 Solution

Configuration count. Choose the end-to-end direction as positive and let $N_R,N_L$ count rightward and leftward links. ThenA configuration is determined by choosing the positions of its leftward links, soThe counts must be nonnegative integers; in particular, allowed signed extensions occur in steps of $2a$. If distance means $|L|$ with neither direction selected, there are two corresponding sets for $L>0$, adding a factor of two. That constant does not affect the tension. There is no excluded-volume penalty in this one-dimensional idealized count.

Entropy from the multiplicity. Put $x=L/(Na)$, so $|x|<1$. Applying the leading Stirling's approximation to all three large factorials givesTerms subleading in $N$, including logarithmic factorial prefactors, have been omitted consistently. Expanding near zero extension givesStretching reduces the number of available configurations and hence the entropy.

Tension and the range of Hooke's law. The model assigns no extension-dependent internal energy. At fixed temperature, the externally applied tension is $\mathcal{F}=(\partial F/\partial L{)}_T=-T(\partial S/\partial L{)}_T$. Differentiating the full leading entropy givesIt is the external force needed to oppose the chain's restoring entropic force. Only when $|L|\ll Na$ can one put $\mathrm{artanh}x\simeq x$, yieldingThe paper's displayed linear law therefore requires this small-extension approximation, which is not explicitly stated in its premise $L<L_0$. It is not the exact force law up to full extension. The entropic spring constant grows with temperature because stronger thermal disorder makes a given reduction in configurational entropy cost more free energy. This is the one-dimensional entropic chain force-extension relation.

A sparse number of backward links. Write $q=N_L\ll N$, so $N_R=N-q$ and $L=a(N-2q)$. Exactly,For fixed $q$ and large $N$, each numerator factor approaches $N$, givingMore precisely, the logarithm of the product is $-q(q-1)/(2N)+O(q^3/N^2)$. A small relative error in the multiplicity requires $q^2/N\ll 1$; merely $q/N\ll 1$ need not make the omitted product close to one. The weaker sparse-link condition does suffice for the leading entropy density and leading high-extension tension. These are sparse backward links near full chain extension.

Tension in that limit. In the overlap regime $1\ll q\ll N$, use Stirling also for $q!$:ThusThe full leading entropy would give $(k_{B}T/(2a))\ln [(N-N_L)/N_L]$, which has the same dominant logarithm when $N_L/N\to 0$.

Approaching full extension. The continuum force law diverges logarithmically as $N_L\to 0$, or $L\to Na$. One must not literally apply Stirling to a zero factorial: at the discrete endpoint there is exactly one directed configuration, $\Omega =1$, and its entropy is zero. A continuous fixed-extension derivative is not defined at that endpoint.

An independent force-ensemble calculation makes the limiting physical statement precise. For applied tension $\mathcal{F}$, a link has partition function $z=2\cosh (\beta \mathcal{F}a)$, giving the exact mean extensionIt approaches $Na$ only as $\mathcal{F}\to \infty$ at positive temperature, and inversion reproduces the logarithmic force law for mean extension. In a real chain, bond extension or failure eventually replaces the ideal inextensible-link model.