# 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 $E$, $\mathcal{X}=\{1,-1\}^n$. 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

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

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 $\nabla Z$?

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The angle brackets threw me off for a second when I tried to imagine an inner product there. I guess other notation might be subject to the same problem though. –  j.c. Oct 8 '10 at 16:53
What's the standard notation for vector written out explicitly? –  Yaroslav Bulatov Oct 8 '10 at 16:55
Maybe if you write $\in \mathbb{R}^2$ ? It's not a big deal though. –  j.c. Oct 8 '10 at 17:49
If you are going to use angle brackets, you should surely use \langle and \rangle rather than the less-than and greater-than symbols. –  JBL Oct 8 '10 at 19:17
Can you explain your notation, please? Does that indicate the correlation function of $\partial Z/\partial J$ and $\partial Z/\partial h$ with respect to the Gibbs measure? –  Tom LaGatta Oct 8 '10 at 19:24