# How would you solve an equation of the form $e^{-x} - \sin(x) = 0$?

I've been trying to do this for ages. I'm worried that it's impossible, but I have heard that it can be done by hand.

As long as I can get $$x$$ by hand I can obviously work out its value via calculator.

• Graph the two functions $e^{-x}$ and $\sin(x)$ and find where they intersect (approximately). – mattos Mar 31 '20 at 4:09
• I should add that I'm looking for an algebraic method if possible! – Jepho55 Mar 31 '20 at 4:11
• I'd be shocked if it's doable since Wolfram can only offer approximations. Not that "proof by Wolfram" is a thing, just a heuristic. – Eevee Trainer Mar 31 '20 at 4:16
• @EeveeTrainer interestingly, with a mild interpretation, these are close enough to analytic answers. Not sure about the first two solutions yet, but the others clearly look like $n\pi$ for all integer $n \ge 2$. – gt6989b Mar 31 '20 at 4:21
• @gt6989b Look here – gen-ℤ ready to perish Mar 31 '20 at 5:38

First of all, that equation is highly nonlinear, as a result analytically we may not (can not) find a solution.

First Observation: Note that when $$x$$ becomes very large $$e^{-x}$$ becomes very small so $$e^{-x}-\sin x\approx0-\sin x$$ and hence the zeros of $$\sin x$$ are the zeros of $$e^{-x}-\sin x$$. This observation also ensures that solution for $$e^{-x}-\sin x=0$$ exists.

Second Observation: For $$x<0$$ $$e^{-x}$$ is an increasing function and dominates $$\sin x$$ (Which can be seen in the graph too) so there are no zeros of the function $$e^{-x}-\sin x$$ for $$x<0$$.

So what's the solution? Well, the solution can only be found by Numerical methods. One easy way to do that by plotting graphs.

This is the graph of $$e^{-x}-\sin x$$ where the first observation is visualized.

This is the graph showing the intersection of $$e^{-x}$$ [Blue line] and $$\sin x$$ [Red line].

In one word the solutions are $$x=0.589,3.096,6.285$$ etc.

Hope this works.

• Would there be a way to show that over any interval before the first solution the function as a whole is increasing? – Jepho55 Mar 31 '20 at 4:26
• You mean decreasing right? So you wanna show that $e^{-x}-\sin x$ is decreasing in $(\infty,0]$? – Sujit Bhattacharyya Mar 31 '20 at 4:29
• Close. I want to show that it's decreasing over (-infinity, 0.589) – Jepho55 Mar 31 '20 at 4:34

You cannot find a closed form for the zeroes, but you can find asymptotic approximations.

Let $$x_n$$ be the zero in $$(\pi n-\frac \pi 2, \pi n+\frac \pi 2)$$. Write $$x_n=\pi n+\varepsilon_n$$ with $$|\varepsilon_n|<\frac \pi 2$$.

Then $$e^{-\pi n -\varepsilon_n}=\sin(\pi n+\varepsilon_n)=(-1)^n\sin(\varepsilon_n)\tag{1}$$ Taking the limit, the left-hand-side converges to $$0$$. So this implies that $$\varepsilon_n\rightarrow 0$$. Thus, using Taylor expansions in $$(1)$$, $$e^{-\pi n}(1+o(1))= (-1)^n \varepsilon_n(1 +o(1))$$ which means that $$\varepsilon_n=(-1)^ne^{-\pi n}(1+o(1))$$

$$\boxed{x_n=\pi n+(-1)^n e^{-\pi n} + o(e^{-\pi n})}$$ The zeroes therefore converge exponentially fast towards multiples of $$\pi$$.

You can keep playing that game and plug the residual back into $$(1)$$ to find the next term in the expression of $$x_n$$, but this is already converging quite fast.

If you build the series expansion of the rhs of $$y=e^{-x}-\sin(x)$$ around $$x=n \pi$$ and later series reversion forcing $$y=0$$, you obtain $$x_n=n\pi+\frac{1}{1+e^{(1+i) \pi n}}+\frac{1}{2 \left(1+e^{(1+i) \pi n}\right)^3}+\cdots$$

$$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 0 & 0.5625000000000000000 & 0.5885327439818610774 \\ 1 & 3.0963808805403588820 & 3.0963639324106461156 \\ 2 & 6.2850492723072431593 & 6.2850492733825865338 \\ 3 & 9.4246972547386088459 & 9.4246972547385212191 \\ 4 & 12.566374101689367670 & 12.566374101689367677 \\ 5 & 15.707963117247215942 & 15.707963117247215942 \\ 6 & 18.849555928051171524 & 18.849555928051171524 \\ 7 & 21.991148574847125823 & 21.991148574847125823 \\ 8 & 25.132741228730507464 & 25.132741228730507464 \\ 9 & 28.274333882307613598 & 28.274333882307613598 \\ 10 & 31.415926535897955096 & 31.415926535897955096 \end{array} \right)$$