explicit solution for transcendental equation

Question

Does anyone knows whether there is an explicit, analytical solution for transcendental equations of the form $A x + B \tanh(C x) + \coth(x) = 0$, where $A, B$, and $C$ are positive real constants?

Answer 1

The Burniston-Siewert approach for anaytical solution of similar transcendental equation is by integral representations.

In :

http://www4.ncsu.edu/~ces/publist.html

You can find a lot of transcendental equations solved by such integrals, therefore you could adapt this method to your custom problem.

?? HINT ??

For a suitable complex path $\gamma$, x(A,B,C) can be written as: $$x(A,b,C)=\frac{1}{2\pi i} \oint_\gamma z\frac{A+\frac{BC}{\cosh(Cz)^2}-\frac{1}{\sinh(z)^2}}{Az+B\tanh(Cz)+\coth(z)}dz$$

Historical Note In : "The $W_t$ Transcendental Function and Quantum Mechanical Applications", V. E. Markushin, R. Rosenfelder, A. W. Schreiber, http://arxiv.org/abs/math-ph/0104019v2

the equation $$x \tan(x) =y$$ has been investigated and the Lambert-like function $W_t(x)$ has been proposed. Such function is linked to the so called Lambert r-$W$ (i.e. extension of Lambert series based on Laguerre polynomials).

Answer 2

This is a comment on GEdgar's ($C=0$) version and the equation

$$ Ax+\coth(x)=0, $$

(i.e. Find the points of intersection of the curve $y=\coth(x)$ and straight lines through the origin. Existence of real solutions requires $A$ to be of an appropriate sign.)

This can be solved using the functions in

I. Mező and A. Baricz,
On the generalization of the Lambert $W$ function with applications in theoretical physics.
arXiv:1408.3999v1.pdf (18 Aug 2014).

The starting point to using their functions is to rewrite the equation as

$$ \exp(2x)= \frac{x-1/A}{x+1/A} $$

and then with $X=2x$ define

$$ f(X)= \frac{X+2/A}{X-2/A}\exp(X). $$

The inverse function for f is defined in the arXiv paper (and earlier papers referenced there), and hence the solution(s) for $f(X)=1$.

Remark. The inverse Langevin function can also be written in terms of the functions in their arXiv paper.

Answer 3

This is a comment in response to giorgiomugnaini's comment about an exact solution. I an making it an answer since entering the MathJax is much easier.

The "solution" to the equation $ 0 = a + bx-x \tanh^{-1}(\frac1{x})$ is given, in equation (16), when $a < 0$, by $\frac1{x} =2+\frac{1-a}{b}-\frac1{\pi} \int_{-1}^1 \tan^{-1}\left( \frac{\pi t}{2(a+bt-t\tanh^{-1}(t))}\right) dt $.

In view of the very non-elementary nature of that integral, this does not seem to me to be much of a help.

Answer 4

Here is a Lagrange Reversion solution for the larger root of a more general equation:

$$x-(a \tanh( b y)+c \coth(y))=y\implies y=x+\sum_{n=1}^\infty\frac{(-1)^n}{n!}\frac{d^{n-1}(a\tanh(b x)+c\coth(x))^n}{dx^{n-1}}\mathop=^{r=\frac ac} x+\sum_{n=1}^\infty\frac{(-c)^n}{n!}\frac{d^{n-1}(r\tanh(b x)+\coth(x))^n}{dx^{n-1}} $$

Compare results here and here. Now to use the binomial theorem:

$$\frac{d^{n-1}(r\tanh(b x)+\coth(x))^n}{dx^{n-1}} =\sum_{k=0}^n \binom nk r^k\frac{d^{n-1}}{dx^{n-1}}\coth^{n-k}(x)\tanh^k(b x)$$

and somehow take the $n$th derivative of the hyperbolic functions using the single exponential definiton. More will be added