Why are numeric methods the only technique available to solving $\ln(x) = \sin(x)$? Is this $x$ transcendental?

Question

I just read this question about finding the solution to the equation $\ln(x) = \sin(x)$. All the answers focus on using a numerical method to approximate the solution. This is interesting in its own regard, but I am interested in why this is the case. Several questions immediately jump out at me:

1. Is it really the case that there is no analytic expression that represents the solution? Are there general techniques one can use to prove this is the case for similar equations?
2. Is there a connection between these types of solutions and transcendental numbers? Is $x$ a transcendental number?
3. Are there additional complex solutions? Are they more easily expressed? I am not very familiar with complex analysis.

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For the purposes of this question, I will follow the wikipedia definition of an Analytic Expression. This allows the following constructions in the expression: Constant, Variable, Elementary arithmetic operation, Factorial, Integer exponent, N-th root, Rational exponent, Irrational exponent, Logarithm, Trigonometric function, Inverse trigonometric function, Hyperbolic function, Inverse hyperbolic function, Gamma function Bessel function, Special function, Continued fraction, Infinite series

If it is more interesting, what about just a 'closed-form', which allows everything up to the Gamma function in the above list.

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EDIT: I've attempted to use PARI to calculate the continued fraction without much luck. I used the command "x = exp(sin(x))", which returns a truncated power series. I can't really find anything in the documentation for how to get this as a continued fraction; simply doing "contfrac(x)" just returns the power series again inside of some square brackets.

Even if I could get it to display a continued fraction, it seems that PARI is resorting to a numerical method to calculate an approximate value and then displaying this approximate value in different representations. I believe this means the continued fraction series in not simply a truncated version of the "real" (probably) infinite continued fraction, but the entire finite sequence representation of the approximate value. If this is the case, then I cannot look at the sequence to search for a pattern...

Thus question 4. Is there a numerical method for generating the continued fraction?

Answer 1

"Closed form" means expressions of allowed functions. If an equation is solvable in closed form depends therefore on the functions you allow.

The elementary standard functions (Wikipedia: Elementary function), e. g. the trigonometric functions, can be represented as compositions of algebraic functions, $\exp$ and/or $\ln$.

It is known, your equation has a solution.

a) Let us allow the algebraic numbers as closed form:

Because $\sin(x)=-\frac{1}{2}ie^{xi}+\frac{1}{2}ie^{-xi}$, your equation $\ln(x)=\sin(x)$ can be transformed to $\ln(x)=-\frac{1}{2}ie^{xi}+\frac{1}{2}ie^{-xi}$. This is an algebraic equation in $\ln(x)$, $e^{xi}$ and $e^{-xi}$. According to transcendental number theory, such algebraic equations cannot be fulfilled by algebraic numbers $x$. Therefore the solution $x$ is a transcendental number.

b) Let us now allow the Elementary functions as closed form:

The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).

The solutions of an equation of only one unknown are, if they exist, numbers. Each number $c$ is the value of an elementary function, of a constant function $x\mapsto c$, with $x,c\in\mathbb{C}$. Therefore each solution of each equation of only one unknown is in closed form.

Solvability of equations by closed-form expressions is related to the question of existence of inverses of the functions which are contained in the equation.

The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function. You can also take the method of Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.

But Ritt's theorem shows that no antiderivatives, no differentiation and no differential fields are needed for defining the Elementary functions and for answering the question on solvability of equations by applying only Elementary functions.

According to Ritt, your equation $\ln(x)=\sin(x)$ has to be transformed to the homogeneous equation in exp-ln form: $\ln(x)+\frac{1}{2}ie^{xi}-\frac{1}{2}ie^{-xi}=0$. The left side is the function term of an elementary function. Let's name this elementary function $f$. Your equation is than: $f(x)=0$. For solving this equation analytically (in closed form), we have to invert $f$. If the inverse, $\phi$, is an elementary function and its function value at the place $x=0$ does exist, the equation is solvable by an elementary function, and its solution is $x=\phi(0)$.

But $f(x)$ is an algebraic function of $\ln(x)$, $e^{xi}$ and $e^{-xi}$. One can show that there cannot exist an exp-ln representation of your $f(x)$ that contains no algebraic function of algebraic independent arguments. According to the theorem of Joseph Fels Ritt in Ritt 1925, your equation is not solvable by elementary functions therefore.

c) You need numerical methods if the solutions of the equation cannot be expressed by closed-form expressions (That are expressions of only algebraic functions and/or known functions (Special functions, Standard functions)).

Answer 2

Complex solutions: to $\ln(x)=\sin(x)$ solve $z=\sin^{-1}(\ln(z))+2n\pi,z=(2n+1)\pi-\sin^{-1}(\ln(z))$. The solutions are $z_n$ and its conjugates. Maybe the Burniston-Siewert method will work.

Real Solution: We solve: $$\ln(x)=\sin(x)\implies x=2.219107\dots$$

$\def\ind{{0\ne n\in\Bbb Z}}$ Center the function at $x=\frac\pi2$:$$f(x)=\ln\left(x+\frac\pi2\right)-\sin\left(x+\frac\pi2\right)-\ln\left(\frac\pi2\right)+1=-\ln\left(\frac\pi2\right)+1 $$

Complex Fourier series of $f^{-1}(x)=h(x)$ on $[0,L],L=f\left(\frac\pi2\right)=\ln(2)+1$ after removing $x=0$ term and substituting $h(t)\to t$

$$h(x)=\frac1{2L}\sum_{n\in\Bbb Z}e^\frac{\pi i n x}L\int_{-L}^L h(t)e^{-\frac{\pi i n t}L}dt=\frac{\pi+1}{2L}-\frac\pi 4+\frac1{2L}\sum_\ind e^\frac{\pi i n x}L \int_0^\frac\pi2t e^{-\frac{\pi i n}L\left(\ln\left(\frac2\pi t+1\right)+\cos(t)+1\right)}\left(\sin(t)+\frac2{2t+\pi}\right)dt$$

Integration by parts gives:

$$\frac\pi 4\frac xL-\sum_\ind e^\frac{\pi i n x}L\frac{ie^{\pi i n\frac{\ln(\pi)-1}L}}{2\pi n}\int_\frac\pi2^\pi (2 t)^{-\frac{i\pi n}L} e^{\frac{\pi i n}L\sin(t)}dt$$

so far, we have $h(x)$:

Plugging in $x=1-\ln\left(\frac\pi2\right)$ simplifies the series hence a new integrand. Finally, use the Jacobi-Anger expansion and generalized incomplete $\Gamma(z,a,b)$ function: $$\int_\frac\pi2^\pi t^{-\frac{i\pi n}L} e^{\frac{\pi i n}L\sin(t)}=J_0\left(\frac{\pi n}L\right)\frac{L\pi^{1-\frac{\pi i n}L}\left(2^{\frac{\pi i n}L-1}-1\right)}{\pi i n-L}+i\sum_{0\ne m\in\Bbb Z}J_m\left(\frac{\pi n}L\right)\frac{(-i m)^\frac{\pi i n}L}m\Gamma\left(1-\frac{\pi i n}L,-\frac{\pi i m}2,-\pi i m\right)$$

The integral, for the Fourier series converging to $2.219107\dots$, does match the above result. Therefore:

$$\bbox[1px,border: 1px solid blue]{\ln(x)=\sin(x)\implies x=\frac\pi2+\frac{2+2\pi-\pi\ln(\pi)}{ \ln(16)+4}+\sum_\ind\frac 1{2\pi n}\left(\sum_{0\ne m\in\Bbb Z}J_m\left(\frac{\pi n}{\ln(2)+1}\right)\frac{(-i m)^\frac{\pi i n}{\ln(2)+1}}m\Gamma\left(1-\frac{\pi i n}{\ln(2)+1},-\frac{\pi i m}2,-\pi i m\right)+iJ_0\left(\frac{\pi n}{\ln(2)+1}\right)\frac{(\ln(2)+1)\pi^{1-\frac{\pi i n}{\ln(2)+1}}\left(2^{\frac{\pi i n}{\ln(2)+1}-1}-1\right)}{\ln(2)-\pi i n+1}\right)}$$

Answer 3

If you use for example $$\log(x)-\sin(x)=\log(a)-\sin(a)-\sum_{n=1}^\infty \alpha_n (x-a)^n$$ where $$\alpha_n=\frac{\sin \left(a+n\frac{\pi }{2}\right)}{n!}+\frac{(-1)^n }{n\,a^{n}}$$ and use power series reversion, you would obtain $$x=a+\sum_{n=1}^\infty \beta_n\,t^n \qquad \text{where} \qquad t= \frac{\sin (a)-\log(a)}{\alpha (1)}$$ All coefficients can be explicitely computed using the explicit formula given by Morse and Feshbach for the $n^{\text{th}}$ term.

Using $a=\frac {2\pi}3$ and truncating to some order $p$, converted to decimals, the values are $$\left( \begin{array}{cc} p & x_{(p)} \\ 1 & 2.224078158 \\ 2 & 2.218589175 \\ 3 & 2.219158811 \\ 4 & 2.219101627 \\ 5 & 2.219107710 \\ 6 & 2.219107098 \\ 7 & 2.219107152 \\ 8 & 2.219107149 \\ \end{array} \right)$$

Another solution is to use the first iterate of any Newton-like method of order $n$ to have a fully explicit result.