= Solution
For one small transverse displacement, the <elastic energy> of the <elastic filament> is
$$
U[h]=\frac A2\int_0^L[h''(x)]^2\,dx,
$$
where $A$ is the <filament bending modulus>. Its <first variation> is
$$
\delta U=A[h''\delta h'-h'''\delta h]_0^L+A\int_0^Lh''''\delta h\,dx.
$$
The <clamped boundary conditions> remove the left boundary terms. A force-free and torque-free tip permits independent $\delta h(L)$ and $\delta h'(L)$, so the <natural boundary conditions for a free endpoint> are \b[$h''(L)=h'''(L)=0$]: zero bending moment and transverse shear.
There is an important distinction between an energy minimum and a <normal mode>. The unloaded <higher-order Euler-Lagrange equation> is $h''''=0$, whose only solution with these four boundary conditions is $h=0$. A general fluctuating shape is a sum of modes, not one of the stated sinusoidal/hyperbolic functions. To obtain the <clamped--free bending modes>, extremize the bending Rayleigh quotient, or equivalently $U[W]-\tfrac12Ak^4\int W^2dx$ with a fixed norm. Its <Euler-Lagrange equation> is
$$
\boxed{W''''=k^4W.}
$$
This is the <constrained variational characterization of bending modes>. Its four characteristic roots are $\pm k,\pm ik$. Clamping gives $F=-D$ and $E=-B$, hence
$$
W=D(\cos kx-\cosh kx)+B(\sin kx-\sinh kx).
$$
The free-tip conditions reduce to
$$
\begin{pmatrix}\cos q+\cosh q&\sin q+\sinh q\\ \sin q-\sinh q&-(\cos q+\cosh q)\end{pmatrix}\binom DB=0,\qquad q=kL.
$$
Its <determinant> is $-2(1+\cos q\cosh q)$, so nontrivial modes require
$$
\boxed{\cos q_n\cosh q_n=-1,\qquad k_n=q_n/L.}
$$
No zero mode exists, since a cubic satisfying the homogeneous clamp/free conditions is zero. The <clamped-free bending spectrum> starts with $q_1\in(\pi/2,2)$; bisection or Newton iteration gives \b[$q_1\simeq1.875104$, so $k_1\simeq1.875104/L$]. Choosing $D=N(\sin q_n+\sinh q_n)$ and $B=-N(\cos q_n+\cosh q_n)$ gives
$$
\boxed{W_n=N\big[(\sin q_n+\sinh q_n)(\cos k_nx-\cosh k_nx)-(\cos q_n+\cosh q_n)(\sin k_nx-\sinh k_nx)\big].}
$$
The second boundary condition follows from the root equation, and $N$ is arbitrary until a normalization is chosen.
The bending operator with these boundary conditions is positive and <self-adjoint>. Twice integrating by parts gives
$$
\int_0^L W_n''W_m''\,dx=k_m^4\int_0^L W_nW_m\,dx.
$$
Symmetry implies <orthogonality> when $n\ne m$. Expand $h(x)=\sum_n a_nW_n(x)$ and use $I_n=L^{-1}\int_0^LW_n^2dx$. Then
$$
U=\frac{AL}2\sum_n k_n^4 I_na_n^2,\qquad \langle a_na_m\rangle=\delta_{nm}\frac{k_BT}{ALk_n^4I_n},
$$
by the <equipartition theorem>. The supplied endpoint identity therefore yields the <thermal bending fluctuations of a clamped filament>:
$$
\langle h(L)^2\rangle=\frac{4k_BT}{AL}\sum_{n=1}^\infty k_n^{-4}=\frac{4k_BTL^3}{A}\sum_{n=1}^\infty q_n^{-4}.
$$
To evaluate the sum, apply a tip force $f$ and minimize $U-fh(L)$. The modified free-end conditions are $h''(L)=0$ and $-Ah'''(L)=f$, and $h''''=0$ in the interior. Integration gives $h(x)=fx^2(3L-x)/(6A)$, so the <tip-force compliance of a cantilever> is $L^3/(3A)$. On the other hand, minimizing the modal energy minus $f\sum_na_nW_n(L)$ gives $a_n=fW_n(L)/(ALk_n^4I_n)$ and hence
$$
\frac{h(L)}f=\frac4{AL}\sum_nk_n^{-4}=\frac{4L^3}{A}\sum_nq_n^{-4}.
$$
Comparison evaluates the <fourth inverse-power sum of the cantilever spectrum> without truncating the modes:
$$
\boxed{\sum_{n=1}^\infty q_n^{-4}=\frac1{12},\qquad \langle h(L)^2\rangle=\frac{k_BTL^3}{3A}.}
$$
The boxed <variance> is for the specified single transverse direction. An independent equilibrium check follows by differentiating the Gaussian partition function with respect to $f$: $\partial\langle h(L)\rangle/\partial f=\langle h(L)^2\rangle/(k_BT)$ at zero load, reproducing the same result from the static compliance. Keeping only the first <normal mode> gives about $0.32356\,k_BTL^3/A$, roughly $97.1\%$ of the exact <variance>. With two independent equivalent transverse directions, their summed <variance> is twice the boxed result. The small-slope model requires $k_BTL/A\ll1$, or $L$ small compared with the usual three-dimensional <persistence length> $A/(k_BT)$.
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