Solving $\sinh x = kx$

Question

Can we solve the equation $\sinh x = kx$ for $x$ in terms of elementary functions? I've tried reexpressing the hyperbolic sine as exponentials and converting the equation into a quadratic in $e^x$, but this doesn't seem to make the problem any easier. I've considered expanding $\sinh x$ as a Taylor series, but this doesn't seem useful either.

Answer 1

As mentioned in the comments, the only solution for $k\le1$ is $x=0$ and for $k>1$ there are 3 solutions: $x=0,\pm x_\star$. Although there is no closed form in terms of special functions such as the Lambert W function known, it is not hard to numerically compute $x_\star$, the positive nonzero solution. For example, we have fixed-point iteration:

$$x_{n+1}=\ln(2kx_n+\exp(-x_n))$$

or Newton's method:

$$x_{n+1}=x_n-\frac{\sinh(x_n)-kx_n}{\cosh(x_n)-k}$$

or any other numerical method.

Answer 2

As said in comments and answers, you will need a numerical method.

In any manner, instead of looking for the zero of $$f(x)=\frac{\sinh (x)}{x}-k$$ it would be better to look for the zero of $$g(x)=\log \left(\frac{\sinh (x)}{x}\right)-\log(k)$$ which is much more linear when $x$ is large.

Newton iterates would be $$x_{n+1}=x_n-\frac{x_n \sinh (x_n) \log \left(\frac{\sinh (x_n)}{x_n}\right)}{x_n \cosh (x_n)-\sinh (x_n)}$$

If $k$ is large and $x$ too, we could approximate $$g(x)\sim\log \left(\frac{e^x}{2x}\right)-\log(k)\implies x\sim -W_{-1}\left(-\frac{1}{2 k}\right)$$ where appears the second branch of Lambert function (which is not elementary but simple to use).

Trying for $k=123.456789$, starting with the above estimate as $x_0$, Newton iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 7.5276168997015533341 \\ 1 & 7.5276172335102603321 \\ 2 & 7.5276172335102591983 \end{array} \right)$$

When $k$ is small, a very good approximation of $g(x)$ could be obtained using the $[4,4]$ Padé approximant of it. This will be $$g(x)\sim \frac{21 x^2 \left(31 x^2+570\right)}{58 x^4+6300 x^2+71820}- \log(k)$$ which leads to a quadratic equation in $x^2$.

For $k=3$, the above gives $x=2.83971$ while the exact solution would be $x=2.83845$.

Answer 3

$0$ is one solution.

1)

$$\sinh(x)=kx$$ $$\frac{1}{2}e^x-\frac{1}{2}e^{-x}=kx$$ $$\frac{1}{2}(e^x)^2-\frac{1}{2}=kxe^x$$ $$\frac{1}{2}(e^x)^2-kxe^x-\frac{1}{2}=0\tag{1}$$

We see, equation (1) is a polynomial equation of more than one algebraically independent monomials ($x,e^x$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

We see, equation (1) is an irreducible algebraic equation of $x$ and $e^x$ simultaneously if $k$ is an algebraic number.
According to the theorems in [Lin 1983] and [Chow 1999], such kind of equations cannot have solutions except $0$ that are elementary numbers or explicit elementary numbers respectively in this case. Therefore $x$ cannot be an elementary number except $0$ in this case and equation (1) cannot have partial inverses that are elementary functions over non-discrete domains.

We see, because equation (1) is a polynomial equation of $x$ and $e^x$ of a degree greater than $1$, the equation is not in a form for applying Lambert W, Generalized Lambert W or Hyper Lambert W.

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

2)

$$\sinh(x)=kx$$ $$\sinh(x)-kx=0$$ $x\to it$: $$\sinh(it)-ikt=0$$ $$\sin(t)-kt=0$$ $$\frac{1}{k}\sin(t)-t=0$$ $$t-\frac{1}{k}\sin(t)=0$$

We see, this is Kepler's equation.

For $t\neq 0$, Kepler's equation can be solved by Hyper Lambert W:

$$t-\frac{1}{k}\sin(t)=0$$ $$t\ e^{\ln(1-\frac{1}{k}\frac{\sin(t)}{t})}=0$$ $$t=HW\left(\left\{\ln(1-\frac{1}{k}\frac{\sin(x)}{x})\right\}_1;0\right)$$

So we have a closed form for $t$, and the series representations of Hyper Lambert W give some hints for calculating $t$.

[Galidakis 2005] Galidakis , I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997

Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119

[Dubinov/Galidakis 2007] Dubinov, A.; Galidakis, Y.: Explicit solution of the Kepler equation. Physics of Particles and Nuclei Letters 4 (2007) 213-216

[Masson] Masson, Paul: Analytic Physics - An Exact Solution of the Complex Kepler Equation