 Solution

Use the infinite translation-invariant chain first. Set $\rho =e^{-\kappa }$ and Fourier-transform the exchange matrix:For positive $J,\kappa$ this is strictly positive. Its reciprocal is a linear polynomial in $\cos q$, proving the nearest-neighbour form of the exponentially coupled Ising inverse kernel:Expanding the squared difference givesSince $(\cosh \kappa -1)/\sinh \kappa =\tanh (\kappa /2)$, substitution into part (a) yieldsThis is the exact bulk identity. A finite periodic version uses the periodized coupling $J_{ij}=J{\sum }_{\ell \in \mathbb{Z}}{e}^{-\kappa |i-j+\ell N|}$; its discrete Fourier symbols are the same and the squared differences wrap around the ring. Simply truncating the exponential at open ends gives a different inverse at those ends.

For an open chain of $N\ge 2$ sites with the literal finite matrix $J{\rho }^{|i-j|}$, the open-chain endpoint corrections for an exponential Ising kernel are explicit:The inverse has endpoint diagonal entries $1/[J(1-{\rho }^2)]$, interior diagonal entries $(1+{\rho }^2)/[J(1-{\rho }^2)]$ and adjacent entries $-\rho /[J(1-{\rho }^2)]$. Multiplication by the original geometric matrix verifies every entry. These boundary terms affect boundary amplitudes, but not the bulk free-energy density in the thermodynamic limit at fixed $\kappa >0$.