= Solution
Use the <arc length> $s$ and the unit tangent $\mathbf t=\partial_s\mathbf r$ of an <inextensible filament>. In the high-force state choose its positive longitudinal branch. The constraint gives
$$
t_z=\sqrt{1-|\mathbf t_\perp|^2}=1-\frac12|\mathbf t_\perp|^2+O(|\mathbf t_\perp|^4).
$$
The extension is $z=\int_0^L t_zds=L-\frac12\int_0^L|\mathbf t_\perp|^2ds+\cdots$. The <curvature of a space curve> obeys $\kappa^2=|\partial_s\mathbf t|^2$. Since $\partial_st_z=-\mathbf t_\perp\cdot\partial_s\mathbf t_\perp+\cdots$ is second order in small fluctuations, its squared contribution is fourth order. Thus the <worm-like chain> energy through quadratic order is
$$
\boxed{\mathcal E=-fL+\frac12\int_0^L\left[A|\partial_s\mathbf t_\perp|^2+f|\mathbf t_\perp|^2\right]ds.}
$$
Positive force suppresses transverse orientation fluctuations, while the <filament bending modulus> penalizes their spatial variation.
To apply the <equipartition theorem> without missing either component or a mode-counting factor, expand each component in a real orthonormal basis of eigenfunctions of $-\partial_s^2$, $t_\alpha(s)=\sum_q a_{\alpha q}w_q(s)$, with $\int w_qw_{q'}ds=\delta_{qq'}$ and $\alpha=x,y$. The quadratic energy is $\frac12\sum_{\alpha,q}(Aq^2+f)a_{\alpha q}^2$. Each real quadratic degree of freedom contributes $k_BT/2$, so
$$
\langle a_{\alpha q}a_{\beta q'}\rangle=\frac{k_BT}{Aq^2+f}\delta_{\alpha\beta}\delta_{qq'}.
$$
For a chain long compared with the force-dependent <correlation length>, the interior is translation invariant and the mode sum per length becomes $\int dq/(2\pi)$. Equivalently, for the <Fourier transform> $\widetilde t_\alpha(q)=\int e^{-iqs}t_\alpha(s)ds$,
$$
\langle\widetilde t_\alpha(q)\widetilde t_\beta(q')\rangle=2\pi\delta(q+q')\delta_{\alpha\beta}\frac{k_BT}{Aq^2+f}.
$$
The negative and positive wavenumbers here represent the real-field mode count consistently, rather than two independent complex modes. Summing both transverse components gives the <high-force worm-like chain elasticity>
$$
\langle|\mathbf t_\perp|^2\rangle=2k_BT\int_{-\infty}^{\infty}\frac{dq}{2\pi}\frac1{Aq^2+f}=\frac{k_BT}{\sqrt{Af}},
$$
where the integral follows by setting $q=\sqrt{f/A}\,u$ and using $\int_{-\infty}^{\infty}(1+u^2)^{-1}du=\pi$. Therefore
$$
\boxed{\frac{\langle z\rangle}{L}=1-\frac{k_BT}{2\sqrt{Af}}=1-\sqrt{\frac{k_BT}{4L_pf}},\qquad L_p=\frac{A}{k_BT}.}
$$
Here $L_p$ is the three-dimensional zero-force <persistence length>. The result requires $fL_p\gg k_BT$, negligible backbone stretching, and a long-chain bulk approximation. It does not describe the eventual bond-stretching regime. For a continuum polymer description, the force-dependent length should also exceed the microscopic cutoff.
For comparison, a three-dimensional <freely jointed chain> has contour length $L=Nb$ and independent link orientations. At force $f$, a link of polar angle $\vartheta$ has Boltzmann factor $e^{x\cos\vartheta}$ with $x=fb/(k_BT)$. Its angular <canonical partition function> is
$$
Z_1=2\pi\int_{-1}^1e^{xu}du=4\pi\frac{\sinh x}{x}.
$$
The total <canonical partition function> is $Z_1^N$. Differentiating with respect to force gives the <force-extension of a three-dimensional freely jointed chain>:
$$
\frac{\langle z\rangle}{Nb}=\frac{d\log Z_1}{dx}=\coth x-\frac1x.
$$
Consequently
$$
\boxed{1-\frac{\langle z\rangle}{L}\sim\frac{k_BT}{fb}\quad\text{for a freely jointed chain},\qquad 1-\frac{\langle z\rangle}{L}\sim\frac{k_BT}{2\sqrt{Af}}\quad\text{for a worm-like chain}.}
$$
The extension deficits scale as $f^{-1}$ and $f^{-1/2}$ respectively. Inverting them gives force divergences proportional to the first and second inverse powers of the deficit. Freely jointed link rotations cost no bending energy; the <worm-like chain> instead has a continuum of coupled bending modes whose contributing wavelengths change with force. Setting a long-wavelength coarse-grained link length $b=2L_p$ can match weak-force statistics, but does not make these high-force laws identical.
For the <transverse tangent correlation of a stretched worm-like chain>, use a separation $r$ consistently: the source writes $C(y)$ but integrates a separation denoted $r$. In the translation-invariant bulk the spatial average is redundant and
$$
C(r)=\langle\mathbf t_\perp(s)\cdot\mathbf t_\perp(s+r)\rangle=2k_BT\int_{-\infty}^{\infty}\frac{dq}{2\pi}\frac{e^{iqr}}{Aq^2+f}.
$$
To evaluate the integral without an assumed decay law, let $G(r)$ be the inverse <Fourier transform> of $(Aq^2+f)^{-1}$. Then $(-A\partial_r^2+f)G=\delta(r)$. Away from zero, decay at both infinities requires $G=B e^{-|r|/\xi}$ with $\xi=\sqrt{A/f}$. Integrating across zero yields $-A[G'(0^+)-G'(0^-)]=1$, so $B=1/(2\sqrt{Af})$. Thus
$$
\boxed{C(r)=\frac{k_BT}{\sqrt{Af}}e^{-|r|/\xi},\qquad\xi=\sqrt{\frac Af}=\sqrt{\frac{k_BT L_p}{f}}.}
$$
At $r=0$ this agrees with the two-component <equipartition theorem> calculation. The <correlation length> decreases with force and is distinct from the zero-force <persistence length>. The full tangent correlation has a nonzero aligned background; the exponentially decaying quantity here is specifically the transverse correlation.
For a finite open chain, the source's integral samples $s+r$ beyond $L$ and supplies no endpoint tangent conditions. An exact finite-length correlation is therefore not uniquely specified. The result above is the intended bulk form when $L\gg\xi$. As one explicit finite convention, periodically extending the tangent fluctuations gives the <periodic finite-length tangent correlation of a stretched worm-like chain>
$$
C_L(r)=\sum_{j\in\mathbb Z}C(r+jL)=\frac{k_BT}{\sqrt{Af}}\frac{\cosh[(L/2-d_L(r))/\xi]}{\sinh[L/(2\xi)]},
$$
where $d_L(r)$ is the distance from zero around the periodic interval. Then $C_L(0)=(k_BT/\sqrt{Af})\coth[L/(2\xi)]$, and the extension formula acquires this same finite-length factor. The bulk law is recovered as $L/\xi\to\infty$. This makes the endpoint assumption explicit rather than claiming an unspecified finite-chain correlation is exactly translationally invariant.
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