 Solution

Let a segment of fixed length $b$ make angle $\theta$ with an applied force $\mathbf{f}=f\hat{\mathbf{z}}$. Its energy contribution is $-fb\cos \theta$. Independent freely jointed orientations give the single-segment partition function and chain partition functionThe mean end-to-end projection is the force derivative of $k_{B}T\ln Z_N$:For $|u|\ll 1$, the Langevin function is $u/3+O(u^3)$. Isotropy therefore gives the vector responseThe covariance at zero force provides an independent route: $⟨R_i{R}_j{⟩}_0=N{b}^2{\delta }_{ij}/3$, and the Boltzmann weight's first-order term yields $⟨R_i⟩=f_j⟨R_i{R}_j{⟩}_0/(k_{B}T)$. This is a relation for the mean vector or its projection. The zero-force mean vector vanishes even though the chain has finite root mean square size $b\sqrt{N}$.

The controlled weak-force condition is $fb\ll k_{B}T$, not merely that a short part of a measured curve looks straight. With the supplied $b$, $k_{B}T/b\simeq 10.3\mathrm{pN}$ at $25^{\circ }\mathrm{C}$. A raw force near $30\mathrm{pN}$, as in the figure's labelled low-extension portion, would give $u\simeq 2.9$ and is not asymptotically weak for those literal segment parameters. A force offset or different effective model might change that interpretation, but no such correction should be invented from the plot.