Interval of convergence of Lagrange's infinite series

Question

I am reading a book on Orbital Mechanics for Engineering Students by Howard D. Curtis. In that book it was mentioned (in page 119) that there is no closed form solution for $E$ as a function of the eccentricity $e$ in the equation $E-e\sin E=M_e$ but there exists infinite series solutions one of which is given by $$E = M_e+ \sum_{i=0}^{\infty} a_n e^n$$ where $$a_n= 2^{1-n} \sum_{k=0}^{\lfloor n/2\rfloor} \frac{(-1)^k}{(n-k)!k!} (n-2k)^{n-1} \sin((n-2k)M_e)$$ It was also given that the above series converges for $|e|<0.662743419$.

Now my questions are

1. How can we derive an infinite series solutions given such transcendental equation to solve? (I have no idea.)

2. How can we find interval of convergence of this infinite series? (I have tried using Ratio tests etc., to find interval of convergence but they are not that helpful.)

Answer 1

I shall follow the book [KT].

How can we derive an infinite series solutions given such transcendental equation to solve?

In Subsection 3.3.1 they look for a solution of Kepler’s equation $E-e\sin E=M$ in a form of a series of powers of the eccentricity $e$ with the coefficients dependent of $M$, as follows. Represent $E$ in a standard form of a Maclaurin series

$$E=\sum_{k=0}^\infty a_k(M)e^k,\mbox{ where }a_k=\frac{1}{k!}\frac{d^k E}{de^k}{\huge|}_{e=0}.\tag{3.21}$$

Recall that $M$ is considered to be a real parameter. When $e=0$ we have $E=M$, so $a_0=M.$ The remaining coefficients $a_k$ periodically depend on $M$, so (3.21) usually is written as

$$E-M=\sum_{k=0}^\infty a_k(M)e^k.$$

Differentiating Kepler’s equation, we obtain $(1-e \cos E) dE-\sin E de = 0,$ so $$\frac {dE}{de}=\frac{\sin E}{1-e\cos E}.$$

Since $M$ is considered as a parameter, we have $dM=0.$

Calculate the second derivative. $$\frac {d^2E}{de^2}=\left(\frac{\partial}{\partial e}+\frac {dE}{de}\frac{\partial}{\partial E}\right) \frac {\sin E}{1-e \cos E} =\frac{-5e\sin E + 4 \sin 2E - e \sin 3E}{4(1-e \cos E)^3}.$$

Assume that $$\frac{d^k E}{de^k} =\frac{\Phi_k(e,E)}{4^{k-1}(1-e \cos E)^{2k-1}},\tag{3.24}$$ where $\Phi_k$ is a Fourier sine polynomial of order at most $2k-1$, whose coefficients are polynomials of $e$ of degree at most $k-1$ with integer coefficients. This hold for $k = 1, 2$:

$$\Phi_1 = \sin E,\,\Phi_2 = -5e \sin E + 4 \sin 2E - e \sin 3E.$$

The assumption is proved by induction taking into account a recurrence $$\Phi_{k+1} = 4(2k-1)(\cos E-e)\Phi_k + (4\sin E-2e\sin 2E)\frac{\partial\Phi_k}{\partial E}+ (4 + 2e^2-8e\cos E + 2e^2\cos 2E) \frac{\partial\Phi_k}{\partial e},\tag{3.25}$$ which is can easily derived differentiating (3.24).

Formula (3.25) allows to find $\Phi_k$ up to any required order $k$. Then it remains to put $$a_k = \frac 1{4^{k-1}k!}\Phi_k(0,M).$$

How can we find interval of convergence of this infinite series?

They wrote that they do not provide the complete proofs because the latter are too hard. As I understood, in Subsection 3.8.2 they consider the series $E(e)$ as a function of a complex variable $e$ and look for its singularities. The essential those turn out to be solutions of Equation (3.166) $1- E\cos e=0$. It follows

$$e=\frac{E-M}{\sin E}=\frac 1{\cos E}.$$

$\newcommand{\ch}{\operatorname{ch}}$ $\newcommand{\sh}{\operatorname{sh}}$ $\newcommand{\cth}{\operatorname{cth}}$

Let $E=u+iv$. It follows $$(u-M) \cos u \ch v + v \sin u \sh v = \sin u \ch v,$$ $$- (u- M) \sin u \sh v + v \cos u \ch v = \cos u \sh v,$$ $$|e| = (\ch^2 v -\sin^2 u)^{-1/2}.\tag{3.180}$$

When $\sin u = 0 \Leftrightarrow u = k\pi$, the first two equations (3.180) transform to $$ (u-M) \ch v = 0,\, v \ch v = \sh v \Rightarrow v = 0,\, M = u = k\pi,\, |e| = 1.$$

When $\cos u = 0 \Leftrightarrow u = 2k\pi\pm \pi/2$, the first two equations (3.180) transform to $$v \sh v = \ch v,\, (u-M) \sh v = 0 =0 \Rightarrow u = M = 2k\pi\pm \pi/2.$$

In this case $v = 1.199678640$ is a root of the equation $v = \cth v$, $|e|= 1/\sh v = 0.662743419.$

Thus, the convergence radius $R(M)$ of the expansions with respect to the powers of the eccentricity is equal to $1$ when $M = k\pi$ and to $R_0=R(\pi/2)=0.662743419$ when $M=2k\pi\pm \pi/2$. It can be shown that $R_0\le R(M)\le 1$. The number $R_0$ is called Laplace limit. When $|e|<R_0$ then the series of powers of the eccentricity converge for all $M$. When $|e|>R_0$, the series diverge for some values of $M$.

References

[KT] K.V. Kholshevnikov, V.B. Titov, Two-body problem, SPb. State University, Saint-Petersburg, 2007. (The book is in Russian but you can look at the equations).

Answer 2

Part 1

The solution can be obtained by the Lagrange inversion theorem [1].

The equation $E - e \sin E = M_e$ is written as $$\frac{E - M_e}{\sin E} = e.$$ Let $f(E) = \frac{E - M_e}{\sin E}$. Clearly, $f$ is analytic at $M_e$, $f(M_e) = 0$, and $f'(M_e) = \frac{1}{\sin M_e} \ne 0$.

By the Lagrange inversion theorem, we have $$E = M_e + \sum_{n=1}^\infty b_n e^n$$ where $$b_n = \frac{1}{n!} \lim_{w\to M_e} \frac{\mathrm{d}^{n-1}}{\mathrm{d} w^{n-1}} (\sin w)^n.$$

Clearly, $a_0 = 0$, $a_1 = b_1 = \sin M_e$. Let us prove that $a_n = b_n$ for $n \ge 2$.

It suffices to prove that, for $n\ge 2$, $$2^{1-n} \sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} (-1)^k (n-2k)^{n-1}\sin ((n-2k)w) = \frac{\mathrm{d}^{n-1}}{\mathrm{d} w^{n-1}}(\sin w)^n.$$

If $n$ is odd, by using the identity [2] $$\sin^n w = \frac{2}{2^n}\sum_{k=0}^{(n-1)/2} (-1)^{\frac{n-1}{2}-k}\binom{n}{k}\sin ((n-2k)w),$$ we have \begin{align} \frac{\mathrm{d}^{n-1}}{\mathrm{d} w^{n-1}}(\sin w)^n &= \frac{2}{2^n}\sum_{k=0}^{(n-1)/2} (-1)^{\frac{n-1}{2}-k}\binom{n}{k} (n-2k)^{n-1}(-1)^{(n-1)/2}\sin ((n-2k)w)\\ &= \frac{2}{2^n}\sum_{k=0}^{(n-1)/2} (-1)^k\binom{n}{k} (n-2k)^{n-1}\sin ((n-2k)w). \end{align} The desired result follows.

If $n$ is even, by using the identity [2] $$\sin^n w = \frac{1}{2^n}\binom{n}{n/2} + \frac{2}{2^n}\sum_{k=0}^{n/2 - 1}(-1)^{n/2-k}\binom{n}{k}\cos ((n-2k)w),$$ we have \begin{align} \frac{\mathrm{d}^{n-1}}{\mathrm{d} w^{n-1}}(\sin w)^n &= \frac{2}{2^n}\sum_{k=0}^{n/2 - 1}(-1)^{n/2-k}\binom{n}{k}(n-2k)^{n-1}(-1)^{n/2} \sin ((n-2k)w)\\ &= \frac{2}{2^n}\sum_{k=0}^{n/2 - 1}(-1)^k \binom{n}{k}(n-2k)^{n-1}\sin ((n-2k)w). \end{align} The desired result follows.

We are done.

$\phantom{2}$

Part 2

By root test, the series converges if $$e < \frac{1}{\limsup_{n\to \infty} \sqrt[n]{|a_n|}}.$$

We have \begin{align} \sqrt[n]{|a_n|} &= \left|\frac{1}{n!} 2^{1-n} \sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} (-1)^k (n-2k)^{n-1}\sin ((n-2k)M_e)\right|^{1/n}\\[5pt] &\le \left(\frac{1}{n!} 2^{1-n} \sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} (n-2k)^{n-1}\right)^{1/n}\\[5pt] &= \left(\frac{1}{n!} 2^{1-n}n^{n-1}\right)^{1/n}\left(\sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{1/n}. \end{align} By using $\lim_{n\to\infty} (\frac{1}{n!} 2^{1-n}n^{n-1})^{1/n} = \frac{1}{2}\mathsf{e}$, we have $$\frac{1}{\limsup_{n\to \infty} \sqrt[n]{|a_n|}} \ge \liminf_{n\to \infty} \frac{2}{\mathsf{e}}\left(\sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{-1/n} \triangleq \lambda.$$ Then, the series converges if $$e < \lambda.$$

I $\color{blue}{\textrm{GUESS}}$ (note: here $\lim_{n\to \infty}$ rather than $\liminf_{n\to \infty}$) $$\lim_{n\to \infty} \underbrace{\frac{2}{\mathsf{e}}\left(\sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{-1/n}} _{B_n} = 0.662743419... $$

I did some numerical experiments by Maple. For example, it seems $B_n$ is non-increasing; When $n=1000$ (Maple can not easily evaluate $B_n$ for larger $n$),
$$\frac{2}{\mathsf{e}}\left(\sum_{k=0}^{\mathrm{floor}(n/2)} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{-1/n} = 0.6627434531...$$

Reference

[1] https://en.wikipedia.org/wiki/Lagrange_inversion_theorem

[2] https://en.wikipedia.org/wiki/List_of_trigonometric_identities