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For Ising models on finite graphs, is the gradient of Z ' (w/r/t coupling and field) easier to compute than Z?

Suppose we have a graph $G$ with $n$ vertices, edgeset $E$, $\mathcal{X}=\{1,-1\}^n$\mathcal{X}=\{1,-1\}^n$. The partition function of the spin-1/2 Ising model on $G$ is

$$Z(J,h)=\sum_{x\in \mathcal{X}} \exp\left(J \sum_{(i,j)\in E} x_i x_j + h \sum_i x_i\right)$$

and its gradient with respect to the coupling and applied field is

$$Z'(J,h)=< $\nabla Z(J,h)=\left ( \frac{\partial}{\partial J} Z, \frac{\partial}{\partial h} Z>$$Z \right ).$$

We are interested in computing these quantities to some pre-determined finite precision. Computing $Z$ is hard in general, but easy in special cases, like when $|J|$ is small relative to average degree of the graph.

What can we say about relative hardness of computing $Z'$?\nabla Z$?

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For Ising models on finite graphs, is Z' easier to compute than Z?

Suppose we have a graph $G$ with $n$ vertices, edgeset $E$, $\mathcal{X}=\{1,-1\}^n$

$$Z(J,h)=\sum_{x\in \mathcal{X}} \exp\left(J \sum_{(i,j)\in E} x_i x_j + h \sum_i x_i\right)$$

$$Z'(J,h)=< \frac{\partial}{\partial J} Z, \frac{\partial}{\partial h} Z>$$

We are interested in computing these quantities to some pre-determined finite precision. Computing $Z$ is hard in general, but easy in special cases, like when $|J|$ is small relative to average degree of the graph.

What can we say about relative hardness of computing $Z'$?