Does closed-form solution of $e^{-x}=-\sin(x)$ exist?

Question

Recently I've been developing some electrical engineering equations and I've come across an equation of this form: $$e^{-x}=-\sin(x)$$ I've tried to develop both terms using Taylor Series but I'm still unable to solve it.

I've done a little bit of research and it appears that the only way to solve this equation is by using a numerical method such as Newton-Raphson.

I know there are multiple solutions for this equation but I'm only interested in the first one. Is there any way to avoid using a numerical method? Or at least is there a good approximation to avoid getting a numerical solution?

Answer 1

Notice that solution will happen only when sine comes back up to be positive. By that point the exponential will already be very small so the solution will happen very close to $\pi$. Instead of Taylor expanding about $0$, do it around $\pi$. Around $\pi$ the negative sine will look like a positive sine so that Taylor expansion is easy. For the exponential:

$$e^{-x} = e^{-\pi}e^{-(x-\pi)}$$

$$e^{-\pi}\left(1-(x-\pi)+\frac{(x-\pi)^2}{2} + \cdots\right) = (x-\pi) - \frac{(x-\pi)^3}{6} + \cdots$$

So go as far as you need. Here are the first few solutions as a sequence, where the term in the sequence is the order of the solution ($n$th order meaning including terms only up to that degree):

$$\{ \pi, \frac{1}{1+e^{\pi}}+\pi, 1+e^{\pi} - \sqrt{e^{2\pi} + 2e^{\pi} - 1} + \pi, \cdots \}$$

where we take the minus root because the quantity in the square root has a smaller absolute value than the first term, so the minus gives us the positive root that is closest to $\pi$.

From Wolfram, the first solution is $\approx 3.18306$ while the approximated first and second order solutions to that are $\approx 3.18302$ and $\approx 3.18305$, respectively. I would personally call that good enough.

Answer 2

There's no closed-form solution, I believe, but a very good simple approximation. Note that the solutions lie close to multiples of $\pi$. We can approximate the solution near $n \pi$ with first-order Taylor approximations $$\begin{align*} -\sin x &\approx (-1)^{n+1}(x-n \pi) \\ e^{-x} &\approx -e^{-n\pi} (x - n \pi) + e^{-n\pi} \end{align*} $$ which give solutions $$x \approx n \pi + \frac{(-1)^{n+1}}{e^{n \pi} + 1}.$$ This method is basically single-iteration Newton–Raphsom with an initial guess of $x = n\pi$.

The approximations are very accurate and get exponentially more so with increasing $n$: the $n = 1$ and $n = 2$ approximations agree with the true solution to four decimal places, the $n = 3$ approximation to twelve. Accuracy could be improved further with some perturbative method that takes higher-order terms in the Taylor series into account, but there's probably no need.

Answer 3

Instead of Taylor, we could use the simplest $[1,1]$ Padé approximant of function $$f(x)=e^{-x}+\sin(x)$$ around $n \pi$.

This will give $$x_1=\pi+\frac{2 \left(1+e^{\pi }\right)}{1+4 e^{\pi }+2 e^{2 \pi }}\approx 3.183052$$ $$x_2=2\pi+\frac{2(1- e^{2 \pi })}{1-4 e^{2 \pi }+2 e^{4 \pi }}\approx 6.281314$$ $$x_3=3\pi+\frac{2 \left(1+e^{3 \pi }\right)}{1+4 e^{3 \pi }+2 e^{6 \pi }}\approx 9.424859$$ $$x_4=4\pi+\frac{2(1- e^{4 \pi })}{1-4 e^{4 \pi }+2 e^{8 \pi }}$$ $$x_5=5\pi+\frac{2 \left(1+e^{5 \pi }\right)}{1+4 e^{5 \pi }+2 e^{10 \pi }}$$ while the exact solutions for $x_1,x_2,x_3$ would be respectively $3.183063$, $6.281315$ and $9.424859$.

Now, there are nice patterns to exploit. As an approximation, $$\color{blue}{x_n=n \pi +\frac{2(1-(-1)^n e^{\pi n} )}{1-4(-1)^n e^{\pi n} +2 e^{2 \pi n}}}$$

Edit

If we just focus on the first solution, we could use the $[1,n]$ Padé approximant to still obtain explicit formulae with no radical. This is almost equivalent to use Newton type methods of higher orders.

For example $n=2$ would give $$x=\pi+\frac{3+12 e^{\pi }+6 e^{2 \pi }}{1+11 e^{\pi }+18 e^{2 \pi }+6 e^{3 \pi }}$$

As a function of $n$, the decimal representation of the first root is $$\left( \begin{array}{cc} n & x_{(n)} \\ 0 & \color{red}{3.1830}164857561560653 \\ 1 & \color{red}{3.1830}520565530713217 \\ 2 & \color{red}{3.18306}29833004084833 \\ 3 & \color{red}{3.18306301}00338332755 \\ 4 & \color{red}{3.1830630119}234281754 \\ 5 & \color{red}{3.183063011933}0809849 \\ 6 & \color{red}{3.18306301193336}10391 \\ \cdots & \cdots \\ \infty & \color{red}{3.1830630119333635919} \end{array} \right)$$

Answer 4

I think the the nearest thing to a closed form solution is given by the Lagrange inversion formula near $x = 3$. If $$f(x) = e^{-x}+\sin(x) = y$$ then: $$x = f^{-1}(y) = 3 + \sum_{n=1}^\infty\frac{1}{n!}\lim_{t\to 3}\frac{d^{n-1}}{dt^{n-1}}\left(\frac{t - 3}{f(t) - f(3)}\right)^n(y - f(3))^n.$$