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When we want to solve $x = \cos\left(\,x\,\right)$, usually we use Matlab. Here is my matlab code from Zhihu, a Chinese academic discussion website like Quora (And I run it on my laptop):

x = 1; %The code is from Zhihu. 本代码来自知乎

delta = inf;


count = 0;


while abs(delta) > 1e-10


delta = (x - cos(x))/(1 + sin(x));


x = x - delta;%Newton's method, 牛顿迭代法


count = count + 1;


disp([num2str(count),':',num2str(x， 10)]);%Output the answer


end

And its output:

1:0.7503638678
2:0.7391128909
3:0.7390851334
4:0.7390851332
5:0.7390851332

It converges fastly.

However a paper, published by $2$ professors in North Carolina State University and was cited $30$ times, claims that they have got a complicated $\textbf{exact}$ analytical solution to Kepler's equation: $$ e\sin\left(\,E\,\right)= E - M $$

Suppose $x = \cos\left(\,x\,\right)$. Let $x = y - \pi/2$, then it is reduced to Kepler's equation $\sin\left(\,y\,\right) = y - \pi/2$ with $$ e = 1\,,\ E = y\,,\ M = {\pi \over 2} $$ According to the paper, $x = \cos\left(\,x\,\right)$ can be solved exactly ( That's incredible ). How do you look on this paper ? I fail to catch its steps while reading it.

Muses_China
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2 Answers2

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This is a misunderstanding caused by the fact that the word "analytic" is used two different ways with two completely different meanings. It's a matter of the same English word being used in two different ways.

The authors of the paper are using the word "analytic" (or perhaps "analytical") in the sense of an analytic function, i.e., a function that is locally given by a convergent power series.

They are not using the word in the sense of an analytic expression, i.e. a mathematical expression constructed using well-known operations that lend themselves readily to calculation.

mweiss
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Forsooth! If $F(z)$ is an analytic function on a region $G$ of the complex plane, and the only zero of $F(z)$ in $G$ is at $z=p$, and $\Gamma$ is a positively oriented closed contour around $p$ (with winding number $1$) inside $G$, then by the Residue Theorem $$ p = \frac{1}{2\pi i m} \oint_\Gamma \dfrac{z F'(z)}{F(z)}\; dz $$ where $m$ is the multiplicity of the zero at $z=p$. Thus solutions to equations $F(z) = 0$ for analytic functions $F$ can always be expressed "analytically" as integrals.

In this case you could take $F(z) = \cos(z) - z$. There is one simple zero inside the unit circle, and so it is

$$ \frac{1}{2\pi i} \oint_\Gamma \dfrac{z (-\sin(z) - 1)}{\cos(z)-z} \; dz$$

which using the parametrization $z = e^{i\theta}$ becomes

$$ \frac{1}{2\pi} \int_0^{2\pi} \dfrac{e^{2i\theta}\left(1 + \sin\left(e^{i\theta}\right)\right)}{e^{i\theta} - \cos\left(e^{i\theta}\right)}\; d\theta $$

I haven't bothered to check if this is the same integral these authors find. Of course, many other choices of contour and parametrization could be used. But no formula of this type would be worth publishing in the 20th or 21st centuries.

Robert Israel
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