How can I solve the equation $y=e^{\cos(x)}\sin(x)$?

Question

I was reading about the Lambert W function, and I want to know if it is possible to extend the ideas to solve the given equation for real values of x. $$y=\sin(x)e^{\cos(x)} $$ I know that the W function is only for equations of this form: $$W(x)e^{W(x)}=x$$ I would like any help or ideas regarding this problem, or about the impossibility of solving it. Thanks in advance. [Note: Wolfram|Alpha is unable to solve the equation]

Specifications:

I would like an explicit inverse function to $$x\rightarrow{\sin(x)e^{\cos(x)}}$$ not a numerical solution. The special function $W$ can be used in the explicit form. The original function has the restricted domain of $[-\arccos(a),\arccos(a)]$ for $a=\frac{\sqrt{5}-1}{2}$ and the inverse should have the domain $[-\sqrt{a}e^a,\sqrt{a}e^a]$ for the same $a$

[This question was migrated from MathOverflow]

Answer 1

$$y=\sin(x)e^{\cos(x)}$$ $$\sin(x)e^{\cos(x)}=y$$

As the comments propose:

$x\to\arccos(z)$: $$\sin(\arccos(z))e^z=y$$ $$(1-z^2)(e^z)^2=y^2$$ $$(1-z^2)(e^z)^2-y^2=0\tag{1}$$

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

We see, for algebraic general $y$, equation (1) is an algebraic equation of both $z$ and $e^z$ with no univariate factor. The equation cannot have solutions except $0$ that are elementary numbers therefore.

$$(1-z^2)e^{2z}=y^2$$ $z\to\frac{t}{2}$: $$-\frac{1}{4}(-4+t^2)e^t=y^2$$ $$(t^2-4)e^t=-4y^2$$

As the other answer proposes:
We see, we cannot solve this equation in terms of Lambert W, but in terms of Generalized Lambert W. See page 3 in [Mező/Baricz 2017].

$$(t+2)(t-2)e^t=-4y^2$$ $$t=W\left(^{-2,+2}_{};-4y^2\right)$$ $$z=\frac{1}{2}W\left(^{-2,+2}_{};-4y^2\right)$$ $$x=\arccos\left(\frac{1}{2}W\left(^{-2,+2}_{};-4y^2\right)\right)$$

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

A closed-form expression is a placeholder for its analytic representations, and furthermore, we can use its algebraic properties and special values.

[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

Answer 2

As @Zoe Allen commented, squaring and letting $z=\cos(x)$, we have to solve for $z$ the equation $$y^2=(1-z^2)\,e^{2z}\qquad \implies\qquad e^{-2z}=\frac {1-z^2} {y^2}$$ which has a solution in terms of the generalized Lambert function (have a look at equation $(4)$).

Answer 3

For an approximate solution of the equation.

We want to inverse $$y=\sin(x)\,e^{\cos(x)}$$ for $$0 \leq x \leq \cos^{-1} (a)\qquad\qquad 0 \leq y\leq \sqrt a\, e^a \qquad \text{where} \qquad a=\frac{\sqrt{5}-1}{2}$$

We have $$e^{\cos(x)}=e \sum_{n=0}^\infty \frac{a_n}{b_n} x^{2n}$$ The coefficients $a_n$ and $b_n$ form respectively sequences $A047689$ and $A047690$ in $OEIS$.

So, we know the series expansion of $y$ $$y=e \sum_{n=0}^\infty(-1)^n\, c_n\,x^{2n+1}$$ the first $c_n$ forming the sequence $$ \left\{1,\frac{2}{3},\frac{31}{120},\frac{379}{5040},\frac{1639}{90720 },\frac{150349}{39916800},\frac{4373461}{6227020800},\frac{39074491 }{326918592000},\cdots\right\}$$

To give an idea of the accuracy, using only the terms given in the above table $$\Phi=\int_0^{\cos^{-1}(a)} \Big(\sin(x)\,e^{\cos(x)}-S_{15}\Big)^2\,dx=1.80\times 10^{-12}$$

Using power series reversion and noting $\color{red}{t=\frac y e}$, is obtained $$x=\sum_{n=0}^\infty c_n\, t^{2n+1}$$ the first coefficients being $$\left\{1,\frac{2}{3},\frac{43}{40},\frac{757}{336},\frac{3091}{576}, \frac{582581}{42240},\frac{22334047}{599040},\frac{1872179}{17920},\frac{52772932189}{175472640},\cdots\right\}$$

Some results obtained with the $c_n$ given above

$$\left( \begin{array}{ccc} y & \text{estimate} & \text{solution} \\ 0.05 & 0.0183981233 & 0.0183981233 \\ 0.10 & 0.0368212081 & 0.0368212081 \\ 0.15 & 0.0552944907 & 0.0552944907 \\ 0.20 & 0.0738437638 & 0.0738437638 \\ 0.25 & 0.0924956771 & 0.0924956771 \\ 0.30 & 0.1112780635 & 0.1112780635 \\ 0.35 & 0.1302203043 & 0.1302203043 \\ 0.40 & 0.1493537459 & 0.1493537459 \\ 0.45 & 0.1687121881 & 0.1687121881 \\ 0.50 & 0.1883324635 & 0.1883324635 \\ 0.55 & 0.2082551405 & 0.2082551405 \\ 0.60 & 0.2285253893 & 0.2285253893 \\ 0.65 & 0.2491940675 & 0.2491940675 \\ 0.70 & 0.2703191012 & 0.2703191082 \\ 0.75 & 0.2919672740 & 0.2919673012 \\ 0.80 & 0.3142165817 & 0.3142166781 \\ 0.85 & 0.3371593815 & 0.3371597014 \\ 0.90 & 0.3609066768 & 0.3609076750 \\ \end{array} \right)$$

Notice that we can make a simpler approximation transforming the series into the coresponding $[2n+1,2n]$ Padé approximant of it $P_n$ such as $$P_2=t\,\frac {1-\frac{559253 }{209478}t^2+\frac{18722873 }{21995190} t^4} {1-\frac{698905 }{209478}t^2+\frac{3911235 }{1955128}t^4 }$$ whose error is $\sim \frac 2 5 t^{11}$.