One-dimensional entropic chain force-extension relation ($\mathcal{F}=(k_{B}T/a)\mathrm{artanh}(L/(Na))$)

For $N$ independent links pointing left or right with length $a$, fixed signed extension $L$ gives $N_R=(N+L/a)/2$ and $N_L=(N-L/a)/2$. The multiplicity is $\left(\genfrac{}{}{0ex}{}{N}{N_L}\right)$ when these counts are nonnegative integers. Leading Stirling's approximation gives $\mathcal{F}=(k_{B}T/(2a))\ln (N_R/N_L)$ as the externally applied tension conjugate to $L$. Its small-extension limit is $k_{B}TL/(N{a}^2)$. At fixed tension, the exact one-link partition function is $2\cosh (\beta \mathcal{F}a)$ and the mean extension is $Na\tanh (\beta \mathcal{F}a)$, confirming the same inverse relation without using Stirling at the full-extension endpoint.

The exact count is $\Omega =\left(\genfrac{}{}{0ex}{}{N}{q}\right)=N^q{\prod }_{j=0}^{q-1}(1-j/N)/q!$. For fixed $q$ and large $N$, this is asymptotic to $N^q/q!$. Uniform relative accuracy requires $q^2/N\ll 1$, since the logarithm of the product starts at $-q(q-1)/(2N)$. The weaker assumption $q/N\ll 1$ gives the leading sparse-link entropy but need not give a small relative multiplicity error. In the continuum range $1\ll q\ll N$, the entropic force behaves as $(k_{B}T/(2a))\ln (N/q)$. The fixed-tension mean extension approaches $Na$ only as the tension diverges; at the discrete endpoint $q=0$ the exact entropy is zero and a continuous fixed-extension derivative is not defined.