Proof. This result follows directly from 2 by setting \(h_i = 0\) for all \(i \in V\). ◻

Proof. The proof follows from 3 by setting \(x(t) = \langle C(\gamma) \rangle\) and \(B = \mu_{\max}/2\pi\), where \(\mu_{\max}\) is determined from 2 and 2. ◻

Proof. We start by expressing the product of cosine functions \(h(\mathbf{w})=\prod_{i=1}^n \cos \left(w_i \theta\right)\) using the product-to-sum identity for cosine functions. The product-to-sum identity for \(n\) cosine terms is given by: \[\prod_{k=1}^n \cos \theta_k=\frac{1}{2^n} \sum_{e \in\{-1,1\}^n} \cos \left(\sum_{i=1}^n e_i \theta_i\right),\] where \(\mathbf{e}=\left(e_1, \ldots, e_n\right)\) is an \(n\)-tuple with each \(e_i\) taking values in \(\{-1,1\}\). Applying this identity to \(h(\mathbf{w})\), we have: \[h(\mathbf{w})=\prod_{i=1}^n \cos \left(w_i \theta\right)=\frac{1}{2^n} \sum_{e \in\{-1,1\}^n} \cos \left(\sum_{i=1}^n e_i w_i \theta\right) .\] The maximum angular frequency component of \(h(\mathbf{w})\) corresponds to the term in the sum where the argument of the cosine function is maximised. This maximum occurs when each \(e_i\) is chosen to match the sign of the corresponding \(w_i\), i.e., \(e_i=\operatorname{sgn}\left(w_i\right)\). Therefore, the term with the maximum angular frequency is: \[\cos \left(\sum_{i=1}^n \operatorname{sgn}\left(w_i\right) w_i \theta\right)=\cos \left(\sum_{i=1}^n\left|w_i\right| \theta\right) .\] Thus, the maximum angular frequency of \(h(\mathbf{w})\) is given by \(\sum_{i=1}^n\left|w_i\right|\). ◻

Proof. From lemma 1, we know that the maximum angular frequency of the product \(\prod_{i=1}^n \cos \left(w_i \theta\right)\) is \(\sum_{i=1}^n\left|w_i\right|\), corresponding to the term \(\cos \left(\left(\sum_{i=1}^n\left|w_i\right|\right) \theta\right)\).

Next, consider the product of this cosine term with \(\sin \left(w_0 \theta\right)\). To analyse the frequency content of this product, we use the product-to-sum identity for sine and cosine functions, which states: \[\sin \theta \cos \varphi=\frac{1}{2}[\sin (\theta+\varphi)+\sin (\theta-\varphi)] .\] Applying this identity with \(\theta=w_0 \theta\) and \(\varphi=\sum_{i=1}^n\left|w_i\right| \theta\), we obtain: \[\sin(w_0 \theta) \cos \! \left(\! \left(\! \sum_{i = 1}^n |w_i|\! \right)\! \theta\! \right)\! = \!\frac{1}{2} \! \left[\! \sin \! \left(\! \left( \! w_0 + \sum_{i = 1}^n |w_i| \! \right) \! \theta \!\right)\! + \sin \! \left( \! \left( \! w_0 - \sum_{i = 1}^n |w_i| \! \right) \! \theta \! \right) \! \right].\] From the above expression, we observe that the maximum angular frequency of the product \(g(\mathbf{w})=\sin \left(w_0 \theta\right) \prod_{i=1}^n \cos \left(w_i \theta\right)\) depends on the sign of \(w_0\). If \(w_0>0\), the first sine term on the RHS dominates with the maximum angular frequency, while for \(w_0<0\), the second sine term attains this maximum angular frequency. In both cases, the maximum angular frequency of \(g(\mathbf{w})\) is given by \(\left|w_0\right|+\) \(\sum_{i=1}^n\left|w_i\right|\). ◻

Proof of 2. For the single spin variable \(i\), the terms involving \(\gamma\) are given by 6 . From lemma 2, the maximum frequency is: \[\omega_{\max} \left[ \left\langle C_i \right\rangle_\gamma \right] = 2 \left(|h_i| + \sum\limits_{k \in \mathcal{N}(i)} |J_{ik}| \right).\] For the two spin variables \(u\) and \(v\), the terms involving \(\gamma\) are: \[\left\langle C_{uv} \right\rangle_\gamma = \frac{J_{uv}}{2} \sin(4\beta) f_{uv}(\mathbf{J}, \mathbf{h}, \gamma) - \frac{J_{uv}}{2} \sin^2(2\beta) g_{uv}(\mathbf{J}, \mathbf{h}, \gamma),\] where \[\begin{align} f_{uv}(\mathbf{J}, \mathbf{h}, \gamma) &= \sin (2 J_{uv} \gamma)\Biggl( \underbrace{\cos (2 h_v \gamma) \prod_{w \in e} \cos (2 J_{wv} \gamma)}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}} + \underbrace{\cos (2 h_u \gamma) \prod_{w \in d} \cos (2 J_{uw} \gamma)}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}} \Biggr) , \\ g_{uv}(\mathbf{J}, \mathbf{h}, \gamma) &= \underbrace{\prod_{\substack{w \in e \\ w \notin F}} \cos (2 J_{wv} \gamma) \prod_{\substack{w \in d \\ w \notin F}} \cos (2 J_{uw} \gamma)}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};}} \Biggl( \underbrace{\cos \left(2\gamma(h_u + h_v)\right) \prod_{f \in F} \cos \left(2\gamma (J_{uf} + J_{vf})\right)}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {4};}} \nonumber \\ & \quad - \underbrace{\cos \left(2\gamma (h_u - h_v)\right) \prod_{f \in F} \cos \left(2\gamma (J_{uf} - J_{vf})\right)}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {5};}} \Biggr). \end{align}\] Using lemma 2, we have: \[\begin{align} \omega_{\max} \left[\sin (2 J_{uv} \gamma) \times \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};} \right] &= 2 \left(|J_{uv}| + |h_v| + \sum_{w \in e} |J_{wv}| \right), \\ \omega_{\max} \left[\sin (2 J_{uv} \gamma) \times \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};} \right] &= 2 \left(|J_{uv}| + |h_u| + \sum_{w \in d} |J_{uw}| \right). \end{align}\] Using lemma 1, we have: \[\begin{align} \omega_{\max} \left[ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};} \times \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {4};} \right] &= 2 \left( \sum_{\substack{w \in e \\ w \notin F}} |J_{wv}| + \sum_{\substack{w \in d \\ w \notin F}} |J_{uw}| + |h_u + h_v| + \sum_{f \in F} |J_{uf} + J_{vf}| \right),\\ \omega_{\max} \left[ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};} \times \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {5};} \right] &= 2 \left( \sum_{\substack{w \in e \\ w \notin F}} |J_{wv}| + \sum_{\substack{w \in d \\ w \notin F}} |J_{uw}| + |h_u - h_v| + \sum_{f \in F} |J_{uf} - J_{vf}| \right). \end{align}\] Thus, the maximum frequencies of \(f_{uv}(\mathbf{h}, \gamma)\) and \(g_{uv}(\mathbf{h}, \gamma)\) are: \[\begin{align} \omega_{\max} \left[ f_{uv}(\mathbf{J}, \mathbf{h}, \gamma) \right] &= 2 \left( |J_{uv}| + \max \left\{ |h_v| + \sum_{w \in e} |J_{wv}| , |h_u| + \sum_{w \in d} |J_{uw}| \right\} \right), \\ \omega_{\max} \left[ g_{uv}(\mathbf{J}, \mathbf{h}, \gamma) \right]\! &= \! 2 \! \left(\! \sum_{\substack{w \in e \\ w \notin F}} \! |J_{wv}| \!+\! \sum_{\substack{w \in d \\ w \notin F}} \! |J_{uw}| \! + \! \max \! \left\{ \! |h_u \! \pm \! h_v| \! + \! \sum_{f \in F} \! |J_{uf} \! \pm \! J_{vf}| \!\right\}\! \right). \end{align}\] Therefore, the maximum angular frequency of \(\left\langle C_{uv} \right\rangle_\gamma\) is the maximum of these two frequencies. ◻

Proof of 3. Consider the case without external fields. The maximum angular frequency of the \(\gamma\) terms in the expectation value is \[\omega_{\max} \! \left[\! \left\langle C_{uv} \right\rangle_\gamma \right] \! = \! 2 \times \max \! \left\{\! |J_{uv}| \! + \! \max \! \left\{ \! \sum_{w \in e}\! |J_{wv}| ,\! \sum_{w \in d}\! |J_{uw}|\! \right\}\! ,\! \max\! \left\{\! \sum_{f \in F}\! |J_{uf}\! \pm \! J_{vf}| \!\right\} \! \right\}.\] Since the graph is complete, the number of neighbours excluding vertices \(u\) and \(v\) is \(|e| = |d| = n-2\), and the number of triangles is \(|F| = n-2\). Therefore, we have \[\begin{align} \max \left\{ \sum_{w \in e} |J_{wv}| , \sum_{w \in d} |J_{uw}| \right\} &= n - 2, \\ \max \left\{ \sum_{f \in F} |J_{uf} \pm J_{vf}|\right\} & = 2n -4 \label{sk95tri95max}. \end{align}\tag{21}\] The maximum in the summation of the triangle terms in 21 is achieved when all edges forming a triangle with edge \(\{u, v\}\) have the same weights. Specifically, this occurs when \(\operatorname{sgn}(J_{uf}) = -\operatorname{sgn}(J_{vf})\) for all \(f \in F\). Under this condition, the maximum angular frequency of the \(\gamma\) terms scales as \(\mathcal{O}(n)\).

This scaling remains valid for models with external fields of strength \(\pm 1\), where the maximum angular frequency may differ by a constant factor compared to the case without external fields. Consequently, the maximum frequency is given by \[\nu_{\max} = \frac{\mathcal{O}(n)}{2\pi}.\] Since all weights are \(\pm1\) and each \(\gamma\) term is prefactored by 2, the period of the optimisation function is \(T_{\gamma} = \pi\). Utilising the maximum frequency and the period of the function, sampling period is determined from 4. ◻

Proof of 4. For a triangle-free graph, the maximum angular frequency of the \(\gamma\) terms in the expectation value is given by: \[\omega_{\max} \left[ \left\langle C_{uv} \right\rangle_\gamma \right] = 2 |J_{uv}| + 2\times \max \left\{ \sum_{w \in e} |J_{wv}| , \sum_{w \in d} |J_{uw}| \right\} . \label{triangle95free95max95freq}\tag{22}\] Since the graph is \(D\)-regular, each vertex has \(D-1\) neighbours excluding \(u\) and \(v\), so \(|e| = |d| = D-1\). Consequently, \[\max \left\{ \sum_{w \in e} |J_{wv}| , \sum_{w \in d} |J_{uw}| \right\} = D-1,\] which gives the maximum angular frequency as \[\omega_{\max} \left[ \left\langle C_{uv} \right\rangle_\gamma \right] = 2 + 2(D -1) = 2D.\] The corresponding maximum frequency is \(\omega_{\max}/2\pi = D /\pi\). Since each \(\gamma\) term is prefactored by 2 and the weights are \(\pm 1\), the optimisation function has a period of \(T_{\gamma} = \pi\). Using this period and the maximum frequency, the sampling period is derived from 4.

For triangle-free graphs with varying degrees, the maximum angular frequency is determined by the edge \({u, v}\) where at least one vertex has the maximum degree \(D_{\max}\). ◻

Proof of 5. Using the definitions of the coefficients \(A(\gamma)\) and \(B(\gamma)\), the expectation value for an arbitrary Ising model without external fields can be rewritten as: \[\begin{align} \langle\gamma, \beta|H_P| \gamma, \beta\rangle &= A(\gamma) \sin 4\beta - B(\gamma) \sin^2 2\beta \\ &= A(\gamma) \sin 4\beta - B(\gamma) \left(\frac{1 - \cos 4\beta}{2} \right) \\ &= A(\gamma) \sin 4\beta + \frac{B(\gamma)}{2} \cos 4\beta - \frac{B(\gamma)}{2}. \end{align}\] Using the trigonometric identity for a linear combination of sine and cosine, specifically: \[R \cos (x-\alpha)=A \sin x+B \cos x,\] the expression simplifies to: \[\langle\gamma, \beta|H_P| \gamma, \beta\rangle=R(\gamma) \cos (4 \beta-\alpha(\gamma))-\frac{B(\gamma)}{2},\] where \[R(\gamma)= \mathrm{sgn} \left[B(\gamma) \right]\sqrt{A^2(\gamma)+\frac{B^2(\gamma)}{4}}, \quad \alpha(\gamma)=\arctan \left(2 A(\gamma), B(\gamma)\right) .\] The minimum value of \(\langle\gamma, \beta|H_P| \gamma, \beta\rangle\) is thus: \[\underset{\gamma \in \mathbb{R}}{\min} \left(-\sqrt{A^2(\gamma)+\frac{B^2(\gamma)}{4}}-\frac{B(\gamma)}{2} \right),\] which is achieved by setting \(\beta^*=\dfrac{\alpha(\gamma^*)+\pi}{4}\). ◻

Proof of 6. Using the definitions of the coefficients \(A(\gamma), B(\gamma)\), and \(C(\gamma)\), the expectation value for an arbitrary 2-local Ising model with external fields can be rewritten as: \[\langle\gamma, \beta|H_P| \gamma, \beta\rangle = \sin (2 \beta) A(\gamma)+\sin (4 \beta) B(\gamma)+\sin ^2(2 \beta) C(\gamma).\] To minimise this expression with respect to \(\beta\) for a fixed \(\gamma\), consider: \[\left\langle\beta|H_\gamma| \beta\right\rangle = A\sin2 \beta + B \sin4\beta + C \sin^2 2\beta.\] This function is bounded with endpoints \(\left\langle 0\left|H_\gamma\right| 0\right\rangle=\left\langle\pi\left|H_\gamma\right| \pi\right\rangle=0\) and has a period of \(\pi\). To find the global minimum in \((0, \pi)\), let \(b=2 \beta\): \[\left\langle\beta|H_\gamma| \beta\right\rangle = A\sin b + B \sin 2b + C \sin^2 b.\] Taking the derivative with respect to \(\beta\): \[\begin{align} \frac{d \left\langle\beta|H_\gamma| \beta\right\rangle}{d \beta} &= 2 \left[ A \cos b + 2B \cos 2b + 2C \sin b \cos b \right] \\ &= 2 \left[ A \cos b + 2B (2 \cos^2 b - 1) + 2C \sin b \cos b \right] \\ &= 2 \left[ A \cos b + 4B \cos^2 b - 2B + 2C \sin b \cos b \right]. \end{align}\] Setting \(\dfrac{d \left\langle\beta|H_\gamma| \beta\right\rangle}{d \beta} = 0\) to find the minima: \[\begin{align} 0 &= A\cos b + 4B\cos^2 b - 2B + 2 C \sin b \cos b \\ 4C^2 \sin^2 b\cos^2 b &= (A\cos b + 4B\cos^2 b - 2B)^2\\ 4C^2(1-\cos^2 b )\cos^2 b &= (A\cos b + 4B\cos^2 b - 2B)^2. \end{align}\] Letting \(x=\cos b\), we get: \[4 C^2\left(1-x^2\right) x^2=\left(A x+4 B x^2-2 B\right)^2,\] which simplifies to the quartic equation: \[0 = p(x) = 4 B^2 - 4 A B x + (A^2 - 16 B^2 - 4 C^2) x^2 + 8 A B x^3 + (16 B^2 + 4 C^2) x^4. \label{eq:quartic95roots}\tag{23}\] Let \(x_1, x_2, x_3, x_4\) be the roots of \(p(x)\). The minimum of \(\left\langle\beta\left|H_\gamma\right| \beta\right\rangle\) occurs at one of the \(\beta\) values where \(x=\cos (2 \beta)\) is a root of \(p(x)\). Therefore: \[\beta_\gamma^*= \underset{\beta \in \mathcal{B}_\gamma}{\text{argmin} }\left\langle\beta|H_\gamma| \beta\right\rangle,\] where \[\mathcal{B}_\gamma=\left\{\left.\beta= \pm \frac{1}{2} \arccos \left(x_i\right) \right\rvert\, i=1,2,3,4\right\}.\] Finally, the global optimal angles \(\left(\gamma^*, \beta^*\right)\) are given by: \[\left(\gamma^*, \beta^*\right)=\underset{\gamma \in \mathbb{R}}{\text{argmin} }\left\langle\gamma, \beta_\gamma^*|H_P| \gamma, \beta_\gamma^*\right\rangle.\] ◻

Proof of 1. From 2, the maximum angular frequency for an Ising model with external fields is given by: \[\omega_{\max} \left[ \langle H_P \rangle_\gamma \right]\! = \! \max \left\{ \omega_{\max} \left[ \langle C_i \rangle_\gamma \right], \; \omega_{\max} \left[ \langle C_{uv} \rangle_\gamma \right] \! \Big| \forall i \in V, \; \forall \{u,v\} \in E \right\}.\] For a triangle-free graph with bounded maximum degree \(D_{\max}\), it follows that: \[\omega_{\max} \left[ \langle C_i \rangle_\gamma \right] = \omega_{\max} \left[ \langle C_{uv} \rangle_\gamma \right] = 2(D_{\max}+1).\] Applying 4, we derive the upper bound on the permissible spacing between consecutive samples: \[\Delta_{\gamma} < \frac{\pi}{2(D_{\max}+1)}.\] When converting an Ising model with external fields of size \(n\) (i.e., \(n\) spins or vertices) to an equivalent Ising model without external fields, the maximum degree of the resulting graph increases from \(D_{\max}\) to \(n\). According to 2, the maximum angular frequency for an Ising model without external fields can be computed using 14 . Although the resulting graph may contain triangles, the maximum contribution to the angular frequency arises from the non-triangle terms. Therefore, the maximum angular frequency simplifies to: \[\begin{align} \omega_{\max} \left[ \langle C_{uv} \rangle_\gamma \right] &= 2 \left( |J_{uv}| + \max \left\{ \sum_{w \in e} |J_{wv}|, \sum_{w \in d} |J_{uw}| \right\} \right) \\ &= 2 \left( 1 + \sum_{w \in d} |J_{uw}| \right) \\ &= 2 \left( 1 + (n - 1) \right) \\ &= 2n. \end{align}\] Applying 4 again, we obtain the bound on the maximum permissible spacing between consecutive samples for the transformed model: \[\Delta_{\gamma}^{\prime} < \frac{\pi}{2n}.\] Taking the ratio of \(\Delta_{\gamma}\) to \(\Delta_{\gamma}^{\prime}\), we have: \[\frac{\Delta_{\gamma}}{\Delta_{\gamma}^{\prime}} < \frac{n}{D_{\max} + 1}.\] Thus, when an Ising model with external fields, whose underlying graph is triangle-free with maximum degree \(D_{\max}\), is converted to an equivalent model without external fields, the maximum permissible spacing between consecutive samples decreases by a factor of \(\frac{n}{D_{\max} + 1}\). ◻

Proof. The algorithm begins by subdividing the interval \([0, 2\pi]\) into \(M\) subintervals, each of length \(\Delta_{\gamma} = \pi/M\), where \(M > 2N\). Consequently, the initial number of subintervals satisfies \(M = \mathcal{O}(N)\). Although the algorithm prunes many subintervals in each iteration—specifically those whose midpoint values are below a certain threshold—it is possible, in the worst case, to retain up to \(\mathcal{O}(N)\) subintervals.

With each subdivision step, the number of subintervals doubles. Therefore, after \(i\) iterations, the number of subintervals can grow to \(\mathcal{O}\left(N \cdot 2^i\right)\). Since the algorithm only evaluates the midpoint of each subinterval during subdivision, the total number of function evaluations after \(i\) iterations remains bounded by \(\mathcal{O}\left(N \cdot 2^i\right)\).

The stopping criterion for Green’s algorithm is that the quantity \(\tilde{q} \left(\sec(N \Delta_{\gamma}) - 1\right)\) becomes sufficiently small. Here, \(\tilde{q}\) represents the current estimate of the polynomial’s maximum modulus, and \(\Delta_{\gamma}\) is the current width of the subintervals. Given that \(\|p\|_{\infty} = 1\), we approximate \(\tilde{q} \approx 1\). For small \(\Delta{\gamma}\), we can approximate: \[\sec \left(N \Delta_\gamma\right)-1 \approx \frac{\left(N \Delta_\gamma\right)^2}{2}.\] To ensure that the algorithm’s error is below a tolerance \(\varepsilon\), we require: \[\tilde{q}\left(\sec \left(N \Delta_\gamma\right)-1\right) \lesssim \varepsilon \quad \Rightarrow \quad\left(N \Delta_\gamma\right)^2 \lesssim \varepsilon \quad \Rightarrow \Delta_{\gamma} \lesssim \frac{\sqrt{\varepsilon}}{N}.\] Since each iteration halves \(\Delta_{\gamma}\), starting from \(\Delta_{\gamma}^{(0)} = \mathcal{O}\left(\frac{1}{N}\right)\), after \(i\) halvings we have: \[\Delta_{\gamma} \approx \frac{1}{N} \frac{1}{2^i}.\] To satisfy the error tolerance, we set: \[\frac{1}{N \cdot 2^i} \lesssim \frac{\sqrt{\varepsilon}}{N} \quad \Rightarrow \quad 2^{-i} \lesssim \sqrt{\varepsilon} \quad \Rightarrow \quad 2^i \approx \frac{1}{\sqrt{\varepsilon}} .\] Substituting \(2^i \approx \frac{1}{\sqrt{\varepsilon}}\) into the total number of function evaluations, we obtain \(\mathcal{O}(N/\sqrt{\varepsilon})\). ◻

Proof. \[\begin{align} \underset{x \sim p}{\mathbb{E}} \left[ x \sin \frac{ax}{\sqrt{D}}\right] &= \underset{x \sim p}{\mathbb{E}} \left[ x \sum_{\substack{k=1 \\ \text{ odd }}}^{\infty} \frac{(-1)^{\frac{k-1}{2}}}{k!}\left(\frac{a x}{\sqrt{D}}\right)^k \right] \qquad \left(\text{using } \sin x=\sum_{\substack{k=1 \\ \text{ odd }}}^{\infty} \frac{(-1)^{\frac{k-1}{2}}}{k!} x^k\right) \\ & =\underset{x \sim p}{\mathbb{E}} \left[ \sum_{\substack{k=1 \\ \text{odd}}}^{\infty} \frac{(-1)^{\frac{k-1}{2}}}{k!} \frac{a^k}{D^{\frac{k}{2}}} \, x^{k+1} \right] \\ & = \sum_{\substack{k=1 \\ \text{odd}}}^{\infty} \underset{x \sim p}{\mathbb{E}} \left[ \frac{(-1)^{\frac{k-1}{2}}}{k!} \frac{a^k}{D^{\frac{k}{2}}} \, x^{k+1} \right] \\ &= \sum_{\substack{k=1 \\ \text{odd}}}^{\infty} \frac{(-1)^{\frac{k-1}{2}}}{k!} \frac{a^k}{D^{\frac{k}{2}}} \, \underset{x \sim p}{\mathbb{E}} \left[x^{k+1} \right] \\ &= \frac{a}{\sqrt{D}} \underset{x \sim p}{\mathbb{E}} \left[h^2\right] - \frac{1}{3!} \frac{a^3}{D^{\frac{3}{2}}} \underset{x \sim p}{\mathbb{E}} \left[h^4\right] + \mathcal{O}\left(\frac{1}{D^{\frac{5}{2}}} \right) \\ &= \frac{a}{\sqrt{D}} \underset{x \sim p}{\mathbb{E}} \left[h^2\right]+ \mathcal{O}\left(\frac{1}{D^{\frac{3}{2}}}\right). \end{align}\] ◻

Proof. Using the Taylor expansion: \[\ln (1 + x) = x - \frac{x^2}{2} + \mathcal{O}(x^3),\] we expand: \[\begin{align} \ln \left(1+\frac{a}{D}+\frac{b}{D^2}+ \mathcal{O}\left(\frac{1}{D^3}\right)\!\right)\! &= \frac{a}{D}+\frac{b}{D^2}+ \mathcal{O}\!\left(\frac{1}{D^3}\right)\! -\frac{1}{2}\left(\frac{a}{D}+\frac{b}{D^2}+\mathcal{O}\!\left(\frac{1}{D^3}\right)\!\right)^2 +\mathcal{O}\!\left(\frac{1}{D^3}\right)\\ & =\frac{a}{D}+\frac{b}{D^2}-\frac{a^2}{2 D^2}+\mathcal{O}\left(\frac{1}{D^3}\right) \\ & =\frac{a}{D}+\left(b-\frac{a^2}{2}\right) \frac{1}{D^2}+\mathcal{O}\left(\frac{1}{D^3}\right). \end{align}\] Multiplying by \(D\), we find: \[D\ln \left(1+\frac{a}{D}+\frac{b}{D^2}+ \mathcal{O}\left(\frac{1}{D^3}\right)\right) = a+\left(b-\frac{a^2}{2}\right) \frac{1}{D}+\mathcal{O}\left(\frac{1}{D^2}\right).\]

1. For the first result, we compute: \[\begin{align} \left[1+\frac{a}{D}+\frac{b}{D^2} + \mathcal{O}\left(\frac{1}{D^3}\right)\right]^D &=e^{D \ln \left(1+\frac{a}{D}+\frac{b}{D^2}+\mathcal{O}\left(\frac{1}{D^3}\right)\right)} \\ & =e^{a+\left(b-\frac{a^2}{2}\right) \frac{1}{D}+\mathcal{O}\left(\frac{1}{D^2}\right)} \\ & =e^a \cdot e^{\left(b-\frac{a^2}{2}\right) \frac{1}{D}+\mathcal{O}\left(\frac{1}{D^2}\right)} \\ & =e^a\left(1+\left(b-\frac{a^2}{2}\right) \frac{1}{D}+\mathcal{O}\left(\frac{1}{D^2}\right)\right) . \end{align}\] where the Taylor expansion: \[e^x = 1 + x + \frac{x^2}{2} + \mathcal{O}(x^3)\] is applied in the last step.

2. For the second result: \[\begin{align} \left[1+\frac{a}{D}+\frac{b}{D^2}+ \mathcal{O}\left(\frac{1}{D^3}\right)\right]^{D+1} & =e^a\left(1+\left(b-\frac{a^2}{2}\right) \frac{1}{D}+\mathcal{O}\left(\frac{1}{D^2}\right)\right)\left[1+\frac{a}{D}+\mathcal{O}\left(\frac{1}{D^2}\right)\right] \\ & =e^a\left(1+\left(b-\frac{a^2}{2}\right) \frac{1}{D}+\frac{a}{D}+\mathcal{O}\left(\frac{1}{D^2}\right)\right) \\ & =e^a\left(1+\frac{a+b-\frac{a^2}{2}}{D}+\mathcal{O}\left(\frac{1}{D^2}\right)\right). \end{align}\]

 ◻

Proof.

1. For the first result, we compute: \[\begin{align} \underset{x \sim p}{\mathbb{E}} \left[ \cos \frac{a x}{\sqrt{D}} \right] & =\underset{x \sim p}{\mathbb{E}} \left[ \sum_{\substack{k=0 \\ k \text{ even }}}^{\infty} \frac{(-1)^{k / 2}}{k!}\left(\frac{a x}{\sqrt{D}}\right)^k \right] \quad \left( \text{using } \cos x=\sum_{\substack{k=0 \\ k \text{ even }}}^{\infty} \frac{(-1)^{k / 2}}{k!} x^k \right) \\ & = \sum_{\substack{k=0 \\ k \text{ even }}}^{\infty} \underset{x \sim p}{\mathbb{E}} \left[\frac{(-1)^{k / 2}}{k!}\left(\frac{a x}{\sqrt{D}}\right)^k \right] \\ & =\sum_{\substack{k=0 \\ k \text{ even }}}^{\infty} \frac{(-1)^{k / 2}}{k!} \frac{a^k}{D^{k / 2}} \underset{x \sim p}{\mathbb{E}} \left[x^k\right] \\ &= 1-\frac{1}{2} \frac{a^2}{D} \underset{x \sim p}{\mathbb{E}} \left[x^2\right]+\frac{1}{4!} \frac{a^4}{D^2} \underset{x \sim p}{\mathbb{E}} \left[x^4\right]+ \mathcal{O}\left(\frac{1}{D^3}\right). \end{align}\]

2. Using the 27 , we expand its \(D\)-th power: \[\left( \underset{x \sim p}{\mathbb{E}} \left[ \cos \frac{a x}{\sqrt{D}} \right] \right)^D = \left[1-\frac{1}{2} \frac{a^2}{D} \underset{x \sim p}{\mathbb{E}} \left[x^2\right]+\frac{1}{4!} \frac{a^4}{D^2} \underset{x \sim p}{\mathbb{E}} \left[x^4\right]+ \mathcal{O}\left(\frac{1}{D^3}\right)\right]^D\] Let: \[\begin{align} s &=-\frac{1}{2} a^2 \underset{x \sim p}{\mathbb{E}} \left[x^2\right], \tag{30}\\ t &=\frac{1}{4!} a^4 \underset{x \sim p}{\mathbb{E}} \left[x^4\right]. \tag{31} \end{align}\] Then: \[\begin{align} t-\frac{s^2}{2} & =\frac{1}{4!} a^4 \underset{x \sim p}{\mathbb{E}} \left[x^4\right]-\frac{1}{2}\left(-\frac{1}{2} a^2 \underset{x \sim p}{\mathbb{E}} \left[x^2\right]\right)^2 \\ & =\frac{1}{4!} a^4 \underset{x \sim p}{\mathbb{E}} \left[x^4\right]-\frac{1}{8} a^4\left(\underset{x \sim p}{\mathbb{E}} \left[x^2\right]\right)^2 \\ & =\frac{1}{8} a^4\left[\frac{1}{3} \underset{x \sim p}{\mathbb{E}} \left[x^4\right]-\left(\underset{x \sim p}{\mathbb{E}} \left[x^2\right]\right)^2\right]. \end{align}\] Using 25 of lemma 6, we find: \[\left( \underset{x \sim p}{\mathbb{E}} \left[\cos \frac{a x}{\sqrt{D}} \right] \right)^D = e^{-\frac{1}{2}a^2 \underset{x \sim p}{\mathbb{E}} \left[x^2\right]} \! \left[1 + \frac{1}{8} a^4 \! \left(\frac{1}{3} \underset{x \sim p}{\mathbb{E}} \left[x^4\right] \! - \! \left(\underset{x \sim p}{\mathbb{E}} \left[x^2\right]\right)^2\right)\frac{1}{D} + \mathcal{O}\left(\frac{1}{D^2}\right) \right].\]

3. Similarly, for the \((D+1)\)-th power: \[\left( \underset{x \sim p}{\mathbb{E}} \left[\cos \frac{a x}{\sqrt{D}} \right] \right)^{D+1} = \left[1-\frac{1}{2} \frac{a^2}{D} \underset{x \sim p}{\mathbb{E}} \left[x^2\right]+\frac{1}{4!} \frac{a^4}{D^2} \underset{x \sim p}{\mathbb{E}} \left[x^4\right]+\mathcal{O}\left(\frac{1}{D^3}\right)\right]^{D+1}.\] Using the definitions of \(s\) and \(t\) from 30 and 31 , we compute: \[s+t-\frac{s^2}{2}=-\frac{1}{2} a^2 \underset{x \sim p}{\mathbb{E}} \left[x^2\right]+\frac{1}{8} a^4\left[\frac{1}{3} \underset{x \sim p}{\mathbb{E}} \left[x^4\right]-\left(\underset{x \sim p}{\mathbb{E}} \left[x^2\right]\right)^2\right].\] Using the 26 of lemma 6, we get: \[\begin{align} \left( \underset{x \sim p}{\mathbb{E}} \left[\cos \frac{a x}{\sqrt{D}} \right]\right)^{D+1} &= e^{-\frac{1}{2}a^2\underset{x \sim p}{\mathbb{E}}\left[x^2\right]} \Bigg[ \! 1 \! +\! \Bigg(\!-\frac{1}{2} a^2 \underset{x \sim p}{\mathbb{E}} \! \left[x^2\right]\!+\!\frac{1}{4!} a^4 \underset{x \sim p}{\mathbb{E}} \! \left[x^4\right] \\ &\quad -\frac{1}{8} a^4\left(\underset{x \sim p}{\mathbb{E}} \left[x^2\right]\right)^2\Bigg)\frac{1}{D} + \mathcal{O}\left(\frac{1}{D^2}\right) \Bigg]. \end{align}\]

 ◻

Proof. Using 27 of lemma 7, we have: \[\underset{x, x^{\prime} \sim p}{\mathbb{E}} \left[\cos \left(\left(x \pm x^{\prime}\right) \frac{a}{\sqrt{D}}\right) \right]= 1-\frac{1}{2} \frac{a^2}{D} \underbrace{\underset{x, x^{\prime} \sim p}{\mathbb{E}} \left[\left(x \pm x^{\prime}\right)^2 \right]}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}}+ \mathcal{O}\left(\frac{1}{D^2}\right).\] Expanding \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}\), we get: \[\begin{align} \underset{x, x^{\prime} \sim p}{\mathbb{E}} \left[\left(x \pm x^{\prime}\right)^2 \right]&=\underset{x, x^{\prime} \sim p}{\mathbb{E}}\left[x^2+x^{\prime 2} \pm 2 x x^{\prime}\right] \\ & =\underset{x \sim p}{\mathbb{E}} \left[x^2\right]+ \underset{x^{\prime} \sim p}{\mathbb{E}} \left[x^{\prime 2}\right] \pm 2\left(\underset{x \sim p}{\mathbb{E}} \left[x\right] \right)\left(\underset{x^{\prime} \sim p}{\mathbb{E}} \left[x^{\prime}\right]\right) \\ & =2 \underset{x \sim p}{\mathbb{E}} \left[x^2\right] \pm 2 \left(\underset{x \sim p}{\mathbb{E}} \left[x\right] \right)^2. \end{align}\] Substituting \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}\) back, we obtain: \[\begin{align} \underset{x, x^{\prime} \sim p}{\mathbb{E}} \left[\cos \left(\left(x \pm x^{\prime}\right) \frac{a}{\sqrt{D}}\right) \right] &= 1-\frac{1}{2} \frac{a^2}{D}\left[2 \underset{x \sim p}{\mathbb{E}} \left[x^2\right] \pm 2 \left(\underset{x \sim p}{\mathbb{E}} \left[x\right]\right)^2\right]+ \mathcal{O}\left(\frac{1}{D^2}\right)\\ &= 1-\frac{a^2}{D}\left[\underset{x \sim p}{\mathbb{E}} \left[x^2\right] \pm \left(\underset{x \sim p}{\mathbb{E}} \left[x\right] \right)^2\right]+ \mathcal{O}\left(\frac{1}{D^2}\right). \end{align}\] ◻

Proof. Recall from the proof of lemma 6, we know that: \[\ln \left(1+\frac{a}{D}+\frac{b}{D^2}+ \mathcal{O}\left(\frac{1}{D^3}\right)\right) = \frac{a}{D}+\left(b-\frac{a^2}{2}\right) \frac{1}{D^2}+\mathcal{O}\left(\frac{1}{D^3}\right).\] Multiplying by \(c D^\lambda\), we obtain: \[c D^\lambda \ln \left(1+\frac{a}{D}+\frac{b}{D^2}+\mathcal{O}\left(\frac{1}{D^2}\right)\right)=c\left[a D^{\lambda-1}+\left(b-\frac{a^2}{2}\right) D^{\lambda-2}+\mathcal{O}\left(D^{\lambda-3}\right)\right].\] Expanding the LHS of 33 , we get: \[\begin{align} \left[1+\frac{a}{D}+\frac{b}{D^2}+ \mathcal{O}\left(\frac{1}{D^3}\right)\right]^{c D^\lambda} & =\left[1+\frac{a}{D}+\frac{b}{D^2}+\mathcal{O}\left(\frac{1}{D^3}\right)\right]^{c D^\lambda} \\ & =e^{c D^\lambda \ln \left(1+\frac{a}{D}+\frac{b}{D^2}+\mathcal{O}\left(\frac{1}{D^3}\right)\right)} \\ & =e^{c\left[a D^{\lambda-1}+\left(b-\frac{a^2}{2}\right) D^{\lambda-2}+\mathcal{O}\left(D^{\lambda-3}\right)\right]}. \end{align}\] We now handle two cases based on the value of \(\lambda\).

Case 1: \(\lambda = 1\) \[\begin{align} \left[1+\frac{a}{D}+\frac{b}{D^2}+ \mathcal{O}\left(\frac{1}{D^3}\right)\right]^{c D^\lambda} & =e^{c\left[a+\left(b-\frac{a^2}{2}\right) D^{-1}+\mathcal{O}\left(D^{-2}\right)\right]} \\ & =e^{a c} e^{c\left(b-\frac{a^2}{2}\right) D^{-1}+\mathcal{O}\left(D^{-2}\right)} \\ & =e^{a c}\left(1+c\left(b-\frac{a^2}{2}\right) D^{-1}+\mathcal{O}\left(D^{-2}\right)\right). \end{align}\] Case 2: \(\lambda < 1\) \[\begin{align} \left[1+\frac{a}{D}+\frac{b}{D^2}+ \mathcal{O}\left(\frac{1}{D^3}\right)\right]^{c D^\lambda}&=e^{c\left[a D^{\lambda-1}+\left(b-\frac{a^2}{2}\right) D^{\lambda-2}+\mathcal{O}\left(D^{\lambda-3}\right)\right]} \\ & =1+c\left[a D^{\lambda-1}+\left(b-\frac{a^2}{2}\right) D^{\lambda-2}+O\left(D^{\lambda-3}\right)\right]\\ &\quad +c^2\left[a D^{\lambda-1}+\left(b-\frac{a^2}{2}\right) D^{\lambda-2}+O\left(D^{\lambda-3}\right)\right]^2 \! + \mathcal{O}\left( D^{3(\lambda-1)}\right) \\ & =1+a c D^{\lambda-1}+a^2 c^2 D^{2 \lambda-2}+\mathcal{O}\left(D^{\lambda-2}\right) \\ & =1+a c D^{\lambda-1}+\mathcal{O}\left(D^{2 \lambda-2}\right). \end{align}\] ◻

Proof. Raising 27 of lemma 7 to the power \(cD^\lambda\), we have: \[\left(\underset{x \sim p}{\mathbb{E}} \left[ \cos \frac{a x}{\sqrt{D}} \right]\right)^{cD^\lambda}= \left[1-\frac{1}{2} \frac{a^2}{D} \underset{x \sim p}{\mathbb{E}} \left[x^2 \right]+\frac{1}{4!} \frac{a^4}{D^2} \underset{x \sim p}{\mathbb{E}} \left[x^4 \right]+ \mathcal{O}\left(\frac{1}{D^3}\right)\right]^{cD^\lambda}.\] Using lemma 9, we evaluate the expression for the two cases:

Case 1: \(\lambda = 1\) \[\left(\underset{x \sim p}{\mathbb{E}} \left[\cos \frac{a x}{\sqrt{D}}\right]\right)^{cD^\lambda}= e^{-\frac{c}{2}a^2 \underset{x \sim p}{\mathbb{E}} \left[x^2\right]}\left[1 + \frac{1}{8} a^4c\left(\frac{1}{3} \underset{x \sim p}{\mathbb{E}} \left[x^4\right]-\left(\underset{x \sim p}{\mathbb{E}} \left[x^2\right]\right)^2\right)\frac{1}{D} + \mathcal{O}\left(\frac{1}{D^2}\right) \right].\] Case 2: \(0 \geq \lambda < 1\) \[\left(\underset{x \sim p}{\mathbb{E}} \left[\cos \frac{a x}{\sqrt{D}}\right]\right)^{cD^\lambda}= 1-\frac{c}{2} a^2 \underset{x \sim p}{\mathbb{E}} \left[x^2\right] D^{\lambda-1}+ \mathcal{O}\left(D^{2 \lambda-2}\right).\] ◻

Proof. Raising 32 of lemma 8 to the power \(cD^\lambda\), we get: \[\left[\underset{x, x^{\prime} \sim p}{\mathbb{E}} \left[ \cos \left(\left(x \pm x^{\prime}\right) \frac{a}{\sqrt{D}}\right) \right] \right]^{cD^{\lambda}} =\left[1-\frac{a^2}{D}\left[\underset{x \sim p}{\mathbb{E}} \left[ x^2 \right] \pm(\underset{x \sim p}{\mathbb{E}} \left[ x \right])^2\right]+ \mathcal{O}\left(\frac{1}{D^2}\right)\right]^{cD^{\lambda}}.\] Using lemma 9, we obtain the stated result. ◻

Proof. Recall that the QAOA\(_1\) cost function for an Ising model defined on a graph \(G=(V, E)\) with respect to parameters \(\beta, \gamma \in \mathbb{R}\) is \[\xi_G(\gamma, \beta)=\sum_{i \in V} C_i(\gamma, \beta; G)+\sum_{\{u, v\} \in E} C_{u v}^{(1)}(\gamma, \beta ; G)+ \sum_{\{u, v\} \in E}C_{u v}^{(2)}(\gamma, \beta ; G).\] where the functions \(C_i, C_{u v}^{(1)}\), and \(C_{u v}^{(2)}\) are defined as follows: \[\begin{align} C_i(\gamma, \beta: G) &=h_i \sin (2 \beta) \sin \left(2 h_i \gamma\right) \prod_{k \in \mathcal{N}(i)} \cos \left(2 J_{i k} \gamma\right), \\ C_{u v}^{(1)}(\gamma, \beta ; G) &=\frac{J_{u v}}{2} \sin (4 \beta) \sin \left(2 J_{u v} \gamma\right)\left\{ \cos \left( 2 h_v \gamma\right)\prod_{w \in \mathcal{N}_{v \setminus u}}\cos \left( 2 J_{w v} \gamma\right) +\cos \left(2 h_u \gamma\right)\prod_{w \in \mathcal{N}_{u \setminus v}}\cos \left( 2 J_{u w} \gamma\right)\right\}, \\ C_{u v}^{(2)}(\gamma, \beta ; G) &=-\frac{J_{u v}}{2} \sin ^2 (2 \beta)\prod_{w \in N_{v \setminus \!\! \setminus u}}\cos \left( 2 J_{w v} \gamma\right)\prod_{w \in N_{u \setminus \!\! \setminus v}}\cos \left(2 J_{u w} \gamma\right) \\ &\times \left\{\cos \left(2 \left(h_u+h_v\right) \gamma \right)\prod_{f \in N_{u v}}\cos \left(2 \left(J_{u f}+J_{v f}\right) \gamma \right) -\cos \left(2 \left(h_u-h_v\right) \gamma\right)\prod_{f \in N_{u v}}\cos \left(2 \left(J_{u f}-J_{v f}\right) \gamma\right)\right\}. \nonumber \end{align}\] Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) and \(g: \mathbb{R} \rightarrow \mathbb{R}\) be probability density functions. Consider a graph \(G=(V, E)\) where the edge weights \(J_{u v}\) are independently drawn from \(f\) and the vertex weights \(h_i\) are independently drawn from \(g\). The expected QAOA\(_1\) cost function for \(G\) with parameters \(\beta\) and \(\gamma\) is then \[\underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[\xi_G(\gamma, \beta) \right] = \sum_{i \in V} \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[C_i(\gamma, \beta: G)\right] +\sum_{\{u, v\} \in E} \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left\{C_{u v}^{(1)}(\gamma, \beta ; G)+\sum_{\{u, v\} \in E}C_{u v}^{(2)}(\gamma, \beta ; G)\right\}.\] To facilitate the analysis, we decompose the expectation into separate terms: \[\begin{align} T_1 &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[C_i(\gamma, \beta: G)\right], \\ T_2 &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[ C_{u v}^{(1)}(\gamma, \beta ; G) \right], \\ T_3 &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[ C_{u v}^{(2)}(\gamma, \beta ; G) \right]. \end{align}\] We will compute each of these terms individually to understand their contributions to the overall expected cost function.

Step 1: Computing the Term \(T_1\)

We begin by evaluating the expectation \(T_1\) of the term \(C_i(\gamma, \beta ; G)\) over the distributions of the edge weights \(J \sim f^E\) and vertex weights \(h \sim g^V\): \[\begin{align} T_1 &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[C_i(\gamma, \beta: G)\right]\\ &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[ h_i \sin (2\beta) \sin(2h_i\gamma) \prod_{k \in \mathcal{N}(i)} \cos (2J_{ik}\gamma) \right] \\ &= \sin(4\beta) \underset{h \sim g}{\mathbb{E}} \left[h \sin(2h\gamma)\right] \prod_{k \in \mathcal{N}(i)} \underset{J \sim f}{\mathbb{E}} \left[\cos (2J\gamma)\right]\\ &= \sin(4\beta) \underset{h \sim g}{\mathbb{E}} \left[h \sin(2h\gamma)\right] \left( \underset{J \sim f}{\mathbb{E}} \left[\cos (2J\gamma)\right] \right)^{|\mathcal{N}(i)|}. \end{align}\] Since the graph \(G\) is \((D+1)\)-regular, each vertex \(i\) has \(|\mathcal{N}(i)|=D+1\) neighbors. By setting \(\gamma=\frac{\alpha}{\sqrt{D}}\), the expression for \(T_1\) simplifies to: \[T_1 = \sin(4\beta) \underbrace{\underset{h \sim g}{\mathbb{E}} \left[h \sin(2h\gamma)\right]}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}} \underbrace{\left( \underset{J \sim f}{\mathbb{E}} \left[\cos (2J\gamma)\right] \right)^{D+1}}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}}.\] Next, we apply 5 and 7 to approximate the terms \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}\) and \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}\) as follows: \[\begin{align} \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};} &= \underset{h \sim g}{\mathbb{E}} \left[h \sin \frac{2h \alpha}{\sqrt{D}}\right]=\frac{2\alpha}{\sqrt{D}} \underset{h \sim g}{\mathbb{E}} \left[h^2\right]+ \mathcal{O}\left(\frac{1}{D^{3 / 2}}\right),\\ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};} &= \left( \underset{J \sim f}{\mathbb{E}} \left[\cos (2J\gamma)\right] \right)^{D+1}=e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}\left(1+\mathcal{O}\left(\frac{1}{D}\right)\right). \end{align}\] Substituting \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}\) and \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}\) back into the expression for \(T_1\), we obtain: \[\begin{align} T_1 &= \sin(4\beta) \left[ \frac{2\alpha}{\sqrt{D}} \underset{h \sim g}{\mathbb{E}} \left[h^2\right]+ \mathcal{O}\left(\frac{1}{D^{3 / 2}}\right) \right]e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}\left(1+\mathcal{O}\left(\frac{1}{D}\right)\right) \\ &= \sin(4\beta) \underset{h \sim g}{\mathbb{E}} \left[h^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}+\mathcal{O}\left(\frac{1}{D}\right). \end{align}\] Step 2: Computing the Term \(T_2\)

Next, we evaluate the expectation \(T_2\) of the term \(C_{u v}^{(1)}(\gamma, \beta ; G)\) over the distributions of the edge weights \(J \sim f^E\) and vertex weights \(h \sim g^V\): \[\begin{align} T_2 &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[C_{u v}^{(1)}(\gamma, \beta ; G) \right]\\ &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[\frac{J_{u v}}{2} \sin (4 \beta) \sin \left(2 J_{u v} \gamma\right)\left\{\cos \left( 2 h_v \gamma\right)\prod_{w \in \mathcal{N}_{v \setminus u}}\cos \left( 2 J_{w v} \gamma\right) +\cos \left(2 h_u \gamma\right)\prod_{w \in \mathcal{N}_{u \setminus v}}\cos \left( 2 J_{u w} \gamma\right)\right\} \right] \\ &= \frac{1}{2} \sin(4\beta) \underset{J \sim f}{\mathbb{E}}\left[ J \sin(2J \gamma) \right]\Bigg\{\underset{h \sim g}{\mathbb{E}}\left[\cos \left( 2 h \gamma\right)\right]\prod_{w \in \mathcal{N}_{v \setminus u}}\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \gamma\right)\right]+\underset{h \sim g}{\mathbb{E}}\left[\cos \left( 2 h \gamma\right)\right]\prod_{w \in \mathcal{N}_{u \setminus v}}\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \gamma\right)\right] \Bigg\} \\ &= \frac{1}{2} \sin(4\beta) \underset{J \sim f}{\mathbb{E}}\left[ J \sin(2J \gamma) \right] \underset{h \sim g}{\mathbb{E}}\left[\cos \left( 2 h \gamma\right)\right]\left\{\left(\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \gamma\right)\right] \right)^{|\mathcal{N}_{v\setminus u}|}+\left(\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \gamma\right)\right] \right)^{|\mathcal{N}_{u \setminus v}|}\right\}. \end{align}\] Given that the graph \(G\) is \((D+1)\)-regular, each edge \(\{u, v\}\) has neighbourhoods satisfying: \[|\mathcal{N}_{u \setminus v}| = |\mathcal{N}_{v \setminus u}| = (D+1)-1 = D.\] Substituting this into the expression and setting \(\gamma=\frac{\alpha}{\sqrt{D}}\), we further simplify \(T_2\) as: \[T_2 = \sin(4\beta) \underbrace{\underset{J \sim f}{\mathbb{E}}\left[ J \sin(2J \frac{\alpha}{\sqrt{D}}) \right]}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}} \underbrace{\underset{h \sim g}{\mathbb{E}}\left[\cos \left( 2 h \frac{\alpha}{\sqrt{D}}\right)\right]}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}} \underbrace{\left(\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \frac{\alpha}{\sqrt{D}}\right)\right]\right)^D}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};}}.\] Applying 5 and 7, we approximate the terms \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}\), \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}\), and \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};}\) as follows: \[\begin{align} \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};} &= \underset{J \sim f}{\mathbb{E}}\left[ J \sin(2J \frac{\alpha}{\sqrt{D}}) \right] = \frac{2\alpha}{\sqrt{D}} \underset{J \sim f}{\mathbb{E}}\left[J^2\right] \mathcal{O}\left( \frac{1}{D^{\frac{3}{2}}} \right), \\ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};} &= \underset{h \sim g}{\mathbb{E}}\left[\cos \left( 2 h \frac{\alpha}{\sqrt{D}}\right)\right] = 1 + \mathcal{O}\left( \frac{1}{D} \right),\\ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};} &= \left(\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \frac{\alpha}{\sqrt{D}}\right)\right]\right)^D = e^{-2\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[ J^2\right]}\left(1 + \mathcal{O}\left( \frac{1}{D} \right) \right). \end{align}\] Substituting \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}\), \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}\), and \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};}\) back into the expression for \(T_2\), we obtain: \[\begin{align} T_2 &= \sin(4\beta) \left[ \frac{2\alpha}{\sqrt{D}} \underset{J \sim f}{\mathbb{E}} \left[J^2\right]+ \mathcal{O}\left(\frac{1}{D^{3 / 2}}\right) \right]e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}\left(1+\mathcal{O}\left(\frac{1}{D}\right)\right) \\ &= \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}+\mathcal{O}\left(\frac{1}{D}\right). \end{align}\]

Step 3: Computing the Term \(T_3\)

We now evaluate the expectation \(T_3\) of the term \(C_{u v}^{(2)}(\gamma, \beta ; G)\) over the distributions of the edge weights \(J \sim f^E\) and vertex weights \(h \sim g^V\) as follows: \[\begin{align} T_3 &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[ C_{u v}^{(2)}(\gamma, \beta ; G) \right] \\ &= \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left\{ -\frac{J_{u v}}{2} \sin ^2 (2 \beta) \prod_{w \in N_{v \setminus \!\! \setminus u}} \cos \left( 2 J_{w v} \gamma\right) \prod_{w \in N_{u \setminus \!\! \setminus v}} \cos \left(2 J_{u w} \gamma\right) \right. \\ & \quad \times \left\{\cos \left(2 \left(h_u+h_v\right) \gamma \right)\prod_{f \in N_{u v}}\cos \left(2 \left(J_{u f}+J_{v f}\right) \gamma \right) \left. -\cos \left(2 \left(h_u-h_v\right) \gamma\right)\prod_{f \in N_{u v}}\cos \left(2 \left(J_{u f}-J_{v f}\right) \gamma\right)\right\} \right\} \nonumber \\ &= -\frac{1}{2} \sin^2(2\beta) \underset{J \sim f}{\mathbb{E}}\left[ J \right] \prod_{w \in N_{v \setminus \!\! \setminus u}} \underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \gamma\right) \right] \prod_{w \in N_{u \setminus \!\! \setminus v}} \underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \gamma\right) \right] \\ & \quad \times \left\{ \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h+h^{\prime})\gamma \right) \right] \prod_{f \in N_{u v}} \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J+J^{\prime})\gamma \right) \right] \right. \nonumber \\ & \quad \left. - \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h-h^{\prime})\gamma \right) \right] \prod_{f \in N_{u v}} \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J-J^{\prime})\gamma \right) \right] \right\} \nonumber \\ &= -\frac{1}{2} \sin^2(2\beta) \underset{J \sim f}{\mathbb{E}}\left[ J \right] \left(\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \gamma\right) \right]\right)^{|N_{v \setminus \!\! \setminus u}|+|N_{u \setminus \!\! \setminus v}|} \\ & \quad \times \left\{ \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h+h^{\prime})\gamma \right) \right] \left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J+J^{\prime})\gamma \right) \right]\right)^{|\mathcal{N}_{uv}|} \right. \nonumber \\ & \quad \left. - \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h-h^{\prime})\gamma \right) \right] \left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J-J^{\prime})\gamma \right) \right]\right)^{|\mathcal{N}_{uv}|} \right\} . \nonumber \end{align}\] Since \(G\) is \((D+1)\)-regular graph, we have: \[\begin{align} |\mathcal{N}(u)|&=|\mathcal{N}(v)|=D+1\\ \mathcal{N}(u) &= \mathcal{N}_{u \setminus \!\! \setminus v} \cup \mathcal{N}_{uv} \cup \{v\},\\ \mathcal{N}(v) &= \mathcal{N}_{v \setminus \!\! \setminus u} \cup \mathcal{N}_{uv} \cup \{u\}, \end{align}\] which implies: \[\begin{align} D+1&=|\mathcal{N}(u)|=\left|\mathcal{N}_{u \setminus \!\! \setminus v}\right|+\left|\mathcal{N}_{u v}\right|+1, \\ D+1&=|\mathcal{N}(v)|=\left|\mathcal{N}_{v \setminus \!\! \setminus u}\right|+\left|\mathcal{N}_{u v}\right|+1. \end{align}\] Hence, we get: \[\begin{align} & \left|\mathcal{N}_{u \setminus \!\! \setminus v}\right|+\left|\mathcal{N}_{uv}\right|=\left|\mathcal{N}_{v \setminus \!\! \setminus u}\right|+\left|\mathcal{N}_{uv}\right|=D, \\ \Rightarrow \quad & \left|\mathcal{N}_{u \setminus \!\! \setminus v}\right|=\left|\mathcal{N}_{v \setminus \!\! \setminus u}\right|,\left|\mathcal{N}_{uv}\right| \leq D. \end{align}\] Setting \(\gamma = \frac{\alpha}{\sqrt{D}}\) and using \(\left|\mathcal{N}_{u \setminus \!\! \setminus v}\right|=\left|\mathcal{N}_{v \setminus \!\! \setminus u}\right|\), \(T_3\) can be written as follows: \[\begin{align} T_3 &= -\frac{1}{2} \sin^2(2\beta) \underset{J \sim f}{\mathbb{E}}\left[ J \right] \left(\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \frac{\alpha}{\sqrt{D}}\right) \right]\right)^{2|N_{u \setminus \!\! \setminus v}|} \\ & \quad \times \Bigg\{ \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h+h^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right] \left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J+J^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\right)^{|\mathcal{N}_{uv}|} \\ & \quad - \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h-h^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right] \left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J-J^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\right)^{|\mathcal{N}_{uv}|} \Bigg\}. \end{align}\] For some \(a,b \geq 0\) and \(0 \leq \lambda, \mu \leq 1\), we let \[\begin{align} |\mathcal{N}_{u \setminus \!\! \setminus v}| &= aD^{\lambda}, \\ |\mathcal{N}_{uv}| &= bD^{\mu}. \end{align}\] such that \[\left|\mathcal{N}_{u \setminus \!\! \setminus v}\right|+\left|\mathcal{N}_{uv}\right|=D \Rightarrow a D^\lambda+b D^\mu=D.\] Using the above, \(T_3\) can be rewritten as follows: \[\begin{align} T_3 &= -\frac{1}{2} \sin^2(2\beta) \underset{J \sim f}{\mathbb{E}}\left[ J \right] \underbrace{\left(\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \frac{\alpha}{\sqrt{D}}\right) \right]\right)^{2aD^{\lambda}}}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}} \\ & \quad \times \Bigg\{ \underbrace{\underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h+h^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}} \underbrace{\left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J+J^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\right)^{bD^{\mu}}}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};}} \\ & \quad - \underbrace{\underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h-h^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {4};}} \underbrace{\left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J-J^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\right)^{bD^{\mu}}}_{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {5};}} \Bigg\}. \end{align}\] We now compute the individual terms of \(T_3\) below.

1. Computation of \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}\)

   • Case 1: \(a = 0\) (i.e., \(|\mathcal{N}_{u \setminus \!\! \setminus v}| = 0\)) \[\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};} = \left(\underset{J \sim f}{\mathbb{E}}\left[\cos \left( 2 J \frac{\alpha}{\sqrt{D}}\right) \right]\right)^0 = 1.\]

   • Case 2: \(a > 0\) (i.e., \(|\mathcal{N}_{u \setminus \!\! \setminus v}| \neq 0\))

     Using 10 with substitutions \(x \to J\), \(p \to f\), \(a \to 2\alpha\), and \(c \to 2a\), we obtain: \[\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};} = \begin{cases} e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}} \left[ J^2 \right]} \left(1 + \mathcal{O}\left( D^{-1}\right) \right) & \text{if } \lambda = 1,\\[0.25cm] 1- 4a\alpha^2 \underset{J \sim f}{\mathbb{E}} \left[ J^2 \right] D^{\lambda -1} \mathcal{O}\left(D^{2\lambda - 2} \right) & \text{if } 0 \geq \lambda < 1. \end{cases}\]

   Hence, \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};}\) can be written as follows: \[\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {1};} = \begin{cases} 1 & \text{if } a = 0, \\[0.25cm] e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}} \left[ J^2 \right]} \left(1 + \mathcal{O}\left( D^{-1}\right) \right) & \text{if } a > 0, \, \lambda = 1, \\[0.25cm] 1- 4a\alpha^2 \underset{J \sim f}{\mathbb{E}} \left[ J^2 \right] D^{\lambda -1} \mathcal{O}\left(D^{2\lambda - 2} \right) & \text{if } a >0, \, 0 \geq \lambda < 1. \end{cases}\]

2. Computation of \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}\) and \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {4};}\) \[\begin{align} \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2,4};} &= \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h \pm h^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\\ &= 1-\frac{4\alpha^2}{D}\left[\underset{h \sim g}{\mathbb{E}} \left[h^2\right] \pm(\underset{h \sim g}{\mathbb{E}} [h])^2\right]+\mathcal{O}\left(\frac{1}{D^2}\right) \\ & =1+\mathcal{O}\left(\frac{1}{D}\right) \end{align}\] where the second line was obtained using lemma 8.

3. Computation of \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};}\) and \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {5};}\)

   • Case 1: \(b = 0\) (\(|\mathcal{N}_{uv}| = 0\)) \[\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3,5};} = \left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J \pm J^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\right)^{0} = 1.\]

   • Case 2: \(b > 0\) (\(|\mathcal{N}_{uv}| > 0\))
     Using lemma 11, and setting \(x \to J\), \(p \to f\), \(a \to 2\alpha\), \(c \to b\), and \(\lambda \to \mu\), we get: \[\begin{align} \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3,5};} &= \left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J \pm J^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\right)^{bD^{\mu}} \\ &= \begin{cases} e^{-4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] \pm \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)} \left(1 + \mathcal{O}\left(D^{-1}\right) \right) & \text{if } \mu = 1,\\[0.25cm] 1 - 4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] \pm \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)D^{\mu - 1} + \mathcal{O}\left( D^{2\mu - 2}\right) &\text{if } 0 \leq \mu < 1. \end{cases} \end{align}\]

   Hence, \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3,5};}\) can be written as follows: \[\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3,5};} = \begin{cases} 1 & \text{if } b = 0, \\[0.25cm] e^{-4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] \pm \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)} \left(1 + \mathcal{O}\left(D^{-1}\right) \right) &\text{if } b > 0, \, \mu = 1,\\[0.25cm] 1 - 4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] \pm \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)D^{\mu - 1} + \mathcal{O}\left( D^{2\mu - 2}\right) & \text{if } b > 0, \, 0 \leq \mu < 1. \end{cases}\]

4. Computation of \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};} = \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {2};}\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {3};} - \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {4};}\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {5};}\)

   • Case 1: \(b = 0\) (i.e., \(|\mathcal{N}_{uv}| = 0\)) \[\begin{align} \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};} &= \left(1-\frac{4\alpha^2}{D}\left[\underset{h \sim g}{\mathbb{E}} \left[h^2\right] +(\underset{h \sim g}{\mathbb{E}} [h])^2\right]+\mathcal{O}\left(\frac{1}{D^2}\right) \right) \cdot 1 \\ & \quad - \left(1-\frac{4\alpha^2}{D}\left[\underset{h \sim g}{\mathbb{E}} \left[h^2\right] -(\underset{h \sim g}{\mathbb{E}} [h])^2\right]+\mathcal{O}\left(\frac{1}{D^2}\right) \right) \cdot 1 \nonumber \\ &= -\frac{8 \alpha^2}{D} \left( \underset{h \sim g}{\mathbb{E}} [h] \right)^2 + \mathcal{O}\left( \frac{1}{D^2}\right). \end{align}\]

   • Case 2: \(b > 0\) (i.e., \(|\mathcal{N}_{uv}| > 0\)) \[\begin{align} \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};} &= \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h+h^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right] \left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J+J^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\right)^{bD^{\mu}} \\ & \quad - \underset{h,h^{\prime} \sim g}{\mathbb{E}} \left[ \cos \left(2(h-h^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right] \left( \underset{J,J^{\prime} \sim f}{\mathbb{E}} \left[ \cos \left(2(J-J^{\prime})\frac{\alpha}{\sqrt{D}} \right) \right]\right)^{bD^{\mu}}\\ &= \left[1 + \mathcal{O}\left( D^{-1}\right) \right] \times \begin{cases} e^{-4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] + \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)} \left(1 + \mathcal{O}\left(D^{-1}\right) \right) & \text{if } \mu = 1\\ 1 - 4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] + \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)D^{\mu - 1} + \mathcal{O}\left( D^{2\mu - 2}\right) &\text{if } 0 \leq \mu < 1 \end{cases}\\ &- \left[1 + \mathcal{O}\left( D^{-1}\right) \right] \times \begin{cases} e^{-4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] - \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)} \left(1 + \mathcal{O}\left(D^{-1}\right) \right) &\text{if } \mu = 1\\ 1 - 4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] - \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)D^{\mu - 1} + \mathcal{O}\left( D^{2\mu - 2}\right) &\text{if } 0 \leq \mu < 1. \end{cases} \end{align}\] When \(\mu = 1\), \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};}\) can be further simplified as follows: \[\begin{align} \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};} &= e^{-4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] + \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)}\left(1 + \mathcal{O}\left(D^{-1}\right) \right)^2- e^{-4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] - \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)}\left(1 + \mathcal{O}\left(D^{-1}\right) \right)^2\\ &= e^{-4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] + \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)} - e^{-4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] - \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)} + \mathcal{O}\left(D^{-1}\right) \\ &= e^{-4\alpha^2b \underset{J \sim f}{\mathbb{E}}\left[ J^2 \right]} \left[ e^{-4\alpha^2b \left(\underset{J \sim f}{\mathbb{E}}\left[ J \right]\right)^2} - e^{4\alpha^2b \left(\underset{J \sim f}{\mathbb{E}}\left[ J \right]\right)^2} \right] + \mathcal{O}\left(D^{-1}\right) \\ &= -2 e^{-4\alpha^2b \underset{J \sim f}{\mathbb{E}}\left[ J^2 \right]}\sinh\left( 4\alpha^2b \left(\underset{J \sim f}{\mathbb{E}}\left[ J \right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right), \end{align}\] where in the last line, we have used the identity \(\sinh x = \frac{1}{2}\left(e^x - e^{-x} \right)\).

     When \(0 \leq \mu < 1\), \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};}\) can be simplified as follows: \[\begin{align} \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};} &= \left[1 + \mathcal{O}\left( D^{-1}\right) \right] \left[ 1 - 4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] + \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)D^{\mu - 1} + \mathcal{O}\left( D^{2\mu - 2}\right) \right] \\ & \quad - \left[1 + \mathcal{O}\left( D^{-1}\right) \right] \left[ 1 - 4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] - \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)D^{\mu - 1} + \mathcal{O}\left( D^{2\mu - 2}\right) \right] \\ &= 1 - 4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] + \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)D^{\mu - 1} + \mathcal{O}\left( D^{-1}\right) + \mathcal{O}\left( D^{2\mu - 2}\right) \\ & \quad - 1 - 4\alpha^2b \left( \underset{J \sim f}{\mathbb{E}}\left[J^2 \right] - \left( \underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 \right)D^{\mu - 1} + \mathcal{O}\left( D^{-1}\right) + \mathcal{O}\left( D^{2\mu - 2}\right) \\ &= -8 \alpha^2b \left(\underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 + \mathcal{O}\left( D^{\max(-1, 2\mu-2)}\right). \end{align}\]

   In summary, \(\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};}\) can be expressed as follows: \[\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=2pt] (char) {6};} = \begin{cases} -\frac{8 \alpha^2}{D} \left( \underset{h \sim g}{\mathbb{E}} [h] \right)^2 + \mathcal{O}\left( \frac{1}{D^2}\right) &\text{if } b = 0,\\ -2 e^{-4\alpha^2b \underset{J \sim f}{\mathbb{E}}\left[ J^2 \right]}\sinh\left( 4\alpha^2b \left(\underset{J \sim f}{\mathbb{E}}\left[ J \right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right) &\text{if } b > 0, \, \mu = 1,\\ -8 \alpha^2b \left(\underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 + \mathcal{O}\left( D^{\max(-1, 2\mu-2)}\right) &\text{if } b > 0, \, 0 \leq \mu < 1. \end{cases}\]

Having computed all the terms in \(T_3\), we now begin to evaluate the possible expressions for \(T_3\) for different possible values of \(a\), \(b\), \(\lambda\), and \(\mu\). \[\begin{align} T_3 &= -\frac{1}{2} \sin^2(2\beta) \underset{J \sim f}{\mathbb{E}}[J] \times \begin{cases} 1 &\text{if } a = 0,\\[0.25cm] e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}} \left[ J^2 \right]} \left(1 + \mathcal{O}\left( D^{-1}\right) \right) &\text{if } a > 0, \, \lambda = 1,\\[0.25cm] 1- 4a\alpha^2 \underset{J \sim f}{\mathbb{E}} \left[ J^2 \right] D^{\lambda -1} \mathcal{O}\left(D^{2\lambda - 2} \right) &\text{if } a >0, \, 0 \geq \lambda < 1, \end{cases}\\ & \quad \times \begin{cases} -\frac{8 \alpha^2}{D} \left( \underset{h \sim g}{\mathbb{E}} [h] \right)^2 + \mathcal{O}\left( \frac{1}{D^2}\right) &\text{if } b = 0,\\ -2 e^{-4\alpha^2b \underset{J \sim f}{\mathbb{E}}\left[ J^2 \right]}\sinh\left( 4\alpha^2b \left(\underset{J \sim f}{\mathbb{E}}\left[ J \right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right) &\text{if } b > 0, \, \mu = 1,\\ -8 \alpha^2b \left(\underset{J \sim f}{\mathbb{E}}\left[J \right]\right)^2 + \mathcal{O}\left( D^{\max(-1, 2\mu-2)}\right) &\text{if } b > 0, \, 0 \leq \mu < 1. \end{cases} \end{align}\]

Recall that \(aD^{\lambda} + b D^{\mu} = D\). We can divide into four cases, the possible values of \(a\), \(b\), \(\lambda\), and \(\mu\).

1. \(a=0 \quad \Rightarrow \quad D=b D^\mu \quad \Rightarrow \quad b=\mu=1\)

2. \(b=0 \quad \Rightarrow \quad D=a D^\lambda \quad \Rightarrow \quad a=\lambda=1\)

3. \(\lambda<1 \quad \Rightarrow \quad \mu=1\) (otherwise \(aD^{\lambda}+bD^{\mu} < D\) for some \(D\))

4. \(\mu = 1 \quad \Rightarrow \quad \lambda=1\) (otherwise \(aD^{\lambda}+bD^{\mu} < D\) for some \(D\))

The above gives rise to the following cases:

8pt

As indicated above, four of the above cases result in contradictions. We will consider the remaining 5 cases separately.

• Case 1: \(b = 0, \quad a = \lambda = 1\) (\(|\mathcal{N}_{u \setminus \!\! \setminus v}| = D\), \(|\mathcal{N}_{uv}| = 0\), i.e. no triangles) \[\begin{align} T_3 &= - \frac{1}{2} \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \left( 1 + \mathcal{O}\left(D^{-1}\right)\right)\left[-\frac{8\alpha^2}{D} \left(\underset{h \sim g}{\mathbb{E}}\left[h\right]\right)^2 + \mathcal{O}\left(D^{-2}\right) \right] \\ &= 4\alpha^2 \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]}\left(\underset{h \sim g}{\mathbb{E}}\left[h\right]\right)^2 \frac{1}{D} + \mathcal{O}\left(D^{-2}\right). \end{align}\]

• Case 2: \(a = 0, \quad b = \mu = 1\) (\(|\mathcal{N}_{u \setminus \!\! \setminus v}| = 0\), \(|\mathcal{N}_{uv}| = D\), i.e. Complete Graph) \[\begin{align} T_3 &= - \frac{1}{2} \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] \left[ -2 e^{-4\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4 \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right)\right] \\ &= \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}} \left[J\right] e^{-4\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4 \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right). \end{align}\]

• Case 3: \(a, b > 0\), \(\lambda = \mu = 1\), \(a+b = 1\) \[\begin{align} T_3 &= - \frac{1}{2} \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \left( 1 + \mathcal{O}\left(D^{-1}\right)\right) \\ & \quad \times \left[ -2 e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right)\right] \nonumber \\ &= \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]}\sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right) \\ &= \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4(a+b)\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right) \\ &= \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right). \end{align}\]

• Case 4: \(a, b > 0, \quad \mu = 1\), \(0 \leq \lambda < 1\) \[\begin{align} T_3 &= - \frac{1}{2} \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right]\left(1 - 4a\alpha^2\underset{J \sim f}{\mathbb{E}} \left[J^2\right]D^{\lambda-1}+ \mathcal{O}\left(D^{2\lambda-2}\right) \right) \\ & \quad \times \left[ -2 e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]}\sinh\left( 4b \alpha^2\left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right)+\mathcal{O}\left(D^{-1}\right) \right] \nonumber \\ &= \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right)\\ &\quad- 4a\alpha^2 \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] \underset{J \sim f}{\mathbb{E}}\left[J^2\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right)D^{\lambda-1} \nonumber \\ & \quad + \mathcal{O}\left(D^{\max(2\lambda-2, -1)}\right). \nonumber \end{align}\]

• Case 5: \(a, b > 0\), \(\lambda = 1\), \(0 \leq \mu < 1\) \[\begin{align} T_3 &= - \frac{1}{2} \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \left( 1 + \mathcal{O}\left(D^{-1}\right)\right) \\ & \quad \times \left[-8b \alpha^2 \left( \underset{J \sim f}{\mathbb{E}}\left[J\right] \right)^2 D^{\mu - 1} + \mathcal{O}\left(D^{\max(-1, 2\mu - 2)} \right)\right] \nonumber \\ &= 4b \alpha^2 \sin^2(2\beta) \left( \underset{J \sim f}{\mathbb{E}}\left[J\right] \right)^3 e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} D^{\mu - 1} + \mathcal{O}\left(D^{\max(-1, 2\mu - 2)} \right). \end{align}\]

Step 4: Computing the Sum of Terms \(T_2\) and \(T_3\)

Recall that \(T_2\) is computed as follows: \[T_2 = \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}+\mathcal{O}\left(D^{-1}\right).\]

• Case 1: \(b = 0, \quad a = \lambda = 1\) \[\begin{align} T_2 + T_3 &= \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}+\mathcal{O}\left(D^{-1}\right)\\ & \quad + 4\alpha^2 \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]}\left(\underset{h \sim g}{\mathbb{E}}\left[h\right]\right)^2 \frac{1}{D} + \mathcal{O}\left(D^{-2}\right) \nonumber \\ &= \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}+\mathcal{O}\left(D^{-1}\right). \end{align}\]

• Case 2: \(a = 0, \quad b = \mu = 1\) \[\begin{align} T_2 + T_3 &= \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}\\ & \quad + \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4 \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right). \end{align}\]

• Case 3: \(a, b > 0, \quad \lambda = \mu = 1\), i.e. \(a+b = 1\) \[\begin{align} T_2 + T_3 &= \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}\\ & \quad + \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] e^{-4\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{-1}\right). \end{align}\]

• Case 4: \(a, b > 0, \quad \mu = 1, \quad 0 \leq \lambda < 1\) \[\begin{align} T_2 + T_3 &= \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}+ \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right)\\ &\quad- 4a\alpha^2 \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] \underset{J \sim f}{\mathbb{E}}\left[J^2\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right)D^{\lambda-1} \\ & \quad + \mathcal{O}\left(D^{\max(2\lambda-2, -1)}\right). \end{align}\] Case 4a: \(\lambda<\frac{1}{2}\)
  
  Note that \(\lambda<\frac{1}{2} \Leftrightarrow -\frac{1}{2} > \lambda -1\). Hence, \(T_2 + T_3\) can be further simplified as follows: \[\begin{align} T_2 + T_3 &= \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}\\ & \quad + \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) + \mathcal{O}\left(D^{\lambda-1}\right). \end{align}\] Case 4b: \(\lambda=\frac{1}{2}\)
  
  Note that \(\lambda=\frac{1}{2} \Leftrightarrow -\frac{1}{2} = \lambda -1\). Hence, \(T_2 + T_3\) can be further simplified as follows: \[\begin{align} T_2 + T_3 &= \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) \\ &\quad + \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \Bigg\{ 2\alpha \sin(4\beta) e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}\\ &\quad - 4a\alpha^2 \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right) \Bigg\} \frac{1}{\sqrt{D}}\\ &\quad + \mathcal{O}\left(D^{\max(2\lambda-2, -1)}\right). \end{align}\] Case 4c: \(\lambda>\frac{1}{2}\)
  
  Note that \(\lambda>\frac{1}{2} \Leftrightarrow -\frac{1}{2} < \lambda -1\). Hence, \(T_2 + T_3\) can be further simplified as follows: \[\begin{align} T_2 + T_3 &= \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right)\\ &\quad- 4a\alpha^2 \sin^2(2 \beta) \underset{J \sim f}{\mathbb{E}}\left[J\right] \underset{J \sim f}{\mathbb{E}}\left[J^2\right]e^{-4b\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} \sinh\left(4b \alpha^2 \left(\underset{J \sim f}{\mathbb{E}}\left[J\right]\right)^2 \right)D^{\lambda-1} \\ &\quad + \mathcal{O}\left(\frac{1}{\sqrt{D}}\right). \end{align}\]

• Case 5: \(a, b > 0, \quad \lambda = 1, \quad 0 \leq \mu < 1\) \[\begin{align} T_2 + T_3 &= \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]} + \mathcal{O}\left(D^{-1} \right)\\ & \quad + 4b \alpha^2 \sin^2(2\beta) \left( \underset{J \sim f}{\mathbb{E}}\left[J\right] \right)^3 e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} D^{\mu - 1} + \mathcal{O}\left(D^{\max(-1, 2\mu - 2)} \right). \end{align}\] Case 5a: \(\mu > \frac{1}{2}\)
  
  Note that \(\mu > \frac{1}{2} \Rightarrow \mu - 1> -\frac{1}{2}\). Now, \(-\frac{1}{2} > 2\mu-2 \Leftrightarrow \mu < \frac{3}{4}\). Hence, \(T_2 + T_3\) can be further simplified as follows: \[T_2 + T_3 = 4b \alpha^2 \sin^2(2\beta) \left( \underset{J \sim f}{\mathbb{E}}\left[J\right] \right)^3 e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]} D^{\mu - 1} + \mathcal{O}\left(D^{\max(-\frac{1}{2}, 2\mu - 2)} \right).\] Case 5b: \(\mu = \frac{1}{2}\) \[\begin{align} T_2 + T_3 &= \Bigg( 2\alpha\sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]} + 4b \alpha^2 \sin^2(2\beta) \left( \underset{J \sim f}{\mathbb{E}}\left[J\right] \right)^3 e^{-4a\alpha^2 \underset{J \sim f}{\mathbb{E}}\left[J^2\right]}\Bigg)\frac{1}{\sqrt{D}} + \mathcal{O}\left(D^{-1} \right). \end{align}\] Case 5c: \(\mu < \frac{1}{2}\)
  
  Note that \(\mu < \frac{1}{2} \Rightarrow -1 \leq \mu - 1 < -\frac{1}{2}\). Hence, \(T_2 + T_3\) can be further simplified as follows: \[T_2 + T_3 = \sin(4\beta) \underset{J \sim f}{\mathbb{E}} \left[J^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}+\mathcal{O}\left(D^{\mu-1}\right).\]

Step 5: Computing the Scaled Expected QAOA\(_1\) Cost Function

Recall that the expected QAOA\(_1\) cost function for an Ising model on a graph \(G=(V, E)\) with parameters \(\beta, \gamma \in \mathbb{R}\) is given by: \[\begin{align} \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[\xi_G(\gamma, \beta) \right] &= \sum_{i \in V} \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[C_i(\gamma, \beta; G)\right] + \sum_{\{u, v\} \in E} \underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left\{C_{u v}^{(1)}(\gamma, \beta ; G)+\sum_{\{u, v\} \in E}C_{u v}^{(2)}(\gamma, \beta ; G)\right\} \\ &= \sum_{i \in V} T_1 +\sum_{\{u, v\} \in E} \left( T_2 + T_3 \right) . \end{align}\] For a \((D+1)\)-regular graph, the expectation terms \(T_1, T_2\), and \(T_3\) are identical for all vertices \(i\) and edges \(\{u, v\}\). This allows us to simplify the expression as follows: \[\underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[\xi_G(\gamma, \beta) \right] = T_1 |V| + (T_2+T_3)|E|,\] where \(n=|V|\) is the number of vertices and \(|E|\) is the number of edges in the graph \(G\). Since the graph is \((D+1)\)-regular, the number of edges satisfies \(|E|=\frac{n}{2}(D+1)\). Substituting this into the equation, we obtain: \[\underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[\xi_G(\gamma, \beta) \right] = nT_1 + \frac{n}{2}(D+1)(T_2+T_3).\] To facilitate analysis, we introduce a scaled version of the cost function by dividing the expected cost by the number of edges \(|E|\): \[\frac{1}{|E|}\underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[\xi_G(\gamma, \beta) \right] = \frac{|V|}{|E|} T_1 + (T_2+T_3).\] Calculating the ratio \(|V|/|E|\) for a \((D+1)\)-regular graph, we have: \[\frac{|V|}{|E|} = \frac{n}{\frac{n}{2}(D+1)} = \frac{2}{D+1}.\] Next, we evaluate the term \((|V|/|E|)T_1\) to get the following: \[\begin{align} \frac{|V|}{|E|}T_1 &= \frac{2}{D+1}\left[ \sin(4\beta) \underset{h \sim g}{\mathbb{E}} \left[h^2\right] \frac{2\alpha}{\sqrt{D}}e^{-2 \alpha^2 \underset{J \sim f}{\mathbb{E}} \left[J^2\right]}+\mathcal{O}\left(\frac{1}{D}\right)\right] \\ &= \mathcal{O}\left( D^{-\frac{3}{2}}\right). \end{align}\] Finally, substituting this result into the scaled cost function yields: \[\frac{1}{|E|}\underset{\substack{J \sim f^E \\ h \sim g^V}}{\mathbb{E}} \left[\xi_G(\gamma, \beta) \right] = T_2 + T_3 + \mathcal{O}\left( D^{-\frac{3}{2}}\right).\] This scaling demonstrates that as the degree \(D\) of the graph increases, the contribution from \(T_1\) becomes negligible, and the expected scaled cost function is primarily determined by the terms \(T_2\) and \(T_3\), with corrections of order \(D^{-\frac{3}{2}}\). Therefore, the expressions derived in step 4 accurately characterise the scaled expected QAOA\(_1\) cost function.

Step 6: Optimal \(\gamma^*\) Values for Leading-Order Terms

Before proceeding, we note that the subscripts for the expectation value are omitted, as the expressions in the remainder of this proof involve only the terms where \(J \sim f\).

Not all expressions derived for the different cases yield closed-form solutions for the optimal \(\alpha^*\) value. To address this, we focus on the leading-order terms, which simplify the expressions and allow for closed-form solutions for \(\alpha^*\). The leading-order expressions for the scaled QAOA\(_1\) objective are given by: \[\frac{1}{|E|}\mathbb{E} \left[\xi_G(\alpha, \beta) \right] = \begin{cases} \mathcal{C}_1(\alpha, \beta) + \mathcal{O}\left(D^{-1}\right), & b = 0, \, a = \lambda = 1, \\ \mathcal{C}_2(\alpha, \beta, 1, 1) + \mathcal{O}\left(1\right), & a = 0, \, b = \mu = 1, \\ \mathcal{C}_2(\alpha, \beta, 1, b) + \mathcal{O}\left(1\right), & a, b > 0, \, \lambda = \mu = 1, \\ \mathcal{C}_2(\alpha, \beta, b, b) + \mathcal{O}\left(1\right), & a, b > 0, \, \mu = 1, \, \lambda < 1,\\ \mathcal{C}_3(\alpha, \beta) + \mathcal{O}\left(D^{\mu-1}\right), & a, b > 0, \, \lambda = 1, \, \mu > \frac{1}{2}, \\ \mathcal{C}_1(\alpha, \beta) + \mathcal{O}\left(D^{\mu-1}\right), & a, b > 0, \, \lambda = 1, \, \mu < \frac{1}{2}, \end{cases}\] where the functions \(\mathcal{C}_1\), \(\mathcal{C}_2\), and \(\mathcal{C}_3\) are defined as: \[\begin{align} \mathcal{C}_1(\alpha, \beta) &= 2\alpha\sin(4\beta) \mathbb{E} \left[J^2\right] e^{-2 \alpha^2 \mathbb{E} \left[J^2\right]}\frac{1}{\sqrt{D}}, \\ \mathcal{C}_2(\alpha, \beta, \theta_1, \theta_2) &= \sin^2(2 \beta) \mathbb{E}\left[J\right] e^{-4 \theta_1 \alpha^2 \mathbb{E}\left[J^2\right]} \sinh\left(4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 \right),\\ \mathcal{C}_3(\alpha, \beta) &= 4b\alpha^2 \sin^2(2\beta) \mathbb{E}[J]^3 e^{-4a\alpha^2 \mathbb{E}[J^2]} D^{\mu-1}. \end{align}\] For the case \((a, b > 0, \, \lambda = 1, \, \mu = \frac{1}{2})\), the leading-order terms coincide with the original expression, but they do not admit a closed-form solution for \(\alpha^*\). In contrast, for the cases \((b = 0, \, a = \lambda = 1)\), \((a, b > 0, \, \lambda = 1, \, \mu > \frac{1}{2})\), and \((a, b > 0, \, \lambda = 1, \, \mu < \frac{1}{2})\), the leading-order terms also coincide with the original expressions, but they do allow for closed-form solutions for \(\alpha^*\).

We now proceed to compute the optimal closed-form values of the three functions \(\mathcal{C}_1\), \(\mathcal{C}_2\), and \(\mathcal{C}_3\).

Optimising \(\mathcal{C}_1\)

To determine the optimal \(\alpha\) that minimises \(\mathcal{C}_1\), we fix \(\beta = 3\pi/8\), simplifying \(\mathcal{C}_1\) to: \[\mathcal{C}_1(\alpha) = -2\alpha \mathbb{E} \left[J^2\right] e^{-2 \alpha^2 \mathbb{E} \left[J^2\right]}\frac{1}{\sqrt{D}}.\] The optimal \(\alpha\) is found by differentiating \(\mathcal{C}_1\) with respect to \(\alpha\) and setting the derivative to zero: \[\begin{align} \frac{d \, \mathcal{C}_1(\alpha)}{d \alpha} &= -2\mathbb{E} \left[J^2\right] \left( e^{-2 \alpha^2 \mathbb{E} \left[J^2\right]} - 4 \alpha^2 \mathbb{E} \left[J^2\right] e^{-2 \alpha^2 \mathbb{E} \left[J^2\right]} \right) \frac{1}{\sqrt{D}}\\ &=-2\mathbb{E} \left[J^2\right] \left( 1 - 4 \alpha^2 \mathbb{E} \left[J^2\right] \right) e^{-2 \alpha^2 \mathbb{E} \left[J^2\right]} \frac{1}{\sqrt{D}}. \end{align}\] Setting the derivative to zero yields: \[1-4 \alpha^2 \mathbb{E}\left[J^2\right]=0 \quad \Rightarrow \quad \alpha^2=\frac{1}{4 \mathbb{E}\left[J^2\right]} \quad \Rightarrow \quad \alpha= \pm \frac{1}{2 \sqrt{\mathbb{E}\left[J^2\right]}}.\] Since \(\alpha \geq 0\), the optimal value is: \[\alpha^* = \frac{1}{2 \sqrt{\mathbb{E} \left[J^2\right]}}. \label{global95opt95eqn1}\tag{34}\]

Optimising \(\mathcal{C}_2\)

For \(\mathcal{C}_2\), we set \(\beta = \pi/4\), reducing the expression to: \[\mathcal{C}_2(\alpha, \theta_1, \theta_2) = \mathbb{E}\left[J\right] e^{-4 \theta_1 \alpha^2 \mathbb{E}\left[J^2\right]} \sinh\left(4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 \right).\] To facilitate differentiation, we take the natural logarithm: \[\ln \left[\mathcal{C}_2(\alpha, \theta_1, \theta_2)\right] = \ln \left[\mathbb{E}\left[J\right]\right] -4 \theta_1 \alpha^2 \mathbb{E}\left[J^2\right] + \ln \left[ \sinh\left(4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 \right)\right].\] Differentiating with respect to \(\alpha\) gives: \[\begin{align} \frac{d \, \ln \left[\mathcal{C}_2(\alpha, \theta_1, \theta_2)\right]}{d\alpha} &= -8 \theta_1 \alpha \mathbb{E}\left[J^2\right] + 8 \theta_2 \alpha \mathbb{E}\left[J\right]^2 \coth\left(4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 \right) . \end{align}\] Setting this derivative to zero results in: \[-8 \theta_1 \alpha \mathbb{E}\left[J^2\right] + 8 \theta_2 \alpha \mathbb{E}\left[J\right]^2 \coth\left(4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 \right) = 0.\] It is evident that \(\alpha = 0\) is a critical point. To identify other critical points, we divide through by \(8 \alpha\), assuming \(\alpha \neq 0\), and solve for \(\alpha^2\): \[\begin{align} - \theta_1 \mathbb{E}\left[J^2\right] + \theta_2 \mathbb{E}\left[J\right]^2 \coth\left(4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 \right) &= 0\\ \coth\left(4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 \right) - \dfrac{\theta_1 \mathbb{E}\left[J^2\right]}{\theta_2 \mathbb{E}\left[J\right]^2} &= 0 \\ 4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 &= \coth^{-1} \left(\dfrac{\theta_1 \mathbb{E}\left[J^2\right]}{\theta_2 \mathbb{E}\left[J\right]^2}\right) \\ 4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 &= \frac{1}{2}\ln \left( \frac{\frac{\theta_1 \mathbb{E}\left[J^2\right]}{\theta_2 \mathbb{E}\left[J\right]^2} + 1}{\frac{\theta_1 \mathbb{E}\left[J^2\right]}{\theta_2 \mathbb{E}\left[J\right]^2} - 1} \right) \\ 4 \theta_2 \alpha^2 \mathbb{E}\left[J\right]^2 &= \frac{1}{2}\ln \left( \frac{\theta_1 \mathbb{E}\left[J^2\right] + \theta_2 \mathbb{E}\left[J\right]^2}{\theta_1 \mathbb{E}\left[J^2\right] - \theta_2 \mathbb{E}\left[J\right]^2} \right)\\ \alpha^2 &= \frac{1}{8\theta_2 \mathbb{E}\left[J\right]^2} \ln \left( \frac{\theta_1 \mathbb{E}\left[J^2\right] + \theta_2 \mathbb{E}\left[J\right]^2}{\theta_1 \mathbb{E}\left[J^2\right] - \theta_2 \mathbb{E}\left[J\right]^2} \right). \end{align}\] As \(\alpha \geq 0\), the optimal value for \(\alpha > 0\) is: \[\alpha^* = \sqrt{\frac{1}{8\theta_2 \mathbb{E}\left[J\right]^2} \ln \left( \frac{\theta_1 \mathbb{E}\left[J^2\right] + \theta_2 \mathbb{E}\left[J\right]^2}{\theta_1 \mathbb{E}\left[J^2\right] - \theta_2 \mathbb{E}\left[J\right]^2} \right)}. \label{global95opt95eqn2}\tag{35}\]

Optimising \(\mathcal{C}_3\)

For \(\mathcal{C}_3\), we set \(\beta = \pi/4\), reducing the expression to: \[\mathcal{C}_3(\alpha) = 4b\alpha^2 \mathbb{E}[J]^3 e^{-4a\alpha^2 \mathbb{E}[J^2]} D^{\mu-1}.\] Differentiating with respect to \(\alpha\) gives: \[\begin{align} \frac{d \, \mathcal{C}_3(\alpha)}{d\alpha} &= 4b \mathbb{E}[J]^3 \left[ 2\alpha - 4a\alpha^3 \mathbb{E}[J^2] \right] e^{-4a\alpha^2 \mathbb{E}[J^2]} D^{\mu-1}. \end{align}\] Setting the derivative to zero yields: \[2\alpha - 4a\alpha^3 \mathbb{E}[J^2] = 0.\] It is evident that \(\alpha = 0\) is a critical point. To identify other critical points, we divide through by \(2 \alpha\), assuming \(\alpha \neq 0\), and solve for \(\alpha\): \[\begin{align} 1 - 2a\alpha^2 \mathbb{E}[J^2] &= 0 \\ \alpha &= \pm \frac{1}{\sqrt{2a\mathbb{E}[J^2]}}. \end{align}\] As \(\alpha \geq 0\), the optimal value for \(\alpha > 0\) is: \[\alpha = \frac{1}{\sqrt{2a\mathbb{E}[J^2]}}. \label{global95opt95eqn3}\tag{36}\]

Optimal \(\alpha^*\) Values

The closed-form expressions for the optimal \(\alpha^* \in \mathbb{R}^+\) values for the different cases are given by: \[\alpha^* = \begin{cases} \frac{1}{2\sqrt{\mathbb{E}\left[J^2\right]}} , & b = 0, \, a = \lambda = 1, \\ \sqrt{\frac{1}{8 \mathbb{E}\left[J\right]^2} \ln \left( \frac{\mathbb{E}\left[J^2\right] + \mathbb{E}\left[J\right]^2}{\mathbb{E}\left[J^2\right] - \mathbb{E}\left[J\right]^2} \right)} , & a = 0, \, b = \mu = 1, \\ \sqrt{\frac{1}{8 b \mathbb{E}\left[J\right]^2} \ln \left( \frac{ \mathbb{E}\left[J^2\right] + b \mathbb{E}\left[J\right]^2}{ \mathbb{E}\left[J^2\right] - b \mathbb{E}\left[J\right]^2} \right)} , & a, b > 0, \, \lambda = \mu = 1, \\ \sqrt{\frac{1}{8b \mathbb{E}\left[J\right]^2} \ln \left( \frac{ \mathbb{E}\left[J^2\right] + \mathbb{E}\left[J\right]^2}{ \mathbb{E}\left[J^2\right] - \mathbb{E}\left[J\right]^2} \right)} , & a, b > 0, \, \mu = 1, \, \lambda < 1,\\ \frac{1}{\sqrt{2a\mathbb{E}\left[J^2\right]}} , & a, b > 0, \, \lambda = 1, \, \mu > \frac{1}{2}, \\ \frac{1}{2\sqrt{\mathbb{E}\left[J^2\right]}}, & a, b > 0, \, \lambda = 1, \, \mu < \frac{1}{2}. \end{cases}\] Using the relation \(\gamma = \alpha / \sqrt{D}\), the closed-form expressions for the optimal \(\gamma^* \in \mathbb{R}^+\) values can be readily obtained.

Step 7: Globally Optimal \(\alpha^*\) Coincides with First Local Optimum

In this final part of the proof, we demonstrate that the derived globally optimal value \(\alpha^*\) coincides with the first local optimum of the functions \(\mathcal{C}_1\), \(\mathcal{C}_2\), and \(\mathcal{C}_3\).

Demonstrating Global Optimum Coincides with First Local Optimum for \(\mathcal{C}_1\)

We want to show that the globally optimal parameter 34 for \(\mathcal{C}_1\) coincides with the first local optimum. Recall that the derivative of \(\mathcal{C}_1\) is given by: \[\frac{d \, \mathcal{C}_1(\alpha)}{d \alpha} = -2\mathbb{E} \left[J^2\right] \left( 1 - 4 \alpha^2 \mathbb{E} \left[J^2\right] \right) e^{-2 \alpha^2 \mathbb{E} \left[J^2\right]} \frac{1}{\sqrt{D}}.\] We begin by analysing the behaviour of \(\mathcal{C}_1^{\prime}(\alpha)\) at \(\alpha = 0\). Substituting \(\alpha = 0\) into the derivative, we find that \[\mathcal{C}_1^{\prime}(0) = -2\mathbb{E} \left[J^2\right] \frac{1}{\sqrt{D}} < 0,\] which implies that \(\mathcal{C}_1(\alpha)\) is decreasing at \(\alpha = 0\). Next, we consider the interval \(\alpha \in\left(0, \alpha^*\right)\). In this range, the term \(1-4 \alpha^2 \mathbb{E}_{\mathrm{J}}\left[J^2\right]\) remains positive because \(\alpha^2<\left(\alpha^*\right)^2\). Consequently, the derivative satisfies: \[-2\mathbb{E} \left[J^2\right] \left( 1 - 4 \alpha^2 \mathbb{E} \left[J^2\right] \right) e^{-2 \alpha^2 \mathbb{E} \left[J^2\right]} \frac{1}{\sqrt{D}} < 0.\] This implies that \(\mathcal{C}_1(\alpha)\) continues to decrease throughout the interval \(\left(0, \alpha^*\right)\). At \(\alpha=\alpha^*\), the term \(1-4 \alpha^2 \mathbb{E}_{\mathbf{J}}\left[J^2\right]\) becomes zero and so \(\mathcal{C}_1^{\prime}\left(\alpha^*\right)\) evaluates to \(0\) thus confirming that \(\alpha^*\) is a stationary point.

Finally, we examine the behaviour of \(\mathcal{C}_1^{\prime}(\alpha)\) for \(\alpha>\alpha^*\). In this region, the term \(1-4 \alpha^2 \mathbb{E}_{\mathbf{J}}\left[J^2\right]\) becomes negative because \(\alpha^2>\left(\alpha^*\right)^2\). As a result, the derivative satisfies: \[-2\mathbb{E} \left[J^2\right] \left( 1 - 4 \alpha^2 \mathbb{E} \left[J^2\right] \right) e^{-2 \alpha^2 \mathbb{E} \left[J^2\right]} \frac{1}{\sqrt{D}} > 0.\] This indicates that \(\mathcal{C}_1(\alpha)\) is increasing for \(\alpha>\alpha^*\).

From this analysis, we observe that \(\mathcal{C}_1(\alpha)\) decreases for \(\alpha \in\left[0, \alpha^*\right)\), reaches a stationary point at \(\alpha=\alpha^*\), and increases for \(\alpha>\alpha^*\). This behaviour establishes that \(\alpha^*\) is the first local minimum of \(\mathcal{C}_1(\alpha)\) for \(\alpha \geq 0\).

Demonstrating Global Optimum Coincides with First Local Optimum for \(\mathcal{C}_2\)

We want to show that the globally optimal parameter 35 for \(\mathcal{C}_2\) coincides with the first local optimum. Recall that the derivative of \(\mathcal{C}_2\) with respect to \(\alpha\) is: \[\mathcal{C}_2^{\prime}(\alpha, \theta_1, \theta_2) = 8 \alpha \mathbb{E}[J] e^{-4 \theta_1 \alpha^2 \mathbb{E}\left[J^2\right]}\left[\theta_2 \mathbb{E}[J]^2 \cosh \left(4 \theta_2 \alpha^2 \mathbb{E}[J]^2\right) -\theta_1 \mathbb{E}\left[J^2\right] \sinh \left(4 \theta_2 \alpha^2 \mathbb{E}[J]^2\right)\right],\] where \((\theta_1, \theta_2)\) can be \((1,1)\), \((1,b)\), or \((b,b)\) with \(0 \leq b \leq 1\). Additionally, since the variance is non-negative, we have \(\mathbb{E}[J^2] \geq \mathbb{E}[J]^2\).

The case \(\mathbb{E}[J] > 0\) is trivial because the global optimum that minimises \(\mathcal{C}_2\) is \(\alpha^* = 0\) which clearly is the first local optimum. Thus, we focus on \(\mathbb{E}[J] < 0\). Consider \(\alpha \in (0, \alpha^*)\). Since \(\alpha^2 < (\alpha^*)^2\), we can approximate the hyperbolic functions for small arguments: \[\cosh (x) \approx 1+\frac{x^2}{2}, \quad \sinh (x) \approx x+\frac{x^3}{6}.\] Applying these approximations, the derivative becomes: \[\begin{align} \mathcal{C}_2^{\prime} &\approx 8 \alpha \mathbb{E}[J] e^{-4 \theta_1 \alpha^2 \mathbb{E}\left[J^2\right]}\left[\theta_2 \mathbb{E}[J]^2\left(1+\frac{\left(4 \theta_2 \alpha^2 \mathbb{E}[J]^2\right)^2}{2}\right)-\theta_1 \mathbb{E}\left[J^2\right]\left(4 \theta_2 \alpha^2 \mathbb{E}[J]^2+\frac{\left(4 \theta_2 \alpha^2 \mathbb{E}[J]^2\right)^3}{6}\right)\right]\\ &\approx 8 \alpha \mathbb{E}[J] e^{-4 \theta_1 \alpha^2 \mathbb{E}\left[J^2\right]}\left[\theta_2 \mathbb{E}[J]^2-4 \theta_1 \theta_2 \alpha^2 \mathbb{E}\left[J^2\right] \mathbb{E}[J]^2\right]\\ &= 8 \alpha \theta_2 \mathbb{E}[J]^3 e^{-4 \theta_1 \alpha^2 \mathbb{E}\left[J^2\right]}\left[1 - 4\theta_2 \alpha^2 \mathbb{E}\left[J^2\right] \right]. \end{align}\] Since \(\alpha^2 < (\alpha^*)^2\), the term \(1 - 4 \theta_2 \alpha^2 \mathbb{E}[J^2]\) remains positive. Given that \(\mathbb{E}[J] < 0\) and \(\theta_2 \geq 0\), the derivative \(\mathcal{C}_2^{\prime} < 0\). This implies that \(\mathcal{C}_2\) is decreasing on the interval \((0, \alpha^*)\).

For \(\alpha > \alpha^*\), assuming \(\alpha\) is still small enough for the above approximations to hold, the term \(1 - 4 \theta_2 \alpha^2 \mathbb{E}[J^2]\) becomes negative because \(\alpha^2 > (\alpha^*)^2\). Consequently, \(\mathcal{C}_2^{\prime} > 0\), indicating that \(\mathcal{C}_2\) is increasing for \(\alpha > \alpha^*\).

For sufficiently large \(\alpha \gg \alpha^*\), we approximate the hyperbolic functions as: \[\cosh (x) \approx \sinh (x) \approx \frac{e^x}{2}.\] Substituting these into the derivative gives: \[\mathcal{C}_2^{\prime} \approx 4 \alpha \mathbb{E}[J] e^{-4 \theta_1 \alpha^2 \mathbb{E}\left[J^2\right]}e^{4 \theta_2 \alpha^2 \mathbb{E}[J]^2}\left(\theta_2 \mathbb{E}[J]^2-\theta_1 \mathbb{E}\left[J^2\right]\right).\] Given that \(\theta_2 \leq \theta_1\) and \(\mathbb{E}[J^2] \geq \mathbb{E}[J]^2\), we analyse the expression \(\theta_2 \mathbb{E}[J]^2 - \theta_1 \mathbb{E}[J^2]\):

• If \(\theta_1 \mathbb{E}[J^2] > \theta_2 \mathbb{E}[J]^2\), then \(\theta_2 \mathbb{E}[J]^2 - \theta_1 \mathbb{E}[J^2] < 0\).

• If \(\theta_1 \mathbb{E}[J^2] = \theta_2 \mathbb{E}[J]^2\), then \(\theta_2 \mathbb{E}[J]^2 - \theta_1 \mathbb{E}[J^2] = 0\).

Thus, for \(\alpha \gg \alpha^*\), the derivative \(\mathcal{C}_2^{\prime} > 0\), confirming that \(\mathcal{C}_2\) continues to increase.

The analysis shows that \(\mathcal{C}_2(\alpha)\) is decreasing for \(\alpha \in (0, \alpha^*)\), attains a stationary point at \(\alpha = \alpha^*\), and increases for \(\alpha > \alpha^*\). This behaviour confirms that \(\alpha^*\) is the first local minimum of \(\mathcal{C}_2(\alpha)\) for \(\alpha \geq 0\). Therefore, the globally optimal parameter 35 coincides with this first local optimum.

Demonstrating Global Optimum Coincides with First Local Optimum for \(\mathcal{C}_3\)

Since the proof that the globally optimal value \(\alpha^*\), defined in 36 , coincides with the first local optimum closely parallels the argument presented for \(\mathcal{C}_1\), we omit it here for the sake of brevity. ◻