Determination of infinite series coefficients? Is it possible to determine the infinite sum coefficients $a_n$ for the following identity:
$$1=\displaystyle \sum_{n=1}^{+\infty}\Big[a_n\Big(\cosh(\lambda_nx)-\frac{h}{k\lambda_n}\sinh(\lambda_nx)\Big)\Big]$$
Where $\lambda_n$ are the roots of the following transcendental equation:
$$\lambda_n=\frac{h}{k}\coth \Big(\frac{\lambda_nL}{2}\Big)$$
In addition:
$0 \leq x \leq L$.
$h$ and $k$ are Real and positive constants.

Background:
$$X_n=a_n\cosh(\lambda_nx)+b_n\sinh(\lambda_nx)$$
This is the solution to a DE with boundary conditions:
$$-kX'(0)=hX(0)$$
And:
$$X_n'(L/2)=0$$
So:
$$X_n'=\lambda_n a_n\sinh(\lambda_nx)+\lambda_n b_n\cosh(\lambda_nx)$$
$$-k\lambda b_n=ha_n$$
$$b_n=-\frac{ha_n}{k\lambda_n}$$
$$X_n=a_n\Big(\cosh(\lambda_nx)-\frac{h}{k\lambda_n}\sinh(\lambda_nx)\Big)$$
And:
$$X_n'(L/2)=a_n\Big(\lambda_n\sinh(\lambda_nL/2)-\frac{h}{k}\cosh(\lambda_nL/2)\Big)=0$$
$$\lambda_n=\frac{h}{k}\coth(\lambda_nL/2)$$
With superposition:
$$X(x)=\displaystyle \sum_{n=1}^{+\infty}\Big[a_n\Big(\cosh(\lambda_nx)-\frac{h}{k\lambda_n}\sinh(\lambda_nx)\Big)\Big]$$
In addition, from an initial condition:
$$1=\displaystyle \sum_{n=1}^{+\infty}\Big[a_n\Big(\cosh(\lambda_nx)-\frac{h}{k\lambda_n}\sinh(\lambda_nx)\Big)\Big]$$

More background:
I'm trying to solve the PDE (convective cooling of a straight, uniform rod, quasi-1D, transient regime):

$$u_t=\kappa u_{xx}-\frac{ph}{A\rho c}u$$
Where $u(x,t)=T(x,t)-T_{\infty}$. ($T_{\infty}$ is a constant)
Domain as above.
Initial condition: $u(x,0)=T_0-T_{\infty}$
Boundaries: $-ku_x(0)=hu(0)$ and $-ku_x(L)=hu(L)$
Symmetry: $u_x(L/2)=0$
Note: $\kappa=\frac{k}{\rho c}$ (where $\rho$ and $c$ are Real, positive constants).
Ansatz: $u(x,t)=X(x)\Gamma(t)$
Separation of variables is easy (hint: keep the source term with the $\Gamma(t)$ side of things!)
I find an easy solution with slightly different boundary conditions: $u_x(0,t)=u_x(L,t)=0$ because $X(x)$ is then purely trigonometric with simple eigenvalues.
However, using the boundary conditions I'm really interested in, I find hyperbolic $X(x)$ is needed. I could be wrong on this.

Further edit:
Like Georg, yesterday I also established that the hyperbolic solution for $X(x)$ is not a solution because $c_1=c_2=0$.
Now I've tried the separation constant $-m^2$ (having moved $\kappa$ to the $\Gamma$ side first):
$$X''+m^2X=0$$
$$X=a\cos mx+b\sin mx$$
$$X'=-ma\sin mx+mb\cos mx$$
Boundary condition:
$$-kX'(0)=hX(0)$$
$$-kmb=ha\implies b=-\frac{ha}{km}$$
$$X=a\cos mx-\frac{ha}{km}\sin mx$$
$$X=a(\cos mx-\frac{h}{km}\sin mx)$$
$$X'=a(-m\sin mx-\frac{h}{k}\cos mx)$$
Symmetry condition:
$$X'(L/2)=a(-m\sin mL/2-\frac{h}{k}\cos mL/2)=0$$
$$-m\sin mL/2=\frac{h}{k}\cos mL/2$$
So the eigenvalues are the roots of the following trigonometric equation:
$$m_n=-\frac{h}{k}\cot\Big(\frac{m_nL}{2}\big)$$
Superposition:
$$\Large{X(x)=\displaystyle \sum_{n=1}^{+\infty}a_n\big[\cos(m_nx)-\frac{h}{km_n}\sin(m_nx)\big]}$$
With the initial condition we get:
$$1=\displaystyle \sum_{n=1}^{+\infty}a_n\big[\cos(m_nx)-\frac{h}{km_n}\sin(m_nx)\big]$$
So the coefficients $a_n$ still need determining.
 A: I'm not sure if this will help or if you've already thought of it:
$$\tanh\left(\frac {\lambda_n L}2\right)=\frac{h}{\lambda_nk}$$
$$1=\sum_n\left[a_n\left(\cosh(\lambda_nx)-\tanh(\frac {\lambda_n L}2)\sinh(\lambda_nx)\right)\right]$$$$=\sum_n\left[\frac{a_n}{\cosh(\frac {\lambda_n L}2)}\left(\cosh(\lambda_nx)\cosh(\frac {\lambda_n L}2)-\sinh(\frac {\lambda_n L}2)\sinh(\lambda_nx)\right)\right]$$
$$=\sum_n a_n\frac{\cosh(\lambda_n(x-\frac L2))}{\cosh\left(\frac {\lambda_n L}2\right)}$$
I don't think there's a single coefficient $a_n$ that satisfies the identity. Maybe some more context could clarify the question.
$$\boxed{\textbf{EDIT}}$$ I felt that $h$ and $k$ were for convection and conduction! I believe that this is what you should do:
$$X(x)\dot \Gamma(t)=\kappa X''(x)\Gamma(t)- CX(x)\Gamma(t)$$
$$\frac {\dot \Gamma(t)}{\Gamma(t)}+C=\frac{\kappa X''(x)}{X(x)}=-A$$
$$X''(x)+\lambda X(x)=0$$
where $\lambda=A/\kappa$
If $\lambda<0$ then $$X(x)=c_1\cosh(\sqrt{-\lambda}x)+c_2\sinh(\sqrt{-\lambda}x)$$
Since there's thermal symmetry $X(0)=X(L)$ and $X'(0)=X'(L)$
$$c_1=c_1\cosh(\sqrt{-\lambda}L)+c_2\sinh(\sqrt{-\lambda}L)$$
$$c_1(1-\cosh(\sqrt{-\lambda}L))-c_2\sinh(\sqrt{-\lambda}L)=0$$
and $$c_2(1-\cosh(\sqrt{-\lambda}L))-c_1\sinh(\sqrt{-\lambda}L)=0$$
This becomes: $$\frac{c_2}{c_1}=\frac{c_1}{c_2}$$
$$c_1^2=c_2^2$$
$$c_1=c_2\neq0 \quad\text{or}\quad c_1=-c_2\neq 0 \quad\text{or}\quad c_1=c_2=0$$
Replacing the first two solutions in either one of the equations yields $$\frac{\sinh(\sqrt{-\lambda}L)}{1-\cosh(\sqrt{-\lambda}L)}=1 \quad\text{or}\quad \frac{\sinh(\sqrt{-\lambda}L)}{1-\cosh(\sqrt{-\lambda}L)}=-1$$
Both of which have no solution for $L$ in the real domain which means that $c_1=c_2=0$ then both hyperbolic terms vanish! For the other cases ($\lambda>0$ and $\lambda=0$), you solve using Fourier!
 $$\boxed{\textbf{Further EDIT}}$$ 
$$1=\sum_n a_n\cos(m_n x)+\sum_n b_n \sin(m_n x)$$
$$\cos(m_k x)=\sum_n a_n\cos(m_n x)\cos(m_k x)+\sum_n b_n \sin(m_n x)\cos(m_k x)$$
$$\int_T \cos(m_k x)dx=\int_T \sum_n a_n\cos(m_n x)\cos(m_k x)dx+\int_T \sum_n b_n \sin(m_n x)\cos(m_k x)dx$$
$$\int_T \cos(m_k x)dx= \sum_n a_n\int_T\cos(m_n x)\cos(m_k x)dx+ \sum_n b_n \int_T\sin(m_n x)\cos(m_k x)dx$$
The last term disappears. The cosine function is orthogonal, that is you can easily prove that 
$$\frac{1}{T}\int_T\cos(m_n x)\cos(m_k x)=0 \text{ if } n\neq k$$
$$\frac{1}{T}\int_T\cos(m_n x)\cos(m_k x)=1 \text{ if } n= k$$
