,G}(\star)=J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star)}{2}}q$.

###### Proof.

This can be verified using an inductive argument based on
$\mathcal{M}(\star)=\\{1,2,3,\cdots\\}$. As the base case, the proposed lemma
holds for $M=1$, where $F=p^{\top}Bp$, $G=q^{\top}Cq$ and $M=2$, where
$\\{F,G\\}=-4p^{\top}BCq=-\\{G,F\\}$. Suppose the lemma holds for any tuples
$\star$ such that $\mathcal{M}(\star)\leq n$, we verify that it also holds for
any tuples $\star$ such that $\mathcal{M}(\star)\leq n+1$. For the sake of
simplicity we only verify the case where $\\{\cdot,F\\}$ is applied to
$I_{p,q}^{F,G}(\star)$ and when $\\{\cdot,f\\}$ is applied to
$I_{u,v}^{f,g}(\star)$. To show the inductive argument, we need to compare
$\\{I_{p,q}^{F,G}(\star),F\\}$ and $\\{I_{u,v}^{f,g}(\star),f\\}$.

If $\star$ ends with $\\{\dots,r_{n},0\\}$ where $r_{n}>0$, the new tuple
$\star^{\prime}$ will be $\\{\dots,r_{n},1\\}$. On the other hand, if $\star$
ends with $\\{\dots,r_{n},s_{n}\\}$ where $r_{n},s_{n}>0$, the new tuple
$\star^{\prime}$ will be $\\{\dots,r_{n},s_{n}+1\\}$. In both cases,
$I_{p,q}^{F,G}(\star^{\prime})={I_{p,q}^{F,G}(\star),F}$. Note that
$\mathcal{M}(\star^{\prime})=\mathcal{M}(\star)+1$.

When $\star\in\mathcal{S}_{uu}(n)$,
$I_{p,q}^{F,G}=J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star)-1}{2}}Bp$ and
$I_{u,v}^{f,g}=J(\star)a^{\frac{\mathcal{M}(\star)+1}{2}}b^{\frac{\mathcal{M}(\star)-1}{2}}u^{2}$.
Thus,

$\displaystyle\\{I_{p,q}^{F,G}(\star),F\\}=\\{I_{u,v}^{f,g}(\star),f\\}=0.$

When $\star\in\mathcal{S}_{vv}(n)$,
$I_{p,q}^{F,G}=J(\star)q^{\top}C(BC)^{\frac{\mathcal{M}(\star)-1}{2}}q$ and
$I_{u,v}^{f,g}=J(\star)a^{\frac{\mathcal{M}(\star)-1}{2}}b^{\frac{\mathcal{M}(\star)+1}{2}}v^{2}$.
In this case,

$\displaystyle\\{I_{u,v}^{f,g}(\star),f\\}$
$\displaystyle=\\{J(\star)a^{\frac{\mathcal{M}(\star)-1}{2}}b^{\frac{\mathcal{M}(\star)+1}{2}}v^{2},au^{2}\\}$
$\displaystyle=4J(\star)a^{\frac{\mathcal{M}(\star)+1}{2}}b^{\frac{\mathcal{M}(\star)+1}{2}}uv$
$\displaystyle=4J(\star)a^{\frac{\mathcal{M}(\star^{\prime})}{2}}b^{\frac{\mathcal{M}(\star^{\prime})}{2}}uv.$

On the other hand,

$\displaystyle\\{I_{p,q}^{F,G}(\star),F\\}$
$\displaystyle=\\{J(\star)q^{\top}C(BC)^{\frac{\mathcal{M}(\star)-1}{2}}q,p^{\top}Bp\\}$
$\displaystyle=4J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star)+1}{2}}q$
$\displaystyle=4J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star^{\prime})}{2}}q.$

Since the constant term match and the power of $a$, $b$ matches with that of
$B$ and $C$, the assumption is verified.

When $\star\in\mathcal{S}_{uv}(n)$,
$I_{p,q}^{F,G}=J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star)}{2}}q$ and
$I_{u,v}^{f,g}=J(\star)a^{\frac{\mathcal{M}(\star)}{2}}b^{\frac{\mathcal{M}(\star)}{2}}uv$.
In this case,

$\displaystyle\\{I_{u,v}^{f,g}(\star),f\\}$
$\displaystyle=\\{J(\star)a^{\frac{\mathcal{M}(\star)}{2}}b^{\frac{\mathcal{M}(\star)}{2}}uv,au^{2}\\}$
$\displaystyle=2J(\star)a^{\frac{\mathcal{M}(\star)+2}{2}}b^{\frac{\mathcal{M}(\star)}{2}}u^{2}$
$\displaystyle=2J(\star)a^{\frac{\mathcal{M}(\star^{\prime})+1}{2}}b^{\frac{\mathcal{M}(\star^{\prime})-1}{2}}u^{2}.$

On the other hand,

$\displaystyle\\{I_{p,q}^{F,G}(\star),F\\}$
$\displaystyle=\\{J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star)}{2}}q,p^{\top}Bp\\}$
$\displaystyle=2J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star)}{2}}Bp$
$\displaystyle=2J(\star)p^{\top}(BC)^{\frac{\mathcal{M}(\star^{\prime})-1}{2}}Bp$

Since the constant term match and the power of $a$, $b$ matches with that of
$B$ and $C$, the assumption is verified. ∎

######  Corollary 3.

If $\star\in\mathcal{S}_{uu}(n)$, then
$I_{p,q}^{F,G}(\star)=J(\star)(Qp)^{\top}\Lambda^{\frac{\mathcal{M}(\star)-1}{2}}QBp$.
If $\star\in\mathcal{S}_{vv}(n)$, then
$I_{p,q}^{F,G}(\star)=J(\star)(QC^{\top}q)^{\top}\Lambda^{\frac{\mathcal{M}(\star)-1}{2}}Qq$.
If $\star\in\mathcal{S}_{uv}(n)$, then
$I_{p,q}^{F,G}(\star)=J(\star)(Qp)^{\top}\Lambda^{\frac{\mathcal{M}(\star)}{2}}Qq$
.

###### Proof.

We note that $(BC)^{\frac{\mathcal{M}(\star)-1}{2}}B=(Q^{-1}\Lambda
Q)^{\frac{\mathcal{M}(\star)-1}{2}}B=Q^{-1}\Lambda^{\frac{\mathcal{M}(\star)-1}{2}}QB$,
$C(BC)^{\frac{\mathcal{M}(\star)-1}{2}}=\\\ C(Q^{-1}\Lambda
Q)^{\frac{\mathcal{M}(\star)-1}{2}}=CQ^{-1}\Lambda^{\frac{\mathcal{M}(\star)-1}{2}}Q$,
and $(BC)^{\frac{\mathcal{M}(\star)}{2}}=(Q^{-1}\Lambda
Q)^{\frac{M(\star)}{2}}=Q^{-1}\Lambda^{\frac{M(\star)}{2}}Q$. ∎

This confirms that if $\star\in\mathcal{S}_{pp}$, then $I_{u,v}^{f,g}(\star)$
is a bilinear product of a matrix with $p,p$. Similar properties hold for
$\mathcal{S}_{qq}$ and $\mathcal{S}_{pq}$.

####  C.4.3 Reordering the series: finding the closed-form and absolute
convergence radius

######  Lemma 13.

The Dynkin form (17) of the modified Hamiltonian in multivariate quadratic
case $\widetilde{H}_{\eta}^{F,G}(p,q)$ can be rearranged into the following:

$\displaystyle\hat{H}_{\eta}^{F,G}(p,q)$
$\displaystyle=(Qp)^{\top}\left(\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{pp}(N)}\eta^{\mathcal{M}(\star)-1}\tau(\star)J(\star)\Lambda^{\frac{\mathcal{M}(\star)-1}{2}}\right)QBp$
$\displaystyle\quad+(QC^{\top}q)^{\top}\left(\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{qq}(N)}\eta^{\mathcal{M}(\star)-1}\tau(\star)J(\star)\Lambda^{\frac{\mathcal{M}(\star)-1}{2}}\right)Qq$
$\displaystyle\quad+(Qp)^{\top}\left(\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{pq}(N)}\eta^{\mathcal{M}(\star)-1}\tau(\star)J(\star)\Lambda^{\frac{\mathcal{M}(\star)}{2}}\right)Qq$
(108)

Note that the equivalence between $\widetilde{H}_{\eta}^{F,G}(p,q)$ and
$\hat{H}_{\eta}^{F,G}(p,q)$ is not guaranteed, because we have not proven the
absolute convergence yet.

###### Proof.

Let us start from the Dynkin form (17) of the modified Hamiltonian:

$\widetilde{H}^{F,G}_{\eta}\left(p,q\right)=\sum_{n=1}^{\infty}{\frac{\left(-1\right)^{n-1}}{n}{\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\vdots\\\
r_{n}+s_{n}>0\end{subarray}}{\frac{\eta^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}-1}\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}(p,q)}{\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}}}}}.$

The summation can be reordered as:

$\hat{H}^{F,G}_{\eta}\left(p,q\right)=\sum_{n=1}^{\infty}\sum_{\star\in\mathcal{S}_{pp}(n)\cup\mathcal{S}_{pq}(n)\cup\mathcal{S}_{pq}(n)}\tau(\star)\eta^{\mathcal{M}(\star)}I_{p,q}^{F,G}(\star).$

We compute the summation for $\mathcal{S}_{pp}(n),\mathcal{S}_{qq}(n)$ and
$\mathcal{S}_{pq}(n)$ separately, and apply the results in Lemma 12 to
conclude. ∎

######  Theorem 7.

The reordered $\widetilde{H}_{\eta}$ series in Lemma 13 has the following
closed form:

$\displaystyle\hat{H}_{\eta}^{F,G}(p,q)$
$\displaystyle=(Qp)^{\top}T^{\text{pure}}_{\eta}(\Lambda)QBp$
$\displaystyle\quad+(QC^{\top}q)^{\top}T^{\text{pure}}_{\eta}(\Lambda)Qq$
$\displaystyle\quad+(Qp)^{\top}T^{\text{cross}}_{\eta}(\Lambda)Qq.$

Whenever each eigenvalue $\lambda$ in $\Lambda$ satisfies that
$|\lambda\eta^{2}|\leq 1$, the original BCH series
$\widetilde{H}_{\eta}^{F,G}(p,q)$ converges absolutely and reordering of
summation is allowed. Since $BC$ is assumed to be diagonalizable, if the
maximal singular value of $BC$ is smaller than $\frac{1}{\eta^{2}}$, then
$\widetilde{H}_{\eta}^{F,G}(p,q)=\hat{H}_{\eta}^{F,G}(p,q)$.

###### Proof.

Theorem 3 says that in univariate case, BCH series
$\widetilde{H}_{\eta}^{f,g}(u,v)$ converges to
$\frac{1}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}\arcsin(\sqrt{ab\eta^{2}})(au^{2}+bv^{2}-2abuv\eta)$.
Furthermore, the convergence is absolute whenever $|ab\eta^{2}|<1$.

Assume without loss of generality that $ab>0$, for the other case is similar.
Absolute convergence allows for the following reordering of the summation:

$\displaystyle\widetilde{H}_{\eta}^{f,g}(u,v)$
$\displaystyle=\left(\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{uu}(N)}\eta^{\mathcal{M}(\star)-1}\tau(\star)J(\star)(ab)^{\frac{\mathcal{M}(\star)-1}{2}}\right)au^{2}$
$\displaystyle\quad+\left(\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{vv}(N)}\eta^{\mathcal{M}(\star)-1}\tau(\star)J(\star)(ab)^{\frac{\mathcal{M}(\star)-1}{2}}\right)bv^{2}$
$\displaystyle\quad+\left(\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{uv}(N)}\eta^{\mathcal{M}(\star)-1}\tau(\star)J(\star)(ab)^{\frac{\mathcal{M}(\star)}{2}}\right)uv$
$\displaystyle=\frac{1}{\sqrt{ab\eta^{2}(1-ab\eta^{2})}}\arcsin(\sqrt{ab\eta^{2}})(au^{2}+bu^{2}-2abuv\eta).$
(109)

We now substitute the variable $ab$ with $\lambda$. Equation (109) above
implies the following:

$\displaystyle\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{uu}(N)}\eta^{\mathcal{M}(\star)-1}\tau(\star)J(\star)\lambda^{\frac{\mathcal{M}(\star)-1}{2}}$
$\displaystyle=\frac{\arcsin(\sqrt{\lambda\eta^{2}})}{\sqrt{\lambda\eta^{2}(1-\lambda\eta^{2})}}=T^{\text{pure}}_{\eta}(\lambda)$
(110a)
$\displaystyle\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{vv}(N)}\eta^{\mathcal{M}(\star)-1}\tau(\star)J(\star)\lambda^{\frac{\mathcal{M}(\star)-1}{2}}$
$\displaystyle=\frac{\arcsin(\sqrt{\lambda\eta^{2}})}{\sqrt{\lambda\eta^{2}(1-\lambda\eta^{2})}}=T^{\text{pure}}_{\eta}(\lambda)$
(110b)
$\displaystyle\sum_{N=1}^{\infty}\frac{(-1)^{N-1}}{N}\sum_{\star\in\mathcal{S}_{uv}(N)}\frac{\eta^{\mathcal{M}(\star)-1}}{\mathcal{M}(\star)}\tau(\star)J(\star)\lambda^{\frac{\mathcal{M}(\star)}{2}}$
$\displaystyle=\frac{-2\sqrt{\lambda}\arcsin(\sqrt{\lambda\eta^{2}})}{\sqrt{(1-\lambda\eta^{2})}}=T^{\text{cross}}_{\eta}(\lambda).$
(110c)

Furthermore, the absolute convergence of original Dynkin series (17) when
$|\lambda\eta^{2}|<1$ implies absolute convergence of
$T^{\text{pure}}_{\eta}(\lambda)$ and $T^{\text{cross}}_{\eta}(\lambda)$ when
$|\lambda\eta^{2}|<1$.

Since $\Lambda$ is a diagonal matrix, taking powers of $\Lambda$ is equivalent
with taking the same powers element-wise. Thus, the summation of matrices in
(108) is equivalent with summing each element along the diagonal separately.
Furthermore, since the terms in (108) are precisely those in (109) and thereby
equal to the respective ones in (110), the closed form should equal to the
application of $T_{\eta}^{\text{pure}}$ and $T_{\eta}^{\text{cross}}$ on the
diagonal elements separately. Finally, note that since
$T^{\text{cross}}_{\eta}(\lambda)=|\lambda|T^{\text{pure}}_{\eta}(\lambda)$
holds for both $\lambda>0$ and $\lambda<0$, we could simplify
$T_{\eta}^{\text{cross}}(\Lambda)$ as
$|\Lambda|T_{\eta}^{\text{pure}}(\Lambda)$. Thus, the closed-form is proven.

Finally, following the convergence criteria in Theorem 3, the convergence
criteria here should be:

$\lambda_{\max}(|\Lambda|)\eta^{2}=\sigma_{\max}(BC)<1.$

Since $\sigma_{\max}(BC)\leq\sigma_{\max}(B)\sigma_{\max}(C)$, one sufficient
convergence criterion is:

$\sigma_{\max}(B)\sigma_{\max}(C)\eta^{2}<1.$

∎

##  Appendix D Absolute convergence of the modified Hamiltonian

Much of the existing literature makes references to the convergence of the
modified Hamiltonian [38, 4, 31], as from a theoretical perspective,
convergence of the Dynkin series (17) is necessary to define a modified
Hamiltonian $\widetilde{H}_{\eta}$ in terms of the Dynkin series which is a
well-defined function from $\mathcal{P}\times\mathcal{Q}$ to $\mathbb{R}$. In
general, it is well-known that if the symplectic Euler method (5) is
nonlinear, the Dynkin formula does not converge for any $\eta$ [68, 31, 38].
However, numerical simulations, e.g., those shown in [38] and [85], suggest
that for even non-smooth, closed, proper convex functions $F$ and $G$, the
orbits generated by the iterations (5) lie on a closed, bounded orbit for
$\eta$ sufficiently small. This at least suggests that the results shown above
on the existence of closed, bounded orbits for small enough $\eta>0$ in the
quadratic and logarithm cases also exist for larger classes of functions.

That being said, the existence of closed, bounded orbits which look like the
level set of some function does not necessarily imply that the BCH formula as
it stands in either its Dynkin or integral form (i.e., without renormalization
or analytic continuation) converges to a function whose level sets match with
those orbits. In fact, there are direct counter-examples to the latter when
$F$ is quadratic and $G$ is a polynomial with order higher than 3, and $d=1$.
Numerical simulations—e.g., those in Figure 3—imply the existence of bounded,
closed orbits for small enough $\eta$ and any initial condition throughout
$\mathbb{R}\times\mathbb{R}$, even though [69, 31] show that the BCH formula
does not converge in such cases for any $\eta>0$ when $G$ is a polynomial of
degree $\geq 3$. Note that the smaller circles in the first plot of Figure 3
of the same color are suborbital trajectories traced out on smaller portions
of the phase space than the larger orbit for a strict subset of the
iterations. For instance, for the trajectory in blue corresponding to
$\left(p_{0},q_{0}\right)=\left(2,3\right)$, there are $26$ smaller circles.
Thus, if we consider the dynamics of every 26th iterate of symplectic Euler
(5), then these are still periodic and lie on one of the smaller circles.

(a) $H(x,y)=x^{1.5}+y^{1.5},\eta=1$

(b) $H(x,y)=x^{2}+y^{2},\eta=1$

(c) $H(x,y)=x^{4}+y^{4},\eta=1$

Figure 3: Trajectories of iterations symplectic Euler discretizations of the
Hamiltonian flows plotted in scatter plot. Different colors correspond to
different initial positions.
(a) $H(x,y)=x^{1.5}+y^{1.5},\eta=1$ (b) $H(x,y)=x^{2}+y^{2},\eta=1$ (c)
$H(x,y)=x^{4}+y^{4},\eta=1$

At least to the authors’ knowledge, no prior study other than [69, 31] has
tried to study which general classes of functions $F$ and $G$ will or will not
lead to a convergent modified Hamiltonian, expressed with the Dynkin formula.
But more generally, in the field of Lie algebra, there are numerous studies
which state and prove sufficient criteria for the Dynkin formula to converge
for two elements $x,y$ in a normed Lie algebra
$\left(L,\left|\left|\cdot\right|\right|\right)$ [8, 7, 70, 21]. These studies
require a coercivity bound of the form
$\left|\left|\left[x,y\right]\right|\right|\leq\mu\left|\left|x\right|\right|\left|\left|y\right|\right|$
for some $\mu\geq 0$. To apply these results directly to the Poisson algebra
over $C^{\infty}\left(\mathcal{P}\times\mathcal{Q}\right)$, we would require
at least something of the form
$\left|\left|\left\\{F,G\right\\}\right|\right|=\left|\left|\nabla_{q}F\cdot\nabla_{p}G-\nabla_{p}F\cdot\nabla_{q}G\right|\right|\leq\mu\left|\left|F\right|\right|\left|\left|G\right|\right|$
for some suitable function norm $\left|\left|\cdot\right|\right|$ and $\mu\geq
0$, e.g., some $L^{p}\left(\mathcal{P}\times\mathcal{Q}\right)$ norm or Hölder
space norm [29, 45, 43]. However, this effectively amounts to enforcing a
reverse Poincaré inequality over the function space in question, or
equivalently, that the derivative map is bounded over the function space in
question, which we would only generally expect to be true over a finite-
dimensional function space [45, 65]. There are a few known examples of this,
e.g., the quadratic case discussed in Section 8 and Appendix C. In such cases,
it would be most straightforward to find a basis for the finite-dimensional
function space $\mathcal{F}$ such that the Poisson bracket can be expressed as
a matrix acting on this basis.

To avoid the aforementioned issues in imposing that the Poisson bracket is
bounded (or similarly, that the derivative operator is bounded) over the
underlying function space, we derive criteria for convergence that rely on
imposing growth conditions on the IPBs of $F$ and $G$ instead of directly
bounding the Poisson bracket or derivative operator. Furthermore, we assume
absolute convergence, since doing so both allows us to reorder the terms in
the Dynkin series (17) as needed [64] and allows us to bound the absolute
value of the IPBs (6) in each term, as shown in the start of Appendix D.1.

We consider the following two growth assumptions on the IPBs in the Dynkin
series (17):

  1. 1.

$\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\right|$ is bounded above by a function of
$k:=r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}$ only, and

  2. 2.

$\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\right|$ is bounded above by the product of a
function of $k$ and $\prod_{i=1}^{n}{r_{i}!s_{i}!}$.

In either case, the bound may also depend on $\left(p,q\right)$, in which case
$\widetilde{H}_{\eta}$ is defined as a pointwise limit function at each
$\left(p,q\right)$ separately but is not necessarily continuous. However, if
the bound holds uniformly for all
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$, then by the Weierstrass
M-test [64], $\widetilde{H}_{\eta}$ is also continuous on all of
$\mathcal{P}\times\mathcal{Q}$.

##### Case 1

In this case, we assume a uniform bound on the IPBs which is a function of the
“order” of the IPB—i.e., we assume that

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\right|\leq
A\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)$ (111)

for all $n\in\mathbb{Z}_{+}$ and pairs $r_{i}+s_{i}>0$ for $i=1,\dots,n$,
where $A:\mathbb{Z}_{+}\rightarrow\mathbb{R}_{\geq 0}$. Doing so in general
gives Lemma 15, from which we can derive two corollaries upon further
assumptions for the function $A$ which use explicitly verifiable bounds on the
size of $\eta$.

######  Theorem 8.

Suppose that there exist mappings
$b,c:\mathcal{P}\times\mathcal{Q}\rightarrow\mathbb{R}_{\geq 0}$ such that

$\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\left(p,q\right)\right|\leq\\\
b\left(p,q\right)\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)^{2}c\left(p,q\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}-1}$
(112)

for all $n\in\mathbb{Z}_{+}$,
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$, and integer pairs
$r_{i}+s_{i}>0$ for $i=1,\dots,n$. Then, for all
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$ such that $\eta<\log
2/\left(2c\left(p,q\right)\right)$,

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq\frac{2b\left(p,q\right)e^{2\eta
c\left(p,q\right)}}{2-e^{2\eta c\left(p,q\right)}}.$ (113)

###### Proof.

Replace $A$ with (112) in Lemma 15. Then, we note that the exponential
generating function of the Fubini numbers is [62]

$\sum_{n=0}^{\infty}{a_{n}\frac{x^{n}}{n!}}=\frac{1}{2-e^{x}},$

which converges only when $x<\log 2$. Simplifying the result, we get (113). ∎

######  Theorem 9.

Suppose that there exist mappings
$b,c:\mathcal{P}\times\mathcal{Q}\rightarrow\mathbb{R}_{\geq 0}$ and some
$t\in\mathbb{R}$ such that

$\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\left(p,q\right)\right|\leq\\\
b\left(p,q\right)\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)^{2-t}c\left(p,q\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}}$
(114)

for all $n\in\mathbb{Z}_{+}$,
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$, and integer pairs
$r_{i}+s_{i}>0$ for $i=1,\dots,n$. Then, for all
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$ such that $\eta<\log
2/\left(2c\left(p,q\right)\right)$, $\widetilde{H}_{\eta}\left(p,q\right)$ is
well-defined and bounded, and moreover, $\widetilde{H}_{\eta}\left(p,q\right)$
is continuous on every open subset of $\mathcal{P}\times\mathcal{Q}$ where the
upper bound (114) holds uniformly for all $\left(p,q\right)$ in that open set.
Moreover, the following bound holds for all
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$:

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq\frac{2b\left(p,q\right)}{\eta}\left(\textrm{Li}_{t}\left(\frac{2\eta
c\left(p,q\right)}{\log 2}\right)-\eta\right),$

where $\textrm{Li}_{t}$ is the polylogarithm of order $t$ [81, 5].

###### Proof.

We replace $A$ with (114) in Lemma 15, thereby giving us

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq
b\left(p,q\right)\left(4\sum_{k=1}^{\infty}{\frac{\eta^{k-1}2^{k-1}c\left(p,q\right)^{k}a_{k-1}}{k^{\top}\left(k-1\right)!}}-2\right).$
(115)

We have the following upper bound for the Fubini numbers from [89]:

$\displaystyle a_{k-1}<\frac{\left(k-1\right)!}{\left(\log 2\right)^{k}},$
(116)

and using (116) to bound (115) from above,

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq
2b\left(p,q\right)\left(\frac{1}{\eta}\sum_{k=1}^{\infty}{\frac{1}{k^{\top}}\left(\frac{2\eta
c\left(p,q\right)}{\log 2}\right)^{k}}-1\right).$ (117)

By the definition of polylogarithm [81, 5], the result follows. ∎

##### Case 2

We now assume that

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\right|\leq
B\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}$
(118)

for all $n\in\mathbb{Z}_{+}$ and pairs $r_{i}+s_{i}>0$ for $i=1,\dots,n$,
where $B:\mathbb{Z}_{+}\rightarrow\mathbb{R}_{\geq 0}$. Doing so gives the
more general convergence criteria stated as Lemma 16, and we can also derive a
corollary from Lemma 16 upon further assumptions on the function $B$.

######  Theorem 10.

Suppose that there exist mappings
$b,c:\mathcal{P}\times\mathcal{Q}\rightarrow\mathbb{R}_{\geq 0}$ and some
$t\in\mathbb{R}$ such that

$\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\left(p,q\right)\right|\leq\\\
b\left(p,q\right)\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)^{1-t}c\left(p,q\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}}\prod_{i=1}^{n}{r_{i}!s_{i}!}$
(119)

for all $n\in\mathbb{Z}_{+}$,
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$, and integer pairs
$r_{i}+s_{i}>0$ for $i=1,\dots,n$. Then, for all
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$ such that
$\eta<\left(2-\sqrt{3}\right)/c\left(p,q\right)$,
$\widetilde{H}_{\eta}\left(p,q\right)$ is well-defined and bounded, and
moreover, $\widetilde{H}_{\eta}$ is continuous on every open subset of
$\mathcal{P}\times\mathcal{Q}$ where the upper bound (119) holds uniformly for
all $\left(p,q\right)$ in that open set. Moreover, the following bound holds
for all $\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$:

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq\frac{b\left(p,q\right)}{\eta}\textrm{Li}_{t}\left(\eta\left(2+\sqrt{3}\right)c\left(p,q\right)\right),$

where $\textrm{Li}_{t}$ is the polylogarithm of order $t$.

###  D.1 Proof of the two convergence criteria

For each case, we re-sum the modified Hamiltonian (17) such that the outermost
summation sums over consecutive orders of $\eta$. Doing so also gives each
order-by-order correction in $\eta$ to the modified Hamiltonian and a way to
truncate the modified Hamiltonian at a given order.

Using the notations introduced in Section 6 and the double summation identity
$\sum_{n=m}^{N}{\sum_{k=n}^{N}{a_{nk}}}=\sum_{k=m}^{N}{\sum_{n=m}^{k}{a_{nk}}}$,
for any $m,N\in\mathbb{Z}$ with $m\leq N$ [33], we have

$\displaystyle\widetilde{H}_{\eta}^{\left(N-1\right)}\left(p,q\right)$
$\displaystyle=\sum_{n=1}^{\infty}{\frac{\left(-1\right)^{n-1}}{n}\sum_{k=n}^{N}{\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\vdots\\\ r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{\eta^{k-1}\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}}{k\prod_{i=1}^{n}{r_{i}!s_{i}!}}}}}$
$\displaystyle=\sum_{k=1}^{N}{\frac{\eta^{k-1}}{k}\sum_{n=1}^{k}{\frac{\left(-1\right)^{n-1}}{n}\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\vdots\\\ r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}}{\prod_{i=1}^{n}{r_{i}!s_{i}!}}}}}.$ (120)

Hence, taking $N\rightarrow\infty$ in (120) and assuming absolute convergence,
(17) is equal to

$\displaystyle\widetilde{H}_{\eta}\left(p,q\right)=\sum_{k=1}^{\infty}{\frac{\eta^{k-1}}{k}\sum_{n=1}^{k}{\frac{\left(-1\right)^{n-1}}{n}\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\vdots\\\ r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}}{\prod_{i=1}^{n}{r_{i}!s_{i}!}}}}}.$ (121)

Since the steps above are reversible, in practice, once establishing absolute
convergence of (121), we can reorder the terms back to establish absolute
convergence for the original modified Hamiltonian (17). Moreover, by looking
at (120), we observe that the order $\left(N-1\right)$ correction in the
modified Hamiltonian is

$\displaystyle\frac{\eta^{N-1}}{N}\sum_{n=1}^{N}{\frac{\left(-1\right)^{n-1}}{n}\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\vdots\\\ r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=N\end{subarray}}{\frac{\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}}{\prod_{i=1}^{n}{r_{i}!s_{i}!}}}}.$ (122)

By applying the triangle inequality to (121), we start out with

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}}{k}\sum_{n=1}^{k}{\frac{1}{n}\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\vdots\\\ r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\right|}{\prod_{i=1}^{n}{r_{i}!s_{i}!}}}}}$ (123)

in either case. From here, the game is to bound the inner two sums (i.e.,
(122)) as tightly as is reasonably possible while still giving us a tractable
expression, and then check the outermost sum for convergence.

##### Case 1

Given the upper bound (111) in case 1, (123) can be bounded from above like

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}A\left(k\right)}{k}\sum_{n=1}^{k}{\frac{1}{n}\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\vdots\\\ r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{1}{\prod_{i=1}^{n}{r_{i}!s_{i}!}}}}}.$
(124)

Rather surprisingly, we can express the innermost sum in (124) in closed form,
as formalized in the lemma below:

######  Lemma 14.

For all $n,k\in\mathbb{Z}_{+}$ such that $k\geq n$,

$\displaystyle\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\ \vdots\\\
r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{k!}{\prod_{i=1}^{n}{r_{i}!s_{i}!}}}=n!2^{k}\genfrac{\\{}{\\}}{0.0pt}{}{k}{n},$
(125)

where $\genfrac{\\{}{\\}}{0.0pt}{}{k}{n}$ is a Stirling number of the second
kind—i.e., the number of ways to partition a set of $k$ objects into $n$ non-
empty subsets [33].

###### Proof.

By the residue theorem [44],
$\oint_{\left|z\right|=1}{\frac{1}{z^{1+p}}\frac{dz}{2\pi i}}=\delta_{p0}$,
where $\delta_{ij}$ is the Kronecker delta [80]. Thus, using properties of the
Kronecker delta, we can simplify the left side of (125) as follows:

$\displaystyle\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\ \vdots\\\
r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{k!}{\prod_{m=1}^{n}{r_{m}!s_{m}!}}}=k!\sum_{\begin{subarray}{c}r_{1}+s_{1}\geq
1\\\ \vdots\\\ r_{n}+s_{n}\geq 1\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{1}{\prod_{m=1}^{n}{r_{m}!s_{m}!}}}$
$\displaystyle=k!\sum_{\begin{subarray}{c}r_{1}+s_{1}\geq 1\\\ \vdots\\\
r_{n}+s_{n}\geq
1\end{subarray}}{\frac{\delta_{\left(k-\sum_{m=1}^{n}{\left(r_{m}+s_{m}\right)}\right)0}}{\prod_{m=1}^{n}{r_{m}!s_{m}!}}}=k!\sum_{\begin{subarray}{c}r_{j}+s_{j}\geq
1\\\
j=1,\dots,n\end{subarray}}{\frac{1}{\prod_{m=1}^{n}{r_{m}!s_{m}!}}\oint_{\left|z\right|=1}{\frac{1}{z^{k+1-\sum_{m=1}^{n}{\left(r_{m}+s_{m}\right)}}}\frac{dz}{2\pi
i}}}$
$\displaystyle=k!\oint_{\left|z\right|=1}{\frac{1}{z^{k+1}}\sum_{\begin{subarray}{c}r_{j}+s_{j}\geq
1\\\
j=1,\dots,n\end{subarray}}{\prod_{m=1}^{n}{\frac{z^{r_{m}}z^{s_{m}}}{r_{m}!s_{m}!}}}\frac{dz}{2\pi
i}}=k!\oint_{\left|z\right|=1}{\frac{1}{z^{k+1}}\left(\sum_{r+s\geq
1}{\frac{z^{r}z^{s}}{r!s!}}\right)^{n}\frac{dz}{2\pi i}}.$ (126)

The sum in (126) can be simplified as follows:

$\displaystyle\sum_{r+s\geq
1}{\frac{z^{r}z^{s}}{r!s!}}=\sum_{r+s=1}^{\infty}{\frac{z^{r+s}}{r!s!}}=\sum_{j=1}^{\infty}{\sum_{r+s=j}{\frac{z^{j}}{r!s!}}}=\sum_{j=1}^{\infty}{\frac{z^{j}}{j!}\sum_{r+s=j}{\frac{j!}{r!s!}}}=\sum_{j=1}^{\infty}{\frac{z^{j}}{j!}\sum_{s=0}^{j}{\binom{j}{s}}}.$
(127)

Then, using the binomial theorem and the exponential series
$e^{x}=\sum_{j=0}^{\infty}{x^{j}/j!}$ [5] to simplify (127),

$\displaystyle\sum_{j=1}^{\infty}{\frac{z^{j}}{j!}2^{j}}=\sum_{j=1}^{\infty}{\frac{\left(2z\right)^{j}}{j!}}=e^{2z}-\frac{\left(2z\right)^{0}}{0!}=e^{2z}-1.$
(128)

Substituting (128) into (126) and applying Cauchy’s differentiation formula
[44], we get that

$\displaystyle\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\ \vdots\\\
r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{\frac{k!}{\prod_{m=1}^{n}{r_{m}!s_{m}!}}}=\frac{k!}{2\pi
i}\oint_{\left|z\right|=1}{\frac{\left(e^{2z}-1\right)^{n}}{z^{k+1}}dz}=\frac{d^{(k)}}{dz^{(k)}}\left[\left(e^{2z}-1\right)^{n}\right]_{z=0}.$
(129)

Finally, using the binomial theorem, exponential series, and then the
definition of Stirling numbers of the second kind [33] to expand (129),

$\displaystyle\left(e^{2z}-1\right)^{n}$
$\displaystyle=\sum_{l=0}^{n}{\binom{n}{l}\left(e^{2z}\right)^{n-l}\left(-1\right)^{l}}=\sum_{l=0}^{n}{\binom{n}{l}e^{2z\left(n-l\right)}\left(-1\right)^{l}}.$

This implies for all $k>0$:

$\displaystyle\frac{d^{(k)}}{dz^{(k)}}\left[\left(e^{2z}-1\right)^{n}\right]_{z=0}$
$\displaystyle=\sum_{l=0}^{n}{\binom{n}{l}\frac{d^{(k)}}{dz^{(k)}}\left.e^{2z\left(n-l\right)}\right|_{z=0}\left(-1\right)^{l}}$
$\displaystyle=\sum_{l=0}^{n}{\binom{n}{l}2^{k}\left(n-l\right)^{k}\left.e^{2z\left(n-l\right)}\right|_{z=0}\left(-1\right)^{l}}$
$\displaystyle=n!2^{k}\frac{1}{n!}\sum_{l=0}^{n}{\left(-1\right)^{l}\binom{n}{l}\left(n-l\right)^{k}}$
$\displaystyle=n!2^{k}\genfrac{\\{}{\\}}{0.0pt}{}{k}{n},$

as we wanted to show. ∎

With Lemma 14 proven, we can now proceed with the proof of Lemma 15.

######  Lemma 15.

Suppose that there exists some
$A:\mathbb{Z}_{+}\times\mathcal{P}\times\mathcal{Q}\rightarrow\mathbb{R}_{\geq
0}$ such that

$\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\left(p,q\right)\right|\leq
A\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n},p,q\right)$

for all $n\in\mathbb{Z}_{+}$,
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$, and integer pairs
$r_{i}+s_{i}>0$ for $i=1,\dots,n$. Then, for all
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$,

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}2^{k+1}A\left(k,p,q\right)a_{k-1}}{k^{2}\left(k-1\right)!}}-2A\left(1,p,q\right),$
(130)

where $a_{k-1}$ is the $(k-1)$-th Fubini number [89, 41] (also known as
ordered Bell number [72]) for all $k\in\mathbb{Z}_{+}$.

###### Proof.

Using Lemma 14 and properties of Stirling numbers of the second kind [33] to
simplify (124),

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|$
$\displaystyle\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}A\left(k\right)}{k!k}\sum_{n=1}^{k}{\frac{n!2^{k}}{n}\genfrac{\\{}{\\}}{0.0pt}{}{k}{n}}}$
$\displaystyle=\sum_{k=1}^{\infty}{\frac{\eta^{k-1}2^{k}A\left(k\right)}{k!k}\left(\left(k-1\right)!\genfrac{\\{}{\\}}{0.0pt}{}{k}{k}+\sum_{n=1}^{k-1}{\left(n-1\right)!\genfrac{\\{}{\\}}{0.0pt}{}{k}{n}}\right)}$
$\displaystyle\stackrel{{\scriptstyle(i)}}{{=}}\sum_{k=1}^{\infty}{\frac{\eta^{k-1}2^{k}A\left(k\right)}{k!k}\left[\left(k-1\right)!+\sum_{n=1}^{k-1}{\left(n-1\right)!\left(n\genfrac{\\{}{\\}}{0.0pt}{}{k-1}{n}+\genfrac{\\{}{\\}}{0.0pt}{}{k-1}{n-1}\right)}\right]}$
$\displaystyle=\sum_{k=1}^{\infty}{\frac{\eta^{k-1}2^{k}A\left(k\right)}{k!k}\left(\sum_{n=1}^{k-1}{n!\genfrac{\\{}{\\}}{0.0pt}{}{k-1}{n}}+\sum_{n=1}^{k}{\left(n-1\right)!\genfrac{\\{}{\\}}{0.0pt}{}{k-1}{n-1}}\right)}$
$\displaystyle\stackrel{{\scriptstyle(ii)}}{{=}}\sum_{k=1}^{\infty}{\frac{\eta^{k-1}2^{k}A\left(k\right)}{k!k}\left(2\sum_{m=0}^{k-1}{m!\genfrac{\\{}{\\}}{0.0pt}{}{k-1}{m}}-\delta_{(k-1)0}\right)}.$
(131)

In the above, the step $(i)$ follows from the identities
$\genfrac{\\{}{\\}}{0.0pt}{}{k}{k}=1$ and
$\genfrac{\\{}{\\}}{0.0pt}{}{k}{n}=n\genfrac{\\{}{\\}}{0.0pt}{}{k-1}{n}+\genfrac{\\{}{\\}}{0.0pt}{}{k-1}{n-1}$;
while the step $(ii)$ follows from
$\genfrac{\\{}{\\}}{0.0pt}{}{k-1}{0}=\delta_{(k-1)0}$. Finally, the Fubini
numbers $a_{n}$ can be computed from the Stirling numbers of the second kind
via the formula [62, 41]

$\displaystyle
a_{n}=\sum_{m=0}^{n}{m!\genfrac{\\{}{\\}}{0.0pt}{}{n}{m}}\quad\forall
n\in\mathbb{N}_{0}.$ (132)

Substituting (132) into (131), we are done. ∎

##### Case 2

######  Lemma 16.

Suppose that there exists some
$B:\mathbb{Z}_{+}\times\mathcal{P}\times\mathcal{Q}\rightarrow\mathbb{R}_{\geq
0}$ such that

$\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\left(p,q\right)\right|\leq
B\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n},p,q\right)\prod_{i=1}^{n}{r_{i}!s_{i}!}$

for all $n\in\mathbb{Z}_{+}$,
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$, and integer pairs
$r_{i}+s_{i}>0$ for $i=1,\dots,n$. Then, for all
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$ such that

$\sum_{k=1}^{\infty}{\frac{\left(\eta\left(2+\sqrt{3}\right)\right)^{k}B\left(k,p,q\right)}{k^{2}}}<\infty,$

the modified Hamiltonian $\widetilde{H}_{\eta}\left(p,q\right)$ converges.
Moreover, the following bounds hold for all
$\left(p,q\right)\in\mathcal{P}\times\mathcal{Q}$:

$\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}B\left(k,p,q\right)}{k^{2}}\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{2\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}}\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}B\left(k,p,q\right)}{k}\left(2+\sqrt{3}\right)^{k}}.$

Before proving Lemma 16, we need to show a few more properties. Assuming the
upper bound (118), (123) can be bounded above like

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}B\left(k\right)}{k}\sum_{n=1}^{k}{\frac{1}{n}\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\vdots\\\ r_{n}+s_{n}>0\\\
r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k\end{subarray}}{1}}}.$ (133)

Simplifying the innermost sum in (133) is equivalent with a combinatorics
problem: counting the number of
$r_{1},\dots,r_{n},s_{1},\dots,s_{n}\in\mathbb{N}_{0}$ such that
$r_{1}+s_{1},\dots,r_{n}+s_{n}>0$ and $r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}=k$.
Analogously, this is equivalent with counting the number of ways that we can
fit $k$ indistinguishable balls into $n$ ordered red and $n$ ordered blue bins
such that at least one ball ends up in the first red or blue bin, at least one
ball ends up in the second red or blue bin, and so on. In turn, this is equal
to the number of ways we can fit $n$ out of $k$ total indistinguishable balls
into $n$ ordered red and $n$ ordered blue bins such that at least one ball
ends up in the first red or blue bin, at least one ball ends up in the second
red or blue bin, and so on, multiplied by the number of ways we can fit the
remaining balls into the bins. The first of these quantities is $2^{n}$, as
for each pair of red and blue bins, we can place one ball into either of the
two bins, and there are a total of $n$ pairs of red and blue bins. Hence, the
second of these is the number of ways to put $\left(k-n\right)$ unlabeled
balls into $2n$ distinct bins—i.e., [30]

$\displaystyle\binom{k-n+2n-1}{2n-1}=\binom{k+n-1}{2n-1}=\frac{\left(k+n-1\right)!}{\left(k-n\right)!\left(2n-1\right)!},$
(134)

but we can further rewrite (134) as the following:

$\displaystyle\frac{\left(k+n-1\right)\left(k+n-2\right)\cdots\left(k-n+1\right)\left(k-n\right)!}{\left(k-n\right)!\left(2n-1\right)!}=\frac{nk^{2}\left(k^{2}-1^{2}\right)\cdots\left(k^{2}-\left(n-1\right)^{2}\right)}{kn^{2}\left(n^{2}-1^{2}\right)\cdot\left(n^{2}-\left(n-1\right)^{2}\right)}.$
(135)

Thus, (133) can be simplified to

$\displaystyle\left|\widetilde{H}_{\eta}\left(p,q\right)\right|$
$\displaystyle\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}B\left(k\right)}{k^{2}}\sum_{n=1}^{k}{2^{n}\frac{\left(k^{2}-0^{2}\right)\left(k^{2}-1^{2}\right)\cdots\left(k^{2}-\left(n-1\right)^{2}\right)}{\left(n^{2}-0^{2}\right)\left(n^{2}-1^{2}\right)\cdot\left(n^{2}-\left(n-1\right)^{2}\right)}}}$
$\displaystyle=\sum_{k=1}^{\infty}{\frac{\eta^{k-1}B\left(k\right)}{k^{2}}\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{2\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}}.$
(136)

As far as the authors know, the inner sum and product in (136) cannot be
expressed in closed form. However, we can use Riemann sums and properties of
the LogSumExp function [19] to produce an upper bound which is saturated as
$k\rightarrow\infty$ as follows:

######  Lemma 17.

For every $r>0$,

$\displaystyle\lim_{k\rightarrow\infty}{\frac{\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}}{\left(\frac{2+r+\sqrt{r\left(4+r\right)}}{2}\right)^{k}}}=1,$
(137)

or equivalently,

$\displaystyle\lim_{k\rightarrow\infty}{\frac{1}{k}\log\left(\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}\right)}=\log\left(\frac{2+r+\sqrt{r\left(4+r\right)}}{2}\right).$
(138)

Furthermore, we have the following upper bound for all $k\in\mathbb{N}$:

$\displaystyle\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}\leq
k\left(\frac{2+r+\sqrt{r\left(4+r\right)}}{2}\right)^{k}.$ (139)

###### Proof.

The equivalence between (137) and (138) and the implication from

$\displaystyle\max\left\\{\int_{0}^{\max\left\\{0,\left\lfloor{kx_{+}}\right\rfloor-1\right\\}/k}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}\right)^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{\left(\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}\right)^{2}}{r}\right),\right.$
$\displaystyle\hskip
71.13188pt\left.\int_{0}^{\left(\left\lceil{kx_{+}}\right\rceil-1\right)/k}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\frac{\left\lceil{kx_{+}}\right\rceil}{k}\right)^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{\left(\frac{\left\lceil{kx_{+}}\right\rceil}{k}\right)^{2}}{r}\right)\right\\}$
$\displaystyle\leq$
$\displaystyle\frac{1}{k}\log\left(\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}\right)$
$\displaystyle\leq$
$\displaystyle\max\left\\{\int_{0}^{\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}\right)^{2}-y^{2}}\right)dy},\int_{0}^{\frac{\left\lceil{kx_{+}}\right\rceil}{k}}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\frac{\left\lceil{kx_{+}}\right\rceil}{k}\right)^{2}-y^{2}}\right)dy}\right\\}+\frac{\log
k}{k}$ $\displaystyle\leq$
$\displaystyle\log\left(\frac{2+r+\sqrt{r\left(4+r\right)}}{2}\right)+\frac{\log
k}{k},$ (140)

where $x_{+}=\sqrt{\frac{r}{4+r}}$, to (139) follow from sequential continuity
and arithmetic properties of the logarithm [64]. Hence, it suffices to prove
(138) and (140); at first, we show the latter. For any
$x_{1},\dots,x_{k}\in\mathbb{R}$, we have the following lower and upper bounds
[19]:

$\displaystyle\max\left\\{x_{1},\dots,x_{k}\right\\}\leq\log\left(\exp\left(x_{1}\right)+\dots+\exp\left(x_{k}\right)\right)\leq\max\left\\{x_{1},\dots,x_{k}\right\\}+\log
k,$ (141)

In particular, if we let
$x_{n}:=\sum_{m=0}^{n-1}{\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)}$
for $n=1,\dots,k$ and divide both sides by $1/k$, then (141) implies that

$\displaystyle\max_{n=1,\dots,k}\sum_{m=0}^{n-1}{\frac{1}{k}\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)}\leq\frac{1}{k}\log\left(\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}\right)$
$\displaystyle\leq\max_{n=1,\dots,k}\sum_{m=0}^{n-1}{\frac{1}{k}\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)}+\frac{\log
k}{k}.$ (142)

The reader might recognize the sum on the leftmost or rightmost sides of (142)
as a Riemann sum for the following improper integral:

$\displaystyle\int_{0}^{x}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy},$
(143)

where $x:=n/k$. Thus, we work along this vein of thought. At first, the
integrand of (143) is increasing in $y$ for $0\leq y\leq x\leq 1$. To show
this, note that

$\displaystyle\frac{\partial}{\partial
y}\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)=\frac{2y\left(1-x^{2}\right)}{\left(1-y^{2}\right)\left(x^{2}-y^{2}\right)},$
(144)

and whenever $0<x<1$, $0<y<1$, and $\left|y\right|<\left|x\right|$, (144) is
positive. Thus, for any $n,m\in\mathbb{N}_{0}$ such that $0\leq m\leq n-1$,

$\displaystyle\frac{1}{k}\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)$
$\displaystyle=\int_{m/k}^{\left(m+1\right)/k}{\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)dy}\leq\int_{m/k}^{\left(m+1\right)/k}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\frac{n}{k}\right)^{2}-y^{2}}\right)dy}$
$\displaystyle\Rightarrow\sum_{m=0}^{n-1}{\frac{1}{k}\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)}$
$\displaystyle\leq\sum_{m=0}^{n-1}{\int_{m/k}^{\left(m+1\right)/k}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\frac{n}{k}\right)^{2}-y^{2}}\right)dy}}=\int_{0}^{x}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy}.$
(145)

Furthermore, if we instead have that $1\leq m\leq n$, then

$\displaystyle\frac{1}{k}\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)$
$\displaystyle=\int_{\left(m-1\right)/k}^{m/k}{\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)dy}\geq\int_{\left(m-1\right)/k}^{m/k}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\frac{n}{k}\right)^{2}-y^{2}}\right)dy}$
$\displaystyle\Rightarrow\sum_{m=1}^{n-1}{\frac{1}{k}\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)}$
$\displaystyle\geq\sum_{m=1}^{n-1}{\int_{\left(m-1\right)/k}^{m/k}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\frac{n}{k}\right)^{2}-y^{2}}\right)dy}}=\int_{0}^{x-1/k}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy}$
$\displaystyle\Rightarrow\sum_{m=0}^{n-1}{\frac{1}{k}\log\left(\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}\right)}$
$\displaystyle\geq\int_{0}^{x-1/k}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{x^{2}}{r}\right).$
(146)

Since (145) and (146) hold for all $n=1,\dots,k$ (or equivalently, for all
$x=1/k,\dots,1$), (142), (145), and (146) imply that

$\max_{x=1/k,\dots,1}\left\\{\int_{0}^{x-1/k}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{x^{2}}{r}\right)\right\\}\leq\frac{1}{k}\log\left(\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}\right)\\\
\leq\max_{x=1/k,\dots,1}\left\\{\int_{0}^{x}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy}\right\\}+\frac{\log
k}{k}.$ (147)

We will see that the leftmost and rightmost sides of (147) are close for $k$
large; there is no known closed-form expression for the leftmost side. We can
maximize the rightmost side and then use the location of the maximum to
approximate the leftmost side—i.e., for all $z=1/k,\dots,1$,

$\int_{0}^{z-1/k}{\log\left(\frac{r\left(1-y^{2}\right)}{z^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{z^{2}}{r}\right)\\\
\leq\max_{x=1/k,\dots,1}\left\\{\int_{0}^{x-1/k}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{x^{2}}{r}\right)\right\\}$
(148)

Thus, to further simplify (147), we should compute
$\int_{0}^{x}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy}$,
maximize the resultant expression over all $x\in\left[0,1\right]$, then see
how this implies the maximum over $x=1/k,\dots,1$. The derivative of
$f(x)=\int_{0}^{x}{\log\left(\frac{r\left(1-y^{2}\right)}{x^{2}-y^{2}}\right)dy}$
(with respect to $x$) is
$f^{\prime}(x)=\log\left(\frac{r\left(1-x^{2}\right)}{4x^{2}}\right)$ which 1)
has two zeroes at $x_{\pm}=\pm\sqrt{\frac{r}{4+r}}$ and 2) second derivative
$f^{\prime\prime}(x)=\frac{2}{x^{3}-x}$. Hence, since $f^{\prime\prime}(x)<0$
for $0<x<1$ and $f^{\prime\prime}(x)>0$ for $-1<x<0$, it is clear that $f$ is
maximized at $x_{+}=\sqrt{\frac{r}{4+r}}<1$ on the domain $\left[0,1\right]$.
Moreover, since $f$ is always concave down on $\left[0,1\right]$, the maximum
on the “grid” $\left\\{1/k,\dots,1\right\\}$ will be either at
$\frac{\left\lceil{kx_{+}}\right\rceil}{k}$ or
$\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}$,
since
$x_{+}\in\left[\frac{\left\lfloor{kx_{+}}\right\rfloor}{k},\frac{\left\lceil{kx_{+}}\right\rceil}{k}\right]$—i.e.,
one of the two multiples of $1/k$ adjacent to $x_{+}$. Thus, (147) and (148)
imply that

$\displaystyle\max\left\\{\int_{0}^{\max\left\\{0,\left\lfloor{kx_{+}}\right\rfloor-1\right\\}/k}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}\right)^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{\left(\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}\right)^{2}}{r}\right),\right.$
$\displaystyle\left.\int_{0}^{\left(\left\lceil{kx_{+}}\right\rceil-1\right)/k}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\frac{\left\lceil{kx_{+}}\right\rceil}{k}\right)^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{\left(\frac{\left\lceil{kx_{+}}\right\rceil}{k}\right)^{2}}{r}\right)\right\\}$
$\displaystyle\leq\frac{1}{k}\log\left(\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}\right)$
$\displaystyle\leq\max\left\\{\int_{0}^{\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}\right)^{2}-y^{2}}\right)dy},\right.$
$\displaystyle\hskip
42.67912pt\left.\int_{0}^{\frac{\left\lceil{kx_{+}}\right\rceil}{k}}{\log\left(\frac{r\left(1-y^{2}\right)}{\left(\frac{\left\lceil{kx_{+}}\right\rceil}{k}\right)^{2}-y^{2}}\right)dy}\right\\}+\frac{\log
k}{k},$ (149)

which establishes part of (140). Furthermore, since
$\lim_{k\rightarrow\infty}{\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}}=\lim_{k\rightarrow\infty}{\frac{\left\lceil{kx_{+}}\right\rceil}{k}}=x_{+}$,
taking limits of (149) gives

$\displaystyle\int_{0}^{x_{+}}{\log\left(\frac{r\left(1-y^{2}\right)}{x_{+}^{2}-y^{2}}\right)dy}-\frac{1}{k}\log\left(\frac{x_{+}^{2}}{r}\right)$
$\displaystyle\leq\frac{1}{k}\log\left(\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}\right)$
$\displaystyle\leq\int_{0}^{x_{+}}{\log\left(\frac{r\left(1-y^{2}\right)}{x_{+}^{2}-y^{2}}\right)dy}+\frac{\log
k}{k}$ (150)

for sufficiently large $k$. With help from Wolfram Mathematica [83], the
integral in (150) evaluates to the following:

$\displaystyle\log\left(\frac{2+r+\sqrt{r(4+r)}}{2}\right).$ (151)

Finally, since
$\lim_{k\rightarrow\infty}{\frac{1}{k}\log\left(\frac{x_{+}^{2}}{r}\right)}=\lim_{k\rightarrow\infty}{+\frac{\log
k}{k}}=0$, by the squeeze theorem [64], (150) and (151) imply (138).

It remains to show the last inequality in (140). Note that
$\frac{\left\lceil{kx_{+}}\right\rceil}{k}$ and
$\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}$
are “close” to $x_{+}$, but
$f\left(\frac{\left\lceil{kx_{+}}\right\rceil}{k}\right)$ and
$f\left(\max\left\\{\frac{1}{k},\frac{\left\lfloor{kx_{+}}\right\rfloor}{k}\right\\}\right)$
can still only be at most $f\left(x_{+}\right)$. Hence, (149) implies that

$\displaystyle\frac{1}{k}\log\left(\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{r\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}\right)\leq\log\left(\frac{2+r+\sqrt{r(4+r)}}{2}\right)+\frac{\log
k}{k},$

as we wanted to show. ∎

####  D.1.1 Proof of Lemma 16

Lemma 17 gives us a simple way to analyze the convergence and boundedness of
the inner sum and product within (136). Hence, we have the proof of Lemma 16:

###### Proof of Lemma 16.

The equation (137) from Lemma 17 implies by the limit comparison test [64]
that the convergence of (136) is equivalent with the convergence of

$\displaystyle\sum_{k=1}^{\infty}{\frac{\left(\eta\left(2+\sqrt{3}\right)\right)^{k}B\left(k\right)}{k^{2}}}.$

Moreover, (139) from Lemma 17 can be used to estimate (136) as follows:

$\displaystyle\sum_{k=1}^{\infty}{\frac{\eta^{k-1}B\left(k\right)}{k^{2}}\sum_{n=1}^{k}{\prod_{m=0}^{n-1}{\frac{2\left(k^{2}-m^{2}\right)}{n^{2}-m^{2}}}}}\leq\sum_{k=1}^{\infty}{\frac{\eta^{k-1}B\left(k\right)}{k}\left(2+\sqrt{3}\right)^{k}},$

thereby proving Lemma 16. ∎

###  D.2 Applications of the convergence criteria

We now explore some basic examples of functions $F$ and $G$ and see if the
criteria above confirm an absolutely convergent modified Hamiltonian,
expressed with Dynkin series in those cases. That being said, we invite
readers to look for other functions $F$ and $G$ which satisfy (or fail to
satisfy) the convergence criteria outlined in Theorems 8, 9, and 10, or more
generally, those from Lemmas 16 and 15.

####  D.2.1 Quadratic case

In Section 8 and Appendix C, we computed the closed-form of the modified
Hamiltonian, expressed with the Dynkin series when $f$ and $g$ are both
quadratic functions. The convergence criteria we proposed in Appendix D are
also satisfied.

######  Proposition 2.

When $F=ap^{2},G=dq^{2}$, we have that

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\left(p,q\right)\right|\leq\max\\{p^{2},q^{2},pq\\}\left(\max\\{4a,4d\\}\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}}$
(152)

Thus, all convergence criteria in Appendix D are satisfied.

###### Proof.

We show (152) by induction. First, for the base case, it is trivial that

$\displaystyle
F=ap^{2}\leq\max\\{p^{2},q^{2},pq\\}\left(\max\\{4a,4d\\}\right)^{1},G=dq^{2}\leq\max\\{p^{2},q^{2},pq\\}\left(\max\\{4a,4d\\}\right)^{1}.$

We now suppose that

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\left(p,q\right)\right|\leq\max\\{p^{2},q^{2},pq\\}\left(\max\\{4a,4d\\}\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}}.$

We may apply $F$ or $G$ to the IPBs. We start with the case of $F$.

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}+1}\right\\}\left(p,q\right)\right|$
$\displaystyle=\left|\left\\{\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\},F\right\\}\left(p,q\right)\right|$
$\displaystyle\leq\left|\max\\{p^{2},q^{2},pq\\}\left(\max\\{4a,4d\\}\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}}\cdot
4a\right|$
$\displaystyle\leq\left|\max\\{p^{2},q^{2},pq\\}\left(\max\\{4a,4d\\}\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}+1}\right|.$

Similarly, for the case of $G$, we have:

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}G\right\\}\left(p,q\right)\right|$
$\displaystyle=\left|\left\\{\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\},G\right\\}\left(p,q\right)\right|$
$\displaystyle\leq\left|\max\\{p^{2},q^{2},pq\\}\left(\max\\{4a,4d\\}\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}}\cdot
4d\right|$
$\displaystyle\leq\left|\max\\{p^{2},q^{2},pq\\}\left(\max\\{4a,4d\\}\right)^{r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}+1}\right|.$

Hence, using (152) in tandem with any of the lemmas or theorems from Appendix
D, the modified Hamiltonian for quadratic case converges absolutely for
sufficiently small, nonzero $\eta$.

For a discussion about the conserved quantity predicted by
$\widetilde{H}_{\eta}$ in this case, see Section 8 or Appendix C. ∎

####  D.2.2 Quartic case

We now present a case in which our lemma does not hold: $F=p^{4},G=q^{4}$.

######  Proposition 3.

When $r_{1}=s_{1}=\cdots=r_{n}=s_{n}=1$,

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\right|=(n+1)(2n)!4^{2n-1}p^{2n+1}q^{2n+1}.$ (153)

Thus, none of the convergence criteria in Appendix D can be satisfied in this
case.

###### Proof.

We show this by induction. The formula is true for $n=1$, for
$|\\{p^{4},q^{4}\\}|=16p^{3}q^{3}=2\cdot 2\cdot 4\cdot p^{3}\cdot p^{3}$.
Suppose that the formula is true for $n$. We calculate the $n+1$-case:

$\displaystyle|\\{\\{(n+1)(2n)!4^{2n-1}p^{2n+1}q^{2n+1},G\\},F\\}|$
$\displaystyle=|(n+1)(2n)!(2n+1)(2n+4)4^{2n+1}p^{2n+3}q^{2n+3}|$
$\displaystyle=|(n+2)(2n)!(2n+1)(2n+2)4^{2n+1}p^{2n+3}q^{2n+3}|$
$\displaystyle=\left((n+1)+1\right)2(n+1)!4^{2(n+1)-1}p^{2(n+1)+1}q^{2(n+1)+1}.$

Hence, by induction, we have shown (153). But since even just the terms with
IPBs that have $r_{1}=s_{1}=\cdots=r_{n}=s_{n}=1$ grow with
$(2n)!=\left(r_{1}+\dots+r_{n}+s_{1}+\dots+s_{n}\right)!$, none of the
convergence criteria in Appendix D can be satisfied. ∎

It is not known whether the modified Hamiltonian $\widetilde{H}_{\eta}$
actually converges in this case, although Figure 3 suggests that it could
converge for sufficiently small $\eta>0$ (or at least that a conserved
quantity exists in that case).

####  D.2.3 Logarithmic case

To further demonstrate the limitations of the current convergence criteria
derived in this paper, we calculate the case where $F=\log(a+p),G=\log(b+q)$.

######  Proposition 4.

When $F=\log(a+p),G=\log(b+q)$,

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}}\right\\}\right|=\frac{(S-2)!}{(a+p)^{S-1}(b+q)^{S-1}},\text{
where $S=\sum_{i=1}^{n}s_{i}+\sum_{i=1}^{n}r_{i}$.}$

Yet again, we claim that none of the convergence criteria in Appendix D are
satisfied.

###### Proof.

As in the quadratic and quartic cases, we show this by induction. Note that
$|\\{F,G\\}|=\frac{1}{(a+p)(b+q)}$, and hence the case where $s_{1}=r_{1}=1$
is verified. Assume now that the formula is true for some $n\in\mathbb{N}$, in
which case

$\displaystyle\left|\left\\{G^{r_{1}}F^{s_{1}}G^{r_{2}}F^{s_{2}}\cdots
G^{r_{n}}F^{s_{n}+1}\right\\}\right|$
$\displaystyle=\frac{(S-1)!}{(a+p)^{S}(b+q)^{S}}.$

The case in which we apply $G$ to the formula is very similar. However, the
Fubini numbers $a_{k-1}\approx(k-1)!/\left(2\left(\log 2\right)^{k}\right)$
[72], and hence, a factorial increase of the IPBs with $k$ is too fast for the
sums in either Lemma 15 or Lemma 16 to converge. ∎

However, there is still a known, absolutely convergent modified Hamiltonian
for sufficiently small $\eta>0$. As shown in [31] and restated in Appendix E,
the MH is of the form
$F\left(\left(\alpha+p\right)\left(\beta+q\right)\right)$ for an original
Hamiltonian of the form
$H\left(p,q\right)=\log\left(\alpha+p\right)+\log\left(\beta+q\right)$. This
shows that these convergence criteria we derived, despite requiring
sophisticated and careful techniques to derive, are sufficient but still not
necessary to ensure that the MH converges.

##  Appendix E List of known modified Hamiltonians

In this section, we review known choices of $F$ and $G$ such that when the
symplectic Euler integrator (5) is applied to $F+G$, the iterations will
conserve a modified Hamiltonian exactly. We remark that the results discussed
here might not be exhaustive—these are only the cases that the authors are
aware of in the literature. Here, we summarize the original Hamiltonian system
and the modified Hamiltonian conserved by symplectic Euler (5) when applied to
each respective system.

##### The quadratic case.

The quadratic case where $(p,q)\in\mathbb{R}^{d\times d}$ and
$H(p,q)=p^{\top}Bp+q^{\top}Cq$ and $BC$ is diagonalizable is well-studied in
this paper. Refer to Theorem 3 or Appendix C for details.

##### The logarithmic case.

In [31], the authors study the log case where $p,q\in\mathbb{R}$ and

$H(p,q)=\log\left(\alpha+p\right)+\log\left(\beta+q\right).$

In this case, when $p,q$ evolve under the symplectic Euler method (5), one can
check the quantity
$L(p,q):=\log\left(\alpha+p\right)+\log\left(\beta+q\right)$ is exactly
conserved.

By using an action-angle argument, [31] derive the following closed form of
the modified Hamiltonian:

$\displaystyle\widetilde{H}_{\eta}(p,q)=\log\left(1-\frac{\eta}{L(p,q)}\right)-\frac{\eta}{L(p,q)}\log\left(1-\frac{\eta}{L(p,q)}\right).$

##### The lattice KdV equation case.

In [4], the authors study the symplectic Euler integration of the following
Hamiltonian system governed by lattice KdV equations: $p,q\in\mathbb{R}$ and
$\epsilon$ is a constant in $\mathbb{R}$,

$H(p,q)=\epsilon\log\left(\epsilon^{2}-p^{2}\right)+\epsilon\log\left(\epsilon^{2}-q^{2}\right).$

In this case, when $p,q$ evolve under the symplectic Euler method (5), the
quantity

$\mathcal{I}=p^{2}q^{2}-\epsilon^{2}(p^{2}+q^{2})-2\epsilon\eta pq$

is exactly conserved. Note that $\eta$ in the above equation corresponds to
the stepsize in the symplectic Euler method (5). By using an action-angle
argument, [4] derive the following closed form of the modified Hamiltonian:

$\tilde{H}_{\eta}(\mathcal{P})=\int^{\mathcal{P}}\int_{0}^{\frac{2\epsilon^{2}\eta\sqrt{-\mathcal{P}^{\prime}}}{\epsilon^{4}+\mathcal{P}^{\prime}}}\frac{1}{2\sqrt{\eta^{2}\epsilon^{2}q^{2}-\left(\epsilon^{2}-q^{2}\right)\left(\epsilon^{2}q^{2}+\mathcal{P}^{\prime}\right)}}dqd\mathcal{P}^{\prime},$

where they use $\mathcal{P}$ to denote
$p^{2}q^{2}-\epsilon^{2}p^{2}-\epsilon^{2}q^{2}$. Note that the outer integral
is an indefinite integral and the inner integral is a definite integral
starting from $0$.

##### The Suris maps.

[69] summarizes all possible Hamiltonian systems in $\mathbb{R}^{1\times 1}$
with $F(p)=\frac{p^{2}}{2}$ such that there exists a conserved quantity to the
iterations of symplectic Euler (5). [69] also assumes that the conserved
quantity, $\hat{H}(p,q)$, must be of the form
$\hat{H}_{0}(p,q)+\eta\hat{H}_{1}(p,q)$, where both $\hat{H}_{0}(p,q)$ and
$\hat{H}_{1}(p,q)$ are not functions of $\eta$.

Note that the existence of a conserved quantity is a necessary condition for
the convergence of the Dynkin series (17) for $\widetilde{H}_{\eta}$, since
$\widetilde{H}_{\eta}$ is itself a conserved quantity by Theorem 2. To show
this, suppose that $\widetilde{H}_{\eta}$ is convergent yet there does not
exist a conserved quantity for the iterations of symplectic Euler (5). Then,
we arrive at a contradiction by Theorem 2.

Under these assumptions, a conserved quantity $\hat{H}$ exists if and only if
$G(q)$ in the decomposition $H(p,q)=F(p)+G(q)$ satisfies the following
conditions, and takes the listed form. In the expressions listed below,
$A,B,C,D,E$ are arbitrary constants in $\mathbb{R}$.

$\displaystyle F(p)=\frac{p^{2}}{2},\quad G(q,\eta)$
$\displaystyle=-\int_{x}\frac{A+Bx+Cx^{2}+Dx^{3}}{\eta-\eta^{2}(E+\frac{C}{3}x+\frac{D}{2}x^{2})}dx,$
$\displaystyle\hat{H}(p,q)$
$\displaystyle=\frac{1}{2}(p-q)^{2}-\eta\left[-\frac{1}{2}A(p+q)-\frac{1}{2}Bpq-\frac{1}{6}Cpq(p+q)-\frac{1}{4}Dp^{2}q^{2}-\frac{1}{2}E(p-q)^{2}\right];$

$\displaystyle F(p)=\frac{p^{2}}{2},\quad G(q,\eta)$
$\displaystyle=-\int_{x}\frac{2}{\omega\eta^{2}}\arctan\left(\frac{(\omega\eta/2)(A\sin\omega
x+B\cos\omega x+C\sin 2\omega x+D\cos 2\omega x)}{1-(\omega\eta/2)(A\cos\omega
x-B\sin\omega x+C\cos 2\omega x-D\sin 2\omega x+E)}\right)dx,$
$\displaystyle\hat{H}$
$\displaystyle=\frac{1-\cos\omega(p-q)}{\omega^{2}}+\frac{\eta}{2\omega}\left[A(\cos\omega
p+\cos\omega q)-B(\sin\omega p+\sin\omega q)\right.$
$\displaystyle\left.\quad+C\cos\omega(p+q)-D\sin\omega(p+q)+E\cos\omega(p-q)\right];$

$\displaystyle F(p)=\frac{p^{2}}{2},\quad G(q,\eta)$
$\displaystyle=-\int_{x}\frac{1}{\alpha\eta^{2}}\ln\left(\frac{1+\alpha\eta(B\exp(-\alpha
x)+D\exp(-2\alpha x)-E)}{1-\alpha\eta(A\exp(\alpha x)+C\exp(2\alpha
x)+E)}\right)dx,$ $\displaystyle\hat{H}$
$\displaystyle=\frac{\cosh\alpha(p-q)-1}{\alpha^{2}}+\frac{\eta}{2\alpha}\Big{[}-A(\exp(\alpha
p)+\exp(\alpha q))+B(\exp(-\alpha p)+\exp(-\alpha q))$
$\displaystyle-C\exp(\alpha(p+q))+D\exp(-\alpha(p+q))-2E\cosh\alpha(p-q)\Big{]}.$

##  Appendix F Partial proof of Conjecture 1

We start by taking the Taylor series of the $N$th modification $H_{N}$:

$\displaystyle
H_{N}(p_{n+1},q_{n+1})=H_{N}(p_{n},q_{n})+\sum_{i=1}^{\infty}\frac{1}{i!}\Big{(}(p_{n+1}-p_{n})\cdot\nabla_{p}+(q_{n+1}-q_{n})\cdot\nabla_{q}\Big{)}^{i}H_{N}(p_{n},q_{n}),$
(154)

where the gradients are to be applied to $H_{N}$. Equivalently, by using the
binomial theorem to expand (154), we have

$H_{N}(p_{n+1},q_{n+1})-H_{N}(p_{n},q_{n})=\\\
\sum_{i=1}^{\infty}\frac{1}{i!}\left\\{\sum_{j=0}^{i}\binom{i}{j}\nabla_{p}^{j}\nabla_{q}^{i-j}H_{N}(p_{n},q_{n})\left[\left(p_{n+1}-p_{n}\right)^{\otimes
j},\left(q_{n+1}-q_{n}\right)^{\otimes\left(i-j\right)}\right]\right\\},$
(155)

where it is implied that the dimensions corresponding to $\nabla_{p}^{j}$
multiply with those corresponding to $p_{n+1}-p_{n}$, and those for
$\nabla_{q}^{i-j}$ multiply with those for $q_{n+1}-q_{n}$.777This is
suggested by the matching indices in $j$ and $i-j$ and the respectively
similar ordering of $\nabla_{p}^{j}\nabla_{q}^{i-j}$ with
$\left[\left(p_{n+1}-p_{n}\right)^{\otimes
j},\left(q_{n+1}-q_{n}\right)^{\otimes\left(i-j\right)}\right]$. We proceed by
plugging the iterates (5) of symplectic Euler into (155):

$\displaystyle H_{N}(p_{n+1},q_{n+1})-H_{N}(p_{n},q_{n})=$ (156)
$\displaystyle\sum_{i=1}^{\infty}\frac{\eta^{i}}{i!}\Big{(}-\nabla_{q}G\Big{|}_{q=q_{n}}\cdot\nabla_{p}+\nabla_{p}F\Big{|}_{p=p_{n+1}}\cdot\nabla_{q}\Big{)}^{i}H_{N}(p_{n},q_{n}),$
(157)

or alternatively,

$H_{N}(p_{n+1},q_{n+1})-H_{N}(p_{n},q_{n})=\\\
\sum_{i=1}^{\infty}\frac{\eta^{i}}{i!}\left\\{\sum_{j=0}^{i}\left(-1\right)^{j}\binom{i}{j}\left[\nabla_{p}^{j}\nabla_{q}^{i-j}H_{N}(p_{n},q_{n}),\nabla_{q}G(q_{n})^{\otimes
j},\nabla_{p}F(p_{n+1})^{\otimes\left(i-j\right)}\right]\right\\}.$ (158)

The fact that $\nabla_{p}F$ is evaluated at $p_{n+1}$ requires us to also
approximate $F(p_{n+1})$ with Taylor series:

$\displaystyle F(p_{n+1})$
$\displaystyle=F\left(p_{n}-\eta\nabla_{q}G(q_{n})\right)$
$\displaystyle=\left.F\left(p_{n}\right)-\eta\left.\left[\nabla_{p}F,\nabla_{q}G\right]\right|_{p_{n},q_{n}}+\frac{\eta^{2}}{2!}\left.\left[\nabla_{p}^{2}F,\nabla_{q}G^{\otimes
2}\right]\right|_{p_{n},q_{n}}-\frac{\eta^{3}}{3!}\left[\nabla_{p}^{3}F,\nabla_{q}G^{\otimes
3}\right]\right|_{p_{n},q_{n}}+\cdots$
$\displaystyle=F\left(p_{n}\right)+\eta\left.\\{F,G\\}\right|_{p_{n},q_{n}}+\frac{\eta^{2}}{2!}\left.\\{\\{F,G\\},G\\}\right|_{p_{n},q_{n}}+\frac{\eta^{3}}{3!}\left.\\{\\{\\{F,G\\},G\\},G\\}\right|_{p_{n},q_{n}}+\cdots$
$\displaystyle=\left.e^{\eta\\{\cdot,G\\}}F\right|_{p_{n},q_{n}}.$ (159)

We proceed by plugging (159) into (157):

$\displaystyle
H_{N}(p_{n+1},q_{n+1})-H_{N}(p_{n},q_{n})=\left.\sum_{i=1}^{\infty}\frac{\eta^{i}}{i!}\left[-\nabla_{q}G\left(q_{n}\right)\cdot\nabla_{p}+\left.\nabla_{p}\left(e^{\eta\\{\cdot,G\\}}F\right)\right|_{p_{n},q_{n}}\cdot\nabla_{q}\right]^{i}H_{N}\right|_{p_{n},q_{n}}.$
(160)

For (160), we note that the evaluations at $p_{n},q_{n}$ both inside and
outside the brackets are intentional. The reader should interpret this
notation in the sense that, when the differential operator inside the brackets
is repeatedly applied to $H_{N}$, the derivatives with respect to $p$ and $q$
do not apply to $\nabla_{q}G\left(q_{n}\right)$ or
$\left.\nabla_{p}\left(e^{\eta\\{\cdot,G\\}}F\right)\right|_{p_{n},q_{n}}$, as
one would normally expect from the product rule.

We proceed by rewriting the terms contained in
$\widetilde{H}_{N}(p_{n+1},q_{n+1})-\widetilde{H}_{N}(p_{n},q_{n})$ as the sum
of $N+1$ Taylor series. Using the exponential operator series expansion, we
can further rewrite (160) in the following form:

$\displaystyle H_{N}(p_{n+1},q_{n+1})-H_{N}(p_{n},q_{n})=$
$\displaystyle\left.\left\\{\exp\left[\eta\left(-\nabla_{q}G\left(q_{n}\right)\cdot\nabla_{p}+\left.\nabla_{p}\left(\exp\left(\eta\left\\{\cdot,G\right\\}\right)F\right)\right|_{p_{n},q_{n}}\cdot\nabla_{q}\right)\right]-1\right\\}H_{N}\right|_{p_{n},q_{n}}.$
(161)

Then, by taking a Taylor series of (161) about $\eta=0$, the
$\mathcal{O}\left(\eta^{j}\right)$ term in (161) is

$\displaystyle C_{j,k}:=\Omega_{j}(H_{k})\quad\forall
j\in\mathbb{N},k\in\mathbb{N}_{0},$ (162)

where
$\Psi_{j}:C^{r+j}\left(\mathcal{P}\times\mathcal{Q},\mathbb{R}\right)\rightarrow
C^{r}\left(\mathcal{P}\times\mathcal{Q},\mathbb{R}\right)$
($r\in\mathbb{N}_{0}$) is the differential operator

$\displaystyle\Omega_{j}\left(f\right):=\frac{1}{j!}\left.\frac{\partial^{(j)}}{\partial\eta^{(j)}}\left\\{\exp\left[\eta\left(-\nabla_{q}G\left(q_{n}\right)\cdot\nabla_{p}+\left.\nabla_{p}\left(\exp\left(\eta\left\\{\cdot,G\right\\}\right)F\right)\right|_{p_{n},q_{n}}\cdot\nabla_{q}\right)\right]-1\right\\}\right|_{\eta=0}f$
(163)

for all $f\in C^{r+j}\left(\mathcal{P}\times\mathcal{Q},\mathbb{R}\right)$ and
$j\in\mathbb{N}$.

To get the conservation error of the $k$th-order truncated modified
Hamiltonian, $\widetilde{H}_{\eta}^{(k)}$, we combine the first $k$
modifications and truncate the Taylor series at
$\mathcal{O}\left(\eta^{k+1}\right)$ using the Lagrange form of the remainder
for multivariate Taylor series. Hence, we have the following lemma.

######  Lemma 18.

Suppose that the domains $\mathcal{P}$ and $\mathcal{Q}$ are convex and
closed, and assume that $F$ and $G$ are sufficiently smooth. If the condition

$\displaystyle\sum_{j=0}^{m-1}\left.C_{m-j,j}\right|_{p,q}=0,\quad\text{ for
all }~{}1\leq m\leq N+1,$ (164)

holds for all $(p,q)\in\mathcal{P}\times\mathcal{Q}$ and that
$|C_{m-j+1,j}|\leq L$ holds for all $(p,q)\in\mathcal{P}\times\mathcal{Q}$,
then we have that
$\widetilde{H}_{\eta}^{(N)}(p_{n+1},q_{n+1})-\widetilde{H}_{\eta}^{(N)}(p_{n},q_{n})\leq\eta^{N+2}(N+1)L$.

###### Proof.

Since $\mathcal{P}$ and $\mathcal{Q}$ are convex and closed, we can apply the
Lagrange form of the remainder in Taylor’s theorem [64] to (161) for a
remainder at any fixed order in $j+k=N$ (i.e., a fixed order of truncation in
$\eta$). Doing so gives

$\displaystyle\left(H_{0}(p_{n+1},q_{n+1})-H_{0}(p_{n},q_{n})\right)$
$\displaystyle=\left[\eta
C_{1,0}+\eta^{2}C_{2,0}+\dots+\eta^{N+1}C_{N+1,0}\right]_{z=z_{n}}+\eta^{N+2}\left.C_{N+2,0}\right|_{z=\zeta_{0}}$
$\displaystyle\eta^{1}\left(H_{1}(p_{n+1},q_{n+1})-H_{1}(p_{n},q_{n})\right)$
$\displaystyle=\left[\eta^{2}C_{1,1}+\eta^{3}C_{2,1}+\dots+\eta^{N+1}C_{N,1}\right]_{z=z_{n}}+\eta^{N+2}\left.C_{N+1,1}\right|_{z=\zeta_{1}}$
$\displaystyle\eta^{2}\left(H_{2}(p_{n+1},q_{n+1})-H_{2}(p_{n},q_{n})\right)$
$\displaystyle=\left[\eta^{3}C_{1,2}+\eta^{4}C_{2,2}+\dots+\eta^{N+1}C_{N-1,2}\right]_{z=z_{n}}+\eta^{N+2}\left.C_{N,2}\right|_{z=\zeta_{2}}$
$\displaystyle\vdots\qquad$ $\displaystyle=\qquad\vdots$
$\displaystyle\eta^{N}\left(H_{N}(p_{n+1},q_{n+1})-H_{N}(p_{n},q_{n})\right)$
$\displaystyle=\eta^{N+1}\left.C_{1,N}\right|_{z=z_{n}}+\eta^{N+2}\left.C_{2,N}\right|_{z=\zeta_{N}}$
(165)

for the order-$N$ truncated modified Hamiltonian, where
$\zeta_{0},\zeta_{1},\dots,\zeta_{N}$ lie on the convex line segment
$L_{n}:=\left\\{z\in\mathbb{R}^{d}\times\mathbb{R}^{d}:z=\left(1-\lambda\right)z_{n+1}+\lambda
z_{n},0\leq\lambda\leq 1\right\\}$ between $z_{n}$ and $z_{n+1}$. Thus,

$\displaystyle\left|\widetilde{H}_{\eta}^{(N)}(p_{n+1},q_{n+1})-\widetilde{H}_{\eta}^{(N)}(p_{n},q_{n})\right|$
$\displaystyle=\left|\sum_{m=1}^{N+1}\eta^{m}\sum_{j=0}^{m-1}\left.C_{m-j,j}\right|_{p_{n},q_{n}}+\eta^{N+2}\sum_{m=0}^{N}C_{N+2-m,m}|_{z=\zeta_{m}}\right|$
$\displaystyle=\left|\sum_{m=1}^{N+1}0+\eta^{N+2}\sum_{m=0}^{N}C_{N+2-m,m}|_{z=\zeta_{m}}\right|$
$\displaystyle\leq\eta^{N+2}\sum_{m=0}^{N}\left|C_{N+2-m,m}|_{z=\zeta_{m}}\right|$
$\displaystyle\leq\eta^{N+2}(N+1)L.$ (166)

∎

It now remains to show that the sum along each diagonal in (165) actually
cancels to zero except for that which corresponds to the last coefficient at
order $\eta^{N+2}$. To partially answer this question, in the next two
subsections, we show that the following lemma holds:

######  Lemma 19.

For $N=0,1,2,3$, if $F:\mathcal{P}\times\mathbb{R}$ and
$G:\mathcal{Q}\times\mathbb{R}$ have derivatives up to order $N+2$ on their
respective domains, then we have that $\sum_{j=0}^{i}C_{i+1-j,j}=0$ for all
$i\leq N$ on all of $\mathcal{P}\times\mathcal{Q}$. Moreover, for
$N=0,1,\dots,10$, when $d=1$ and $F,G$ have derivatives up to order $N+2$ on
their respective domains, then we have $\sum_{j=0}^{i}C_{i+1-j,j}=0$ for all
$i\leq N$ on $\mathcal{P}\times\mathcal{Q}$.

Once Lemma 19 has been proven, we could use that and Lemma 18 to show that
$\widetilde{H}_{\eta}^{(N)}(p_{n+1},q_{n+1})-\widetilde{H}_{\eta}^{(N)}(p_{n},q_{n})=\mathcal{O}(\eta^{N+2})$.
However, at least up to $j+k=5$ (i.e., up to $N=3$), we will show that each
$C_{j,k}$ can be expressed as a linear combination of products of higher-order
derivatives of $F$ and $G$ such that, for each product in this linear
combination, the sum of the orders of derivatives in the product adds up to
$2\left(j+k\right)=2N+4$, no individual derivative is of higher order than
$j+k=N+2$, and there are $N+3$ of such terms. Thus, we can show Conjecture 1
for (in theory) arbitrarily high $N\in\mathbb{N}_{0}$ by assuming that $F$ and
$G$ are both $L$-smooth of orders $1,\dots,N+2$, since in that case, (166) can
be simplified to

$\displaystyle\left|\widetilde{H}_{\eta}^{\left(N\right)}\left(p_{n+1},q_{n+1}\right)-\widetilde{H}_{\eta}^{\left(N\right)}\left(p_{n},q_{n}\right)\right|\leq\Psi\left(N\right)L^{N+3}\eta^{N+2}$
(167)

for some function $\Psi:\mathbb{N}_{0}\rightarrow\mathbb{Q}$ to be determined.

Using the aforementioned steps, we show how to verify (167) for orders
$N\in\\{0,1,2,3\\}$ and for arbitrary $d$ in Appendix F.1 by direct
computation. We also show how this can be verified by computer for orders
$N\in\\{0,1,\dots,10\\}$ and $d=1$ in Appendix F.2.

###  F.1 Verification up to $N=3$ via direct computation

We start by computing $\Psi_{j}$ for $j=1,\dots,5$, since all will be used
when checking Conjecture 1 up to $N=3$ by hand.

##### Calculation of $\Omega_{j}$ up to $j=5$

$\displaystyle\Omega_{1}(\cdot)$
$\displaystyle=-\nabla_{p}(\cdot)\nabla_{q}G+\nabla_{q}(\cdot)\nabla_{p}F$
(168) $\displaystyle\Omega_{2}(\cdot)$
$\displaystyle=\nabla_{q}(\cdot)\nabla_{p}\\{F,G\\}+\frac{1}{2}\nabla_{pp}(\cdot)\left[\nabla_{q}G^{\otimes
2}\right]-\nabla_{pq}(\cdot)[\nabla_{q}G,\nabla_{p}F]+\frac{1}{2}\nabla_{qq}(\cdot)\left[\nabla_{p}F^{\otimes
2}\right]$ (169) $\displaystyle\Omega_{3}(\cdot)$
$\displaystyle=\frac{1}{2}\nabla_{q}(\cdot)\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]+\nabla_{pq}(\cdot)[\nabla_{q}G,\nabla_{pp}F\nabla_{q}G]-\nabla_{qq}(\cdot)[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad-\frac{1}{6}\nabla_{ppp}(\cdot)[\nabla_{q}(G)^{\otimes
3}]+\frac{1}{2}\nabla_{ppq}(\cdot)[\nabla_{q}G^{\otimes 2},\nabla_{p}F]$
$\displaystyle\quad-\frac{1}{2}\nabla_{pqq}(\cdot)[\nabla_{q}G,\nabla_{p}F^{\otimes
2}]+\frac{1}{6}\nabla_{qqq}(\cdot)[\nabla_{p}(F)^{\otimes 3}]$ (170)
$\displaystyle\Omega_{4}(\cdot)$
$\displaystyle=\frac{1}{24}\nabla_{pppp}(\cdot)[\nabla_{q}G^{\otimes
4}]-\frac{1}{6}\nabla_{pppq}(\cdot)[\nabla_{q}G^{\otimes
3},\nabla_{p}F]-\frac{1}{2}\nabla_{ppq}(\cdot)[\nabla_{q}G^{\otimes
2},\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad+\frac{1}{4}\nabla_{ppqq}(\cdot)[\nabla_{q}G^{\otimes
2},\nabla_{p}F^{\otimes
2}]-\frac{1}{2}\nabla_{pq}(\cdot)[\nabla_{q}G,\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]]$
$\displaystyle\quad+\nabla_{pqq}(\cdot)[\nabla_{q}G,\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]-\frac{1}{6}\nabla_{pqqq}(\cdot)[\nabla_{q}G,\nabla_{p}F^{\otimes
3}]$
$\displaystyle\quad-\frac{1}{6}\nabla_{q}(\cdot)\nabla_{pppp}F[\nabla_{q}G^{\otimes
3}]+\frac{1}{2}\nabla_{qq}(\cdot)[\nabla_{p}F,\nabla_{ppp}{F}[\nabla_{q}G^{\otimes
2}]]$ $\displaystyle\quad-\frac{1}{2}\nabla_{qqq}(\cdot)[\nabla_{p}F^{\otimes
2},\nabla_{pp}F\nabla_{q}G]+\frac{1}{2}\nabla_{qq}(\cdot)[(\nabla_{pp}F\nabla_{q}G)^{\otimes
2}]+\frac{1}{24}\nabla_{qqqq}(\cdot)[\nabla_{p}F^{\otimes 4}]$ (171)

$\displaystyle\Omega_{5}(\cdot)$
$\displaystyle=-\frac{1}{120}\nabla_{ppppp}\left(\cdot\right)\left[\nabla_{q}G^{\otimes
5}\right]+\frac{1}{24}\nabla_{ppppq}\left(\cdot\right)\left[\nabla_{q}G^{\otimes
4},\nabla_{p}F\right]+\frac{1}{6}\nabla_{pppq}\left(\cdot\right)\left[\nabla_{q}G^{\otimes
3},\nabla_{pp}F\nabla_{q}G\right]$
$\displaystyle\quad-\frac{1}{12}\nabla_{pppqq}\left(\cdot\right)\left[\nabla_{q}G^{\otimes
3},\nabla_{p}F^{\otimes
2}\right]+\frac{1}{4}\nabla_{ppq}\left(\cdot\right)\left[\nabla_{q}G^{\otimes
2},\nabla_{ppp}F\left[\nabla_{q}G^{\otimes 2}\right]\right]$
$\displaystyle\quad-\frac{1}{2}\nabla_{ppqq}\left(\cdot\right)\left[\nabla_{q}G^{\otimes
2},\nabla_{p}F,\nabla_{pp}F\nabla_{q}G\right]+\frac{1}{12}\nabla_{ppqqq}\left(\cdot\right)\left[\nabla_{q}G^{\otimes
2},\nabla_{p}F^{\otimes 3}\right]$
$\displaystyle\quad+\frac{1}{6}\nabla_{pq}\left(\cdot\right)\left[\nabla_{q}G,\nabla_{pppp}F\left[\nabla_{q}G^{\otimes
3}\right]\right]-\frac{1}{2}\nabla_{pqq}\left(\cdot\right)\left[\nabla_{q}G,\nabla_{p}F,\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]$
$\displaystyle\quad+\frac{1}{2}\nabla_{pqqq}\left(\cdot\right)\left[\nabla_{q}G,\nabla_{p}F^{\otimes
2},\nabla_{pp}F\nabla_{q}G\right]-\frac{1}{2}\nabla_{pqq}\left(\cdot\right)\left[\nabla_{q}G,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]$
$\displaystyle\quad-\frac{1}{24}\nabla_{pqqqq}\left(\cdot\right)\left[\nabla_{q}G,\nabla_{p}F^{\otimes
4}\right]+\frac{1}{24}\nabla_{q}\left(\cdot\right)\nabla_{ppppp}F\left[\nabla_{q}G^{\otimes
4}\right]$
$\displaystyle\quad-\frac{1}{6}\nabla_{qq}\left(\cdot\right)\left[\nabla_{p}F,\nabla_{pppp}F\left[\nabla_{q}G^{\otimes
3}\right]\right]+\frac{1}{4}\nabla_{qqq}\left(\cdot\right)\left[\nabla_{p}F^{\otimes
2},\nabla_{ppp}F\left[\nabla_{q}G^{\otimes 2}\right]\right]$
$\displaystyle\quad-\frac{1}{6}\nabla_{qqqq}\left(\cdot\right)\left[\nabla_{p}F^{\otimes
3},\nabla_{pp}F\nabla_{q}G\right]+\frac{1}{2}\nabla_{qqq}\left(\cdot\right)\left[\nabla_{p}F,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]$
$\displaystyle\quad-\frac{1}{2}\nabla_{qq}\left(\cdot\right)\left[\nabla_{pp}F\nabla_{q}G,\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]+\frac{1}{120}\nabla_{qqqqq}\left(\cdot\right)\left[\nabla_{p}F^{\otimes
5}\right].$ (172)

##### For $N=0$:

Note $C_{1,0}$ equals to $\Omega_{1}(H_{0})$. Hence,

$\displaystyle C_{1,0}$ $\displaystyle=\Psi_{1}(H_{0})$
$\displaystyle=-\nabla_{q}G\nabla_{p}(H_{0})+\nabla_{p}F\nabla_{q}(H_{0})$
$\displaystyle=-\nabla_{q}G\nabla_{p}F+\nabla_{p}F\nabla_{q}G$
$\displaystyle=0.$ (173)

Thus, the case where $N=0$ is verified. And in this case, we do not need to
get an upper bound on the remainder, as that will be taken care of when
dealing with when $N=1$.

##### For $N=1$:

We first calculate $\Omega_{2}(H_{0})$ and $\Omega_{1}(H_{1})$:

$\displaystyle\Omega_{2}(H_{0})$
$\displaystyle=\nabla_{q}(H_{0})\nabla_{p}\\{F,G\\}+\frac{1}{2}\nabla_{pp}(H_{0})\left[\nabla_{q}G^{\otimes
2}\right]-\nabla_{pq}(H_{0})[\nabla_{q}G,\nabla_{p}F]+\frac{1}{2}\nabla_{qq}(H_{0})\left[\nabla_{p}F^{\otimes
2}\right]$ $\displaystyle=-\frac{1}{2}\nabla_{pp}F\left[\nabla_{q}G^{\otimes
2}\right]+\frac{1}{2}\nabla_{qq}G\left[\nabla_{p}F^{\otimes 2}\right],$ (174)
$\displaystyle\Omega_{1}(H_{1})$
$\displaystyle=-\nabla_{p}(H_{1})\nabla_{q}G+\nabla_{q}(H_{1})\nabla_{p}F=\frac{1}{2}\nabla_{pp}F[\nabla_{q}G^{\otimes
2}]-\frac{1}{2}\nabla_{qq}G[\nabla_{p}F^{\otimes 2}].$ (175)

We now sum (174) and (175) up:

$\displaystyle\Omega_{2}(H_{0})+\Omega_{1}(H_{1})$
$\displaystyle=-\frac{1}{2}\nabla_{pp}F\left[\nabla_{q}G^{\otimes
2}\right]+\frac{1}{2}\nabla_{qq}G\left[\nabla_{p}F^{\otimes 2}\right]$
$\displaystyle\quad+\frac{1}{2}\nabla_{pp}F[\nabla_{q}G^{\otimes
2}]-\frac{1}{2}\nabla_{qq}G[\nabla_{p}F^{\otimes 2}]$ $\displaystyle=0.$ (176)

Thus, the case where $N=1$ is verified. Furthermore, to get an upper bound on
the $N=0$ case, assuming that $F$ and $G$ are $L$-smooth of orders $1$ and
$2$, (166) gives us

$\displaystyle\left|\widetilde{H}_{\eta}^{\left(0\right)}\left(p_{n+1},q_{n+1}\right)-\widetilde{H}_{\eta}^{\left(0\right)}\left(p_{n},q_{n}\right)\right|$
$\displaystyle\leq\eta^{2}\max_{z\in
L_{n}}\left|\Psi_{2}\left(H_{0}\right)\left(z\right)\right|$
$\displaystyle\leq\frac{\eta^{2}}{2}\left(\max\left|\nabla_{pp}F\left[\nabla_{q}G^{\otimes
2}\right]\right|+\max\left|\nabla_{qq}G\left[\nabla_{p}F^{\otimes
2}\right]\right|\right)$ $\displaystyle\leq L^{3}\eta^{2}.$

The second inequality above follows from triangle inequality, and the last
from Cauchy-Schwarz inequality.

##### For $N=2$:

We first calculate $\Omega_{3}(H_{0})$, $\Omega_{2}(H_{1})$ and
$\Omega_{1}(H_{2})$:

$\displaystyle\Omega_{3}(H_{0})$
$\displaystyle=\frac{1}{2}\nabla_{q}(H_{0})\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]+\nabla_{pq}(H_{0})[\nabla_{q}G,\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad-\nabla_{qq}(H_{0})[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]-\frac{1}{6}\nabla_{ppp}(H_{0})[\nabla_{q}(G)^{\otimes
3}]$ $\displaystyle\quad+\frac{1}{2}\nabla_{ppq}(H_{0})[\nabla_{q}G^{\otimes
2},\nabla_{p}F]-\frac{1}{2}\nabla_{pqq}(H_{0})[\nabla_{p}G,\nabla_{p}F^{\otimes
2}]+\frac{1}{6}\nabla_{qqq}(H_{0})[\nabla_{p}(F)^{\otimes 3}]$
$\displaystyle=\frac{1}{2}\nabla_{q}G\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]-0-\nabla_{qq}G[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]-\frac{1}{6}\nabla_{ppp}F[\nabla_{q}(G)^{\otimes
3}]$ $\displaystyle\quad+0+0+\frac{1}{6}\nabla_{qqq}G[\nabla_{p}(F)^{\otimes
3}]$ $\displaystyle=\frac{1}{3}\nabla_{ppp}F[\nabla_{q}G^{\otimes
3}]-\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G+\frac{1}{6}\nabla_{qqq}G[\nabla_{p}(F)^{\otimes
3}],$ (177) $\displaystyle\Omega_{2}(H_{1})$
$\displaystyle=\nabla_{q}(H_{1})\nabla_{p}\\{F,G\\}+\frac{1}{2}\nabla_{pp}(H_{1})\left[\nabla_{q}G^{\otimes
2}\right]$
$\displaystyle\quad-\nabla_{pq}(H_{1})[\nabla_{q}G,\nabla_{p}F]+\frac{1}{2}\nabla_{qq}(H_{1})\left[\nabla_{p}F^{\otimes
2}\right]$
$\displaystyle=\frac{1}{2}\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G-\frac{1}{4}\nabla_{ppp}F[\nabla_{q}G^{\otimes
3}]$
$\displaystyle\quad+\frac{1}{2}\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{p}F-\frac{1}{4}\nabla_{qqq}G[\nabla_{p}F^{\otimes
3}]$ $\displaystyle=-\frac{1}{4}\nabla_{ppp}F[\nabla_{q}G^{\otimes
3}]+\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{p}F-\frac{1}{4}\nabla_{qqq}G[\nabla_{p}F^{\otimes
3}],$ (178)

$\displaystyle\Omega_{1}(H_{2})$
$\displaystyle=-\nabla_{p}(H_{2})\nabla_{q}G+\nabla_{q}(H_{2})\nabla_{p}F$
$\displaystyle=-\frac{1}{12}\nabla_{ppp}F[\nabla_{q}G^{\otimes
3}]-\frac{1}{6}\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G+\frac{1}{6}\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G+\frac{1}{12}\nabla_{qqq}G[\nabla_{p}F^{\otimes
3}]$ $\displaystyle=-\frac{1}{12}\nabla_{ppp}F[\nabla_{q}G^{\otimes
3}]+\frac{1}{12}\nabla_{qqq}G[\nabla_{p}F^{\otimes 3}].$ (179)

We now sum (177), (178) and (179):

$\displaystyle\Omega_{3}(H_{0})+\Omega_{2}(H_{1})+\Omega_{1}(H_{2})$
$\displaystyle=\frac{1}{3}\nabla_{ppp}F[\nabla_{q}G^{\otimes
3}]-\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G+\frac{1}{6}\nabla_{qqq}G[\nabla_{p}(F)^{\otimes
3}]$ $\displaystyle\quad-\frac{1}{4}\nabla_{ppp}F[\nabla_{q}G^{\otimes
3}]+\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{p}F-\frac{1}{4}\nabla_{qqq}G[\nabla_{p}F^{\otimes
3}]$ $\displaystyle\quad-\frac{1}{12}\nabla_{ppp}F[\nabla_{q}G^{\otimes
3}]+\frac{1}{12}\nabla_{qqq}G[\nabla_{p}F^{\otimes 3}]$ $\displaystyle=0$
(180)

Thus, the case where $N=2$ is verified. Moreover, to get an upper bound on the
$N=1$ case, assuming that $F$ and $G$ are $L$-smooth of orders $1$, $2$, and
$3$, (166) gives us

$\displaystyle\left|\widetilde{H}_{\eta}^{\left(1\right)}\left(p_{n+1},q_{n+1}\right)-\widetilde{H}_{\eta}^{\left(1\right)}\left(p_{n},q_{n}\right)\right|$
$\displaystyle\leq\eta^{3}\left(\max_{z\in
L_{n}}\left|\Psi_{3}\left(H_{0}\right)\left(z\right)\right|+\max_{z\in
L_{n}}\left|\Psi_{2}\left(H_{1}\right)\left(z\right)\right|\right)$
$\displaystyle\leq\eta^{3}L^{4}\left(\frac{1}{3}+1+\frac{1}{6}+\frac{1}{4}+1+\frac{1}{4}\right)$
$\displaystyle\leq 3L^{4}\eta^{3}.$

##### For $N=3$:

We first calculate $\Omega_{4}(H_{0})$, $\Omega_{3}(H_{1})$,
$\Omega_{2}(H_{2})$ and $\Omega_{1}(H_{3})$:

$\displaystyle\Omega_{4}(H_{0})$
$\displaystyle=\frac{1}{24}\nabla_{pppp}(H_{0})[\nabla_{q}G^{\otimes
4}]-\frac{1}{6}\nabla_{pppq}(H_{0})[\nabla_{q}G^{\otimes
3},\nabla_{p}F]-\frac{1}{2}\nabla_{ppq}(H_{0})[\nabla_{q}G^{\otimes
2},\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad+\frac{1}{4}\nabla_{ppqq}(H_{0})[\nabla_{q}G^{\otimes
2},\nabla_{p}F^{\otimes
2}]-\frac{1}{2}\nabla_{pq}(H_{0})[\nabla_{q}G,\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]]$
$\displaystyle\quad+\nabla_{pqq}(H_{0})[\nabla_{q}G,\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]-\frac{1}{6}\nabla_{pqqq}(H_{0})[\nabla_{q}G,\nabla_{p}F^{\otimes
3}]$
$\displaystyle\quad-\frac{1}{6}\nabla_{q}(H_{0})\nabla_{pppp}F[\nabla_{q}G^{\otimes
3}]+\frac{1}{2}\nabla_{qq}(H_{0})[\nabla_{p}F,\nabla_{ppp}{F}[\nabla_{q}G^{\otimes
2}]]$ $\displaystyle\quad-\frac{1}{2}\nabla_{qqq}(H_{0})[\nabla_{p}F^{\otimes
2},\nabla_{pp}F\nabla_{q}G]+\frac{1}{2}\nabla_{qq}(H_{0})[(\nabla_{pp}F\nabla_{q}G)^{\otimes
2}]+\frac{1}{24}\nabla_{qqqq}(H_{0})[\nabla_{p}F^{\otimes 4}]$
$\displaystyle=\frac{1}{24}\nabla_{pppp}F[\nabla_{q}G^{\otimes
4}]+0-0+0-0+0-0-\frac{1}{6}\nabla_{q}G\nabla_{pppp}F[\nabla_{q}G^{\otimes 3}]$
$\displaystyle\quad+\frac{1}{2}\nabla_{qq}G[\nabla_{p}F,\nabla_{ppp}{F}[\nabla_{q}G^{\otimes
2}]]-\frac{1}{2}\nabla_{qqq}G[\nabla_{p}F^{\otimes
2},\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad+\frac{1}{2}\nabla_{qq}G[(\nabla_{pp}F\nabla_{q}G)^{\otimes
2}]+\frac{1}{24}\nabla_{qqqq}G[\nabla_{p}F^{\otimes 4}]$
$\displaystyle=-\frac{1}{8}\nabla_{pppp}F[\nabla_{q}G^{\otimes
4}]-\frac{1}{2}\nabla_{qqq}G[\nabla_{p}F^{\otimes
2},\nabla_{pp}F\nabla_{q}G]+\frac{1}{24}\nabla_{qqqq}G[\nabla_{p}F^{\otimes
4}]$
$\displaystyle\quad+\frac{1}{2}\nabla_{qq}G[(\nabla_{pp}F\nabla_{q}G)^{\otimes
2}]+\frac{1}{2}\nabla_{qq}G[\nabla_{p}F,\nabla_{ppp}{F}[\nabla_{q}G^{\otimes
2}]],$ (181)

$\displaystyle\Omega_{3}(H_{1})$
$\displaystyle=\frac{1}{2}\nabla_{q}(H_{1})\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]+\nabla_{pq}(H_{1})[\nabla_{q}G,\nabla_{pp}F\nabla_{q}G]-\nabla_{qq}(H_{1})[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad-\frac{1}{6}\nabla_{ppp}(H_{1})[\nabla_{q}G^{\otimes
3}]+\frac{1}{2}\nabla_{ppq}(H_{1})[\nabla_{q}G^{\otimes 2},\nabla_{p}F]$
$\displaystyle\quad-\frac{1}{2}\nabla_{pqq}(H_{1})[\nabla_{q}G,\nabla_{p}F^{\otimes
2}]+\frac{1}{6}\nabla_{qqq}(H_{1})[\nabla_{p}F^{\otimes 3}]$
$\displaystyle=-\frac{1}{4}\nabla_{p}F\nabla_{qq}G\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]-\frac{1}{2}\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G+\frac{1}{2}[\nabla_{p}F^{\otimes
2}]\nabla_{qqq}G\nabla_{pp}F\nabla_{q}G$
$\displaystyle\quad+\frac{1}{12}\nabla_{pppp}F[\nabla_{q}G^{\otimes
4}]-\frac{1}{4}[\nabla_{q}G^{\otimes 2}]\nabla_{ppp}F\nabla_{qq}G\nabla_{p}F$
$\displaystyle\quad+\frac{1}{4}\nabla_{q}G\nabla_{pp}F\nabla_{qqq}G[\nabla_{p}F^{\otimes
2}]-\frac{1}{12}\nabla_{qqqq}G[\nabla_{p}F^{\otimes 4}]$
$\displaystyle=\frac{1}{12}\nabla_{pppp}F[\nabla_{q}G^{\otimes
4}]-\frac{1}{2}[\nabla_{q}G^{\otimes
2}]\nabla_{ppp}F\nabla_{qq}G\nabla_{p}F-\frac{1}{2}\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G$
$\displaystyle\quad+\frac{3}{4}[\nabla_{p}F^{\otimes
2}]\nabla_{qqq}G\nabla_{pp}F\nabla_{q}G-\frac{1}{12}\nabla_{qqqq}G[\nabla_{p}F^{\otimes
4}],$ (182)

$\displaystyle\Omega_{2}(H_{2})$
$\displaystyle=-\nabla_{q}(H_{2})\nabla_{pp}F\nabla_{q}G$
$\displaystyle\quad+\frac{1}{2}\nabla_{pp}(H_{2})\left[\nabla_{q}G^{\otimes
2}\right]-\nabla_{pq}(H_{2})[\nabla_{q}G,\nabla_{p}F]+\frac{1}{2}\nabla_{qq}(H_{2})\left[\nabla_{p}F^{\otimes
2}\right]$
$\displaystyle=-\frac{1}{6}\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G-\frac{1}{12}[\nabla_{p}F^{\otimes
2}]\nabla_{qqq}G\nabla_{pp}F\nabla_{q}G$
$\displaystyle\quad+\frac{1}{24}\nabla_{pppp}F[\nabla_{q}G^{\otimes
4}]+\frac{1}{12}\nabla_{p}F\nabla_{qq}G\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]+\frac{1}{12}\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G$
$\displaystyle\quad-\frac{1}{6}[\nabla_{q}G^{\otimes
2}]\nabla_{ppp}F\nabla_{qq}G\nabla_{p}F-\frac{1}{6}[\nabla_{p}F^{\otimes
2}]\nabla_{qqq}G\nabla_{pp}F\nabla_{q}G$
$\displaystyle\quad+\frac{1}{12}\nabla_{q}G\nabla_{pp}F\nabla_{qqq}G[\nabla_{p}F^{\otimes
2}]+\frac{1}{12}\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G\nabla_{p}F+\frac{1}{24}\nabla_{qqqq}G[\nabla_{p}F^{\otimes
4}]$ $\displaystyle=\frac{1}{24}\nabla_{pppp}F[\nabla_{q}G^{\otimes
4}]-\frac{1}{12}\nabla_{p}F\nabla_{qq}G\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]+\frac{1}{12}\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G\nabla_{p}F$
$\displaystyle\quad-\frac{1}{12}\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G-\frac{1}{6}[\nabla_{p}F^{\otimes
2}]\nabla_{qqq}G\nabla_{pp}F\nabla_{q}G+\frac{1}{24}\nabla_{qqqq}G[\nabla_{p}F^{\otimes
4}],$ (183) $\displaystyle\Omega_{1}(H_{3})$
$\displaystyle=-\nabla_{q}G\nabla_{p}(H_{3})+\nabla_{p}F\nabla_{q}(H_{3})$
$\displaystyle=\frac{1}{12}\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G+\frac{1}{12}[\nabla_{q}G^{\otimes
2}]\nabla_{ppp}F\nabla_{qq}G\nabla_{p}F$
$\displaystyle\quad-\frac{1}{12}\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G\nabla_{p}F-\frac{1}{12}\nabla_{q}G\nabla_{pp}F\nabla_{qqq}G[\nabla_{p}F^{\otimes
2}]$ $\displaystyle=\frac{1}{12}[\nabla_{q}G^{\otimes
2}]\nabla_{ppp}F\nabla_{qq}G\nabla_{p}F+\frac{1}{12}\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}$
$\displaystyle\quad-\frac{1}{12}\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G\nabla_{p}F-\frac{1}{12}\nabla_{q}G\nabla_{pp}F\nabla_{qqq}G[\nabla_{p}F^{\otimes
2}].$ (184)

We now sum (181), (182), (183) and (184):

$\displaystyle\Omega_{4}(H_{0})+\Omega_{3}(H_{1})+\Omega_{2}(H_{2})+\Omega_{1}(H_{3})=$
$\displaystyle=\left(-\frac{1}{8}+\frac{1}{12}+\frac{1}{24}\right)\nabla_{pppp}F[\nabla_{q}G^{\otimes
4}]+\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{12}-\frac{1}{12}\right)\nabla_{p}F\nabla_{qq}G\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]$
$\displaystyle\quad+\left(\frac{1}{2}-\frac{1}{2}-\frac{1}{12}+\frac{1}{12}\right)\nabla_{q}G\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{q}G+\left(\frac{1}{12}-\frac{1}{12}\right)\nabla_{p}F\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G\nabla_{p}F$
$\displaystyle\quad+\left(-\frac{1}{2}+\frac{3}{4}-\frac{1}{6}-\frac{1}{12}\right)[\nabla_{p}F^{\otimes
2}]\nabla_{qqq}G\nabla_{pp}F\nabla_{q}G+\left(\frac{1}{24}-\frac{1}{12}+\frac{1}{24}\right)\nabla_{qqqq}G[\nabla_{p}F^{\otimes
4}]$ $\displaystyle=0.$ (185)

Thus, the case where $N=3$ is verified. To get an upper bound on the $N=2$
case, assuming that $F$ and $G$ are $L$-smooth of orders $1,\dots,4$, (166)
implies

$\displaystyle\left|\widetilde{H}_{\eta}^{\left(2\right)}\left(p_{n+1},q_{n+1}\right)-\widetilde{H}_{\eta}^{\left(2\right)}\left(p_{n},q_{n}\right)\right|$
$\displaystyle\leq\eta^{4}\left(\max_{z\in
L_{n}}\left|\Psi_{4}\left(H_{0}\right)\left(z\right)\right|+\max_{z\in
L_{n}}\left|\Psi_{3}\left(H_{1}\right)\left(z\right)\right|+\max_{z\in
L_{n}}\left|\Psi_{2}\left(H_{2}\right)\left(z\right)\right|\right)$
$\displaystyle\leq\frac{49}{12}L^{5}\eta^{4}.$

##### For $N=4$ (abridged):

Since we are only showing Conjecture 1 up to $N=3$, we do not need to show
that the diagonals of (165) cancel at $N=4$. However, we still need to find
the remainder for $N=3$, which requires computing
$\Psi_{5}\left(H_{0}\right)$, $\Psi_{4}\left(H_{1}\right)$,
$\Psi_{3}\left(H_{2}\right)$, and $\Psi_{2}\left(H_{3}\right)$.

$\displaystyle\Omega_{5}(H_{0})$
$\displaystyle=-\frac{1}{120}\nabla_{ppppp}\left(H_{0}\right)\left[\nabla_{q}G^{\otimes
5}\right]+\frac{1}{24}\nabla_{ppppq}\left(H_{0}\right)\left[\nabla_{q}G^{\otimes
4},\nabla_{p}F\right]+\frac{1}{120}\nabla_{qqqqq}\left(H_{0}\right)\left[\nabla_{p}F^{\otimes
5}\right]$
$\displaystyle\quad+\frac{1}{6}\nabla_{pppq}\left(H_{0}\right)\left[\nabla_{q}G^{\otimes
3},\nabla_{pp}F\nabla_{q}G\right]-\frac{1}{12}\nabla_{pppqq}\left(H_{0}\right)\left[\nabla_{q}G^{\otimes
3},\nabla_{p}F^{\otimes 2}\right]$
$\displaystyle\quad+\frac{1}{4}\nabla_{ppq}\left(H_{0}\right)\left[\nabla_{q}G^{\otimes
2},\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]-\frac{1}{2}\nabla_{ppqq}\left(H_{0}\right)\left[\nabla_{q}G^{\otimes
2},\nabla_{p}F,\nabla_{pp}F\nabla_{q}G\right]$
$\displaystyle\quad+\frac{1}{12}\nabla_{ppqqq}\left(H_{0}\right)\left[\nabla_{q}G^{\otimes
2},\nabla_{p}F^{\otimes
3}\right]+\frac{1}{6}\nabla_{pq}\left(H_{0}\right)\left[\nabla_{q}G,\nabla_{pppp}F\left[\nabla_{q}G^{\otimes
3}\right]\right]$
$\displaystyle\quad-\frac{1}{2}\nabla_{pqq}\left(H_{0}\right)\left[\nabla_{q}G,\nabla_{p}F,\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]+\frac{1}{2}\nabla_{pqqq}\left(H_{0}\right)\left[\nabla_{q}G,\nabla_{p}F^{\otimes
2},\nabla_{pp}F\nabla_{q}G\right]$
$\displaystyle\quad-\frac{1}{2}\nabla_{pqq}\left(H_{0}\right)\left[\nabla_{q}G,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]-\frac{1}{24}\nabla_{pqqqq}\left(H_{0}\right)\left[\nabla_{q}G,\nabla_{p}F^{\otimes
4}\right]$
$\displaystyle\quad+\frac{1}{24}\nabla_{q}\left(H_{0}\right)\nabla_{ppppp}F\left[\nabla_{q}G^{\otimes
4}\right]-\frac{1}{6}\nabla_{qq}\left(H_{0}\right)\left[\nabla_{p}F,\nabla_{pppp}F\left[\nabla_{q}G^{\otimes
3}\right]\right]$
$\displaystyle\quad+\frac{1}{4}\nabla_{qqq}\left(H_{0}\right)\left[\nabla_{p}F^{\otimes
2},\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]-\frac{1}{6}\nabla_{qqqq}\left(H_{0}\right)\left[\nabla_{p}F^{\otimes
3},\nabla_{pp}F\nabla_{q}G\right]$
$\displaystyle\quad+\frac{1}{2}\nabla_{qqq}\left(H_{0}\right)\left[\nabla_{p}F,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]-\frac{1}{2}\nabla_{qq}\left(H_{0}\right)\left[\nabla_{pp}F\nabla_{q}G,\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]$
$\displaystyle=-\frac{1}{120}\nabla_{ppppp}F\left[\nabla_{q}G^{\otimes
5}\right]+0+\frac{1}{120}\nabla_{qqqqq}G\left[\nabla_{p}F^{\otimes
5}\right]+0-0+0-0+0+0-0+0$
$\displaystyle\quad-0-0+\frac{1}{24}\nabla_{q}G\nabla_{ppppp}F\left[\nabla_{q}G^{\otimes
4}\right]-\frac{1}{6}\nabla_{qq}G\left[\nabla_{p}F,\nabla_{pppp}F\left[\nabla_{q}G^{\otimes
3}\right]\right]$
$\displaystyle\quad+\frac{1}{4}\nabla_{qqq}G\left[\nabla_{p}F^{\otimes
2},\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]-\frac{1}{6}\nabla_{qqqq}G\left[\nabla_{p}F^{\otimes
3},\nabla_{pp}F\nabla_{q}G\right]$
$\displaystyle\quad+\frac{1}{2}\nabla_{qqq}G\left[\nabla_{p}F,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]-\frac{1}{2}\nabla_{qq}G\left[\nabla_{pp}F\nabla_{q}G,\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]$
$\displaystyle=\frac{1}{30}\nabla_{ppppp}F\left[\nabla_{q}G^{\otimes
5}\right]+\frac{1}{120}\nabla_{qqqqq}G\left[\nabla_{p}F^{\otimes
5}\right]-\frac{1}{6}\nabla_{qq}G\left[\nabla_{p}F,\nabla_{pppp}F\left[\nabla_{q}G^{\otimes
3}\right]\right]$
$\displaystyle\quad+\frac{1}{4}\nabla_{qqq}G\left[\nabla_{p}F^{\otimes
2},\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right]-\frac{1}{6}\nabla_{qqqq}G\left[\nabla_{p}F^{\otimes
3},\nabla_{pp}F\nabla_{q}G\right]$
$\displaystyle\quad+\frac{1}{2}\nabla_{qqq}G\left[\nabla_{p}F,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]-\frac{1}{2}\nabla_{qq}G\left[\nabla_{pp}F\nabla_{q}G,\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right]\right],$ (186)

$\displaystyle\Omega_{4}(H_{1})$
$\displaystyle=\frac{1}{24}\nabla_{pppp}(H_{1})[\nabla_{q}G^{\otimes
4}]-\frac{1}{6}\nabla_{pppq}(H_{1})[\nabla_{q}G^{\otimes
3},\nabla_{p}F]-\frac{1}{2}\nabla_{ppq}(H_{1})[\nabla_{q}G^{\otimes
2},\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad+\frac{1}{4}\nabla_{ppqq}(H_{1})[\nabla_{q}G^{\otimes
2},\nabla_{p}F^{\otimes
2}]-\frac{1}{2}\nabla_{pq}(H_{1})[\nabla_{q}G,\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]]$
$\displaystyle\quad+\nabla_{pqq}(H_{1})[\nabla_{q}G,\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]-\frac{1}{6}\nabla_{pqqq}(H_{1})[\nabla_{q}G,\nabla_{p}F^{\otimes
3}]$
$\displaystyle\quad-\frac{1}{6}\nabla_{q}(H_{1})\nabla_{pppp}F[\nabla_{q}G^{\otimes
3}]+\frac{1}{2}\nabla_{qq}(H_{1})[\nabla_{p}F,\nabla_{ppp}{F}[\nabla_{q}G^{\otimes
2}]]$ $\displaystyle\quad-\frac{1}{2}\nabla_{qqq}(H_{1})[\nabla_{p}F^{\otimes
2},\nabla_{pp}F\nabla_{q}G]+\frac{1}{2}\nabla_{qq}(H_{1})[(\nabla_{pp}F\nabla_{q}G)^{\otimes
2}]+\frac{1}{24}\nabla_{qqqq}(H_{1})[\nabla_{p}F^{\otimes 4}]$
$\displaystyle=-\frac{1}{48}\nabla_{ppppp}F[\nabla_{q}G^{\otimes
5}]+\frac{1}{12}\nabla_{pppp}F[\nabla_{q}G^{\otimes
3},\nabla_{qq}G\nabla_{p}F]+\frac{1}{4}\nabla_{ppp}F[\nabla_{q}G^{\otimes
2},\nabla_{qq}G\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad-\frac{1}{8}\nabla_{ppp}F[\nabla_{q}G^{\otimes
2},\nabla_{qqq}G\left[\nabla_{p}F^{\otimes
2}\right]]+\frac{1}{4}\nabla_{pp}F[\nabla_{q}G,\nabla_{qq}G\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]]$
$\displaystyle\quad-\frac{1}{2}\nabla_{qqq}G[\nabla_{pp}F\nabla_{q}G,\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]+\frac{1}{12}\nabla_{qqqq}G[\nabla_{pp}F\nabla_{q}G,\nabla_{p}F^{\otimes
3}]$
$\displaystyle\quad+\frac{1}{12}\nabla_{pppp}F[\nabla_{qq}G\nabla_{p}F,\nabla_{q}G^{\otimes
3}]-\frac{1}{4}\nabla_{qqq}G[\nabla_{p}F^{\otimes
2},\nabla_{ppp}{F}[\nabla_{q}G^{\otimes 2}]]$
$\displaystyle\quad+\frac{1}{4}\nabla_{qqqq}G[\nabla_{p}F^{\otimes
3},\nabla_{pp}F\nabla_{q}G]-\frac{1}{4}\nabla_{qqq}G[\nabla_{p}F,(\nabla_{pp}F\nabla_{q}G)^{\otimes
2}]-\frac{1}{48}\nabla_{qqqqq}G[\nabla_{p}F^{\otimes 5}]$
$\displaystyle=-\frac{1}{48}\nabla_{ppppp}F[\nabla_{q}G^{\otimes
5}]+\frac{1}{6}\nabla_{pppp}F[\nabla_{q}G^{\otimes
3},\nabla_{qq}G\nabla_{p}F]$
$\displaystyle\quad+\frac{1}{2}\nabla_{ppp}F[\nabla_{q}G^{\otimes
2},\nabla_{qq}G\nabla_{pp}F\nabla_{q}G]-\frac{3}{8}\nabla_{ppp}F[\nabla_{q}G^{\otimes
2},\nabla_{qqq}G\left[\nabla_{p}F^{\otimes 2}\right]]$
$\displaystyle\quad-\frac{3}{4}\nabla_{qqq}G[\nabla_{p}F,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}]+\frac{1}{3}\nabla_{qqqq}G[\nabla_{p}F^{\otimes
3},\nabla_{pp}F\nabla_{q}G]-\frac{1}{48}\nabla_{qqqqq}G[\nabla_{p}F^{\otimes
5}],$ (187)

$\displaystyle\Omega_{3}(H_{2})$
$\displaystyle=\frac{1}{2}\nabla_{q}\left(H_{2}\right)\nabla_{ppp}F[\nabla_{q}G^{\otimes
2}]+\nabla_{pq}\left(H_{2}\right)[\nabla_{q}G,\nabla_{pp}F\nabla_{q}G]-\nabla_{qq}\left(H_{2}\right)[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad-\frac{1}{6}\nabla_{ppp}\left(H_{2}\right)[\nabla_{q}(G)^{\otimes
3}]+\frac{1}{2}\nabla_{ppq}\left(H_{2}\right)[\nabla_{q}G^{\otimes
2},\nabla_{p}F]$
$\displaystyle\quad-\frac{1}{2}\nabla_{pqq}\left(H_{2}\right)[\nabla_{q}G,\nabla_{p}F^{\otimes
2}]+\frac{1}{6}\nabla_{qqq}\left(H_{2}\right)[\nabla_{p}(F)^{\otimes 3}]$
$\displaystyle=\frac{1}{12}\nabla_{ppp}F\left[\nabla_{qq}G\nabla_{pp}F\nabla_{q}G,\nabla_{q}G^{\otimes
2}\right]+\frac{1}{24}\nabla_{ppp}F\left[\nabla_{qqq}G\left[\nabla_{p}F^{\otimes
2}\right],\nabla_{q}G^{\otimes 2}\right]$
$\displaystyle\quad+\frac{1}{6}\nabla_{ppp}F\left[\nabla_{qq}G\nabla_{pp}F\nabla_{q}G,\nabla_{q}G^{\otimes
2}\right]+\frac{1}{6}\nabla_{qqq}G\left[\nabla_{p}F,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]$
$\displaystyle\quad-\frac{1}{6}\nabla_{qqq}G\left[\nabla_{p}F,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]-\frac{1}{6}\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad-\frac{1}{12}\nabla_{qqqq}G\left[\nabla_{p}F^{\otimes
3},\nabla_{pp}F\nabla_{q}G\right]-\frac{1}{72}\nabla_{ppppp}F\left[\nabla_{q}G^{\otimes
5}\right]-\frac{1}{36}\nabla_{pppp}F\left[\nabla_{qq}G\nabla_{p}F,\nabla_{q}G^{\otimes
3}\right]$
$\displaystyle\quad-\frac{1}{12}\nabla_{ppp}F\left[\nabla_{qq}G\nabla_{pp}F\nabla_{q}G,\nabla_{q}G^{\otimes
2}\right]+\frac{1}{12}\nabla_{pppp}F\left[\nabla_{q}G^{\otimes
3},\nabla_{qq}G\nabla_{p}F\right]$
$\displaystyle\quad+\frac{1}{12}\nabla_{qqq}G\left[\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2},\nabla_{p}F\right]+\frac{1}{12}\nabla_{qqq}G\left[\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right],\nabla_{p}F^{\otimes 2}\right]$
$\displaystyle\quad-\frac{1}{12}\nabla_{qqq}G\left[\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right],\nabla_{p}F^{\otimes
2}]\right]-\frac{1}{12}\nabla_{ppp}F\left[\nabla_{q}G,\left(\nabla_{qq}G\nabla_{p}F\right)^{\otimes
2}\right]$
$\displaystyle\quad-\frac{1}{12}\nabla_{qqqq}G\left[\nabla_{pp}F\nabla_{q}G,\nabla_{p}F^{\otimes
3}\right]+\frac{1}{36}\nabla_{qqqq}G\left[\nabla_{pp}F\nabla_{q}G,\nabla_{p}F^{\otimes
3}\right]$
$\displaystyle\quad+\frac{1}{12}\nabla_{qqq}G\left[\nabla_{pp}F\nabla_{qq}G\nabla_{p}F,\nabla_{p}F^{\otimes
2}\right]+\frac{1}{72}\nabla_{qqqqq}G\left[\nabla_{p}F^{\otimes 5}\right]$
$\displaystyle=\frac{1}{6}\nabla_{ppp}F\left[\nabla_{qq}G\nabla_{pp}F\nabla_{q}G,\nabla_{q}G^{\otimes
2}\right]+\frac{1}{24}\nabla_{ppp}F\left[\nabla_{qqq}G\left[\nabla_{p}F^{\otimes
2}\right],\nabla_{q}G^{\otimes 2}\right]$
$\displaystyle\quad-\frac{1}{6}\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G]$
$\displaystyle\quad-\frac{5}{36}\nabla_{qqqq}G\left[\nabla_{p}F^{\otimes
3},\nabla_{pp}F\nabla_{q}G\right]-\frac{1}{72}\nabla_{ppppp}F\left[\nabla_{q}G^{\otimes
5}\right]$
$\displaystyle\quad+\frac{1}{18}\nabla_{pppp}F\left[\nabla_{q}G^{\otimes
3},\nabla_{qq}G\nabla_{p}F\right]$
$\displaystyle\quad+\frac{1}{12}\nabla_{qqq}G\left[\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2},\nabla_{p}F\right]-\frac{1}{12}\nabla_{ppp}F\left[\nabla_{q}G,\left(\nabla_{qq}G\nabla_{p}F\right)^{\otimes
2}\right]$
$\displaystyle\quad+\frac{1}{12}\nabla_{qqq}G\left[\nabla_{pp}F\nabla_{qq}G\nabla_{p}F,\nabla_{p}F^{\otimes
2}\right]+\frac{1}{72}\nabla_{qqqqq}G\left[\nabla_{p}F^{\otimes 5}\right],$
(188)

$\displaystyle\Omega_{2}(H_{3})$
$\displaystyle=\nabla_{q}\left(H_{3}\right)\nabla_{p}\\{F,G\\}+\frac{1}{2}\nabla_{pp}\left(H_{3}\right)\left[\nabla_{q}G^{\otimes
2}\right]$
$\displaystyle\quad-\nabla_{pq}\left(H_{3}\right)[\nabla_{q}G,\nabla_{p}F]+\frac{1}{2}\nabla_{qq}\left(H_{3}\right)\left[\nabla_{p}F^{\otimes
2}\right]$
$\displaystyle=\frac{1}{12}\nabla_{qqq}G\left[\nabla_{p}F,\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2}\right]+\frac{1}{12}\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G\left[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G\right]$
$\displaystyle\quad-\frac{1}{8}\nabla_{ppp}F\left[\nabla_{qq}G\nabla_{pp}F\nabla_{q}G,\nabla_{q}G^{\otimes
2}\right]-\frac{1}{24}\nabla_{pppp}F\left[\nabla_{qq}G\nabla_{p}F,\nabla_{q}G^{\otimes
3}\right]$
$\displaystyle\quad+\frac{1}{12}\nabla_{qqq}G\left[\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2},\nabla_{p}F\right]+\frac{1}{12}\nabla_{qqq}G\left[\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right],\nabla_{p}F^{\otimes 2}\right]$
$\displaystyle\quad+\frac{1}{12}\nabla_{pp}F\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G[\nabla_{q}G,\nabla_{p}F]+\frac{1}{12}\nabla_{ppp}F\left[\nabla_{q}G,\left(\nabla_{qq}G\nabla_{p}F\right)^{\otimes
2}\right]$
$\displaystyle\quad-\frac{1}{24}\nabla_{qqqq}G\left[\nabla_{pp}F\nabla_{q}G,\nabla_{p}F^{\otimes
3}\right]-\frac{1}{8}\nabla_{qqq}G\left[\nabla_{pp}F\nabla_{qq}G\nabla_{p}F,\nabla_{p}F^{\otimes
2}\right]$
$\displaystyle=\frac{1}{6}\nabla_{qqq}G\left[\left(\nabla_{pp}F\nabla_{q}G\right)^{\otimes
2},\nabla_{p}F\right]+\frac{1}{6}\nabla_{qq}G\nabla_{pp}F\nabla_{qq}G\left[\nabla_{p}F,\nabla_{pp}F\nabla_{q}G\right]$
$\displaystyle\quad-\frac{1}{8}\nabla_{ppp}F\left[\nabla_{qq}G\nabla_{pp}F\nabla_{q}G,\nabla_{q}G^{\otimes
2}\right]-\frac{1}{24}\nabla_{pppp}F\left[\nabla_{qq}G\nabla_{p}F,\nabla_{q}G^{\otimes
3}\right]$
$\displaystyle\quad+\frac{1}{12}\nabla_{ppp}F\left[\nabla_{q}G,\left(\nabla_{qq}G\nabla_{p}F\right)^{\otimes
2}\right]+\frac{1}{12}\nabla_{qqq}G\left[\nabla_{ppp}F\left[\nabla_{q}G^{\otimes
2}\right],\nabla_{p}F^{\otimes 2}\right]$
$\displaystyle\quad-\frac{1}{24}\nabla_{qqqq}G\left[\nabla_{pp}F\nabla_{q}G,\nabla_{p}F^{\otimes
3}\right]-\frac{1}{8}\nabla_{qqq}G\left[\nabla_{pp}F\nabla_{qq}G\nabla_{p}F,\nabla_{p}F^{\otimes
2}\right].$ (189)

Hence, from (186), (187), (188), and (189), we can read off the upper bound on
the $N=3$ case as follows, assuming that $F$ and $G$ are $L$-smooth of orders
$1,\dots,5$:

$\displaystyle\left|\widetilde{H}_{\eta}^{\left(3\right)}\left(p_{n+1},q_{n+1}\right)-\widetilde{H}_{\eta}^{\left(3\right)}\left(p_{n},q_{n}\right)\right|$
$\displaystyle\leq\eta^{5}\left(\max_{z\in
L_{n}}\left|\Psi_{5}\left(H_{0}\right)\left(z\right)\right|+\max_{z\in
L_{n}}\left|\Psi_{4}\left(H_{1}\right)\left(z\right)\right|+\max_{z\in
L_{n}}\left|\Psi_{3}\left(H_{2}\right)\left(z\right)\right|+\max_{z\in
L_{n}}\left|\Psi_{2}\left(H_{3}\right)\left(z\right)\right|\right)$
$\displaystyle\leq\eta^{5}L^{6}\left(\frac{13}{8}+\frac{13}{6}+\frac{61}{72}+\frac{5}{6}\right)=\frac{197}{36}\eta^{5}L^{6}.$

In particular, from our work above, we have confirmed that
$\Psi\left(0\right)=1$, $\Psi\left(1\right)=3$, $\Psi\left(2\right)=49/12$,
and $\Psi\left(3\right)=197/36$ for arbitrary dimension $d\in\mathbb{N}$.

###  F.2 Verification up to $N=10$ via symbolic calculation of the
coefficients

Unlike in Appendix F.1, the following approach only works for $d=1$ dimensions
since SymPy [51] does not currently have support for higher-order derivative
tensors. To compute the successive terms of the modified Hamiltonian in SymPy,
we implement the recursive form of the BCH formula as stated in [21] and [76],
and proved as Lemma 2.15.3 in [76]. When applied to the setting of symplectic
Euler and Alternating Mirror Descent, we recover the following for any
$N\in\mathbb{N}$:

$\displaystyle H_{0}$ $\displaystyle=F+G,$ (190a) $\displaystyle NH_{N}$
$\displaystyle=\frac{1}{2}\left\\{H_{N-1},G-F\right\\}+\sum_{p=1}^{\lfloor(N-1)/2\rfloor}{\frac{B_{2p}}{\left(2p\right)!}\left(\mathrm{ad}_{\widetilde{H}}^{2p}\left(F+G\right)\right)_{N}},$
(190b)

where

$\displaystyle\left(\mathrm{ad}_{\widetilde{H}}^{2p}\left(F+G\right)\right)_{N}=\sum_{\begin{subarray}{c}k_{1}+\dots+k_{2p}=N-1\\\
k_{1}\geq 1,\dots,k_{2p}\geq
1\end{subarray}}{\left\\{\left(F+G\right)H_{k_{2p}-1}\cdots
H_{k_{1}-1}\right\\}}$ (191)

denotes the projection of $\mathrm{ad}_{\widetilde{H}}^{2p}\left(F+G\right)$
onto the homogeneous subspace $\mathcal{L}\left(F,G\right)_{N+1}$ consisting
of degree $N+1$ elements from $\mathcal{L}\left(F,G\right)$, as defined in
(42).