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.
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$x=0$ is always a solution. – Ninad Munshi Dec 11 '19 at 19:02
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2Also since the slope of $\sinh$ at zero is one and it is larger than one at any other point the only solution is zero for $k\le 1$. For $k>1$ there are three solutions that cannot be found in elementary terms. – GReyes Dec 11 '19 at 19:13
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3No. It's even hard to solve with with the Lambert $W$ function, which sometimes can be used to solve similar problems. – Thomas Andrews Dec 11 '19 at 19:14
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All the above comments are easily seen by looking at the graph of $\sinh x$ and rotating a line around the origin. – WhatsUp Dec 11 '19 at 19:22
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Not all the comments - the fact that the equation can't be solved in elementary functions is impossible to see by looking at the graph. @WhatsUp – Thomas Andrews Dec 11 '19 at 19:26
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@GReyes Actually, when $k>1$ then there are three solutions, one of which is $0$ and two others which cannot be represented in terms of elementary functions. (The two others are negatives of each other, so there is one positive solution.) – Thomas Andrews Dec 11 '19 at 19:30
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Sure, the trivial solution $x=0$ is an elementary function of $k$... – GReyes Dec 11 '19 at 19:47
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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 $x=2.83845$.
Claude Leibovici
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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.
Simply Beautiful Art
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A good starting point for such numerical methods is $1+\frac{x_0^2}{6}=k$ for $k$ only slightly larger than $1$. For $k\gg 1$, try $x_0=\ln(2k)$. – J.G. Dec 12 '19 at 06:54
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I would think it more beneficial to begin the Taylor expansion about $x=1$ for $k\approx1$ no? – Simply Beautiful Art Dec 12 '19 at 14:55
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If $\frac{\sinh x}{x}=k\approx1$, the fraction should be approximated as $1+\frac16x^2$. – J.G. Dec 12 '19 at 15:01
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But it'd be more accurate to use $kx=\sinh(x)\simeq\sinh(1)+\cosh(1)(x-1)$ (or another term if desired), since $x$ should be near $1$. Your point gives $x_0\approx0$, which will converge to the trivial solution instead. – Simply Beautiful Art Dec 12 '19 at 15:03
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1Ah nevermind, I visualized the graph in my head wrong lol. – Simply Beautiful Art Dec 12 '19 at 15:21
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Has it been proven that there doesn't exist a closed form solution, or has it just been very difficult? – Nick A. Dec 12 '19 at 22:37
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@NickA. The proof is a modification of the proof that the Lambert W function is not elementary. – Simply Beautiful Art Dec 13 '19 at 15:10
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$0$ is one solution.
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$$\sinh(x)=kx$$ $$\frac{1}{2}e^x-\frac{1}{2}e^{-x}=kx$$ $$\frac{1}{2}(e^x)^2-\frac{1}{2}=kxe^x$$ $$\frac{1}{2}(e^x)^2-kxe^x-\frac{1}{2}=0\tag{1}$$
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.
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
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$$\sinh(x)=kx$$ $$\sinh(x)-kx=0$$ $x\to it$: $$\sinh(it)-ikt=0$$ $$\sin(t)-kt=0$$ $$\frac{1}{k}\sin(t)-t=0$$ $$t-\frac{1}{k}\sin(t)=0$$
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 $t$.
[Masson] Masson, Paul: Analytic Physics - An Exact Solution of the Complex Kepler Equation
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