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When I was a child, I kept pressing $\cos\texttt{ans}$ and found it becomes the number 0.9998477415...

Now I know it's the root of $\displaystyle x=\cos \frac{\pi x}{180}$.

Can we find a series expressions for this number?

Is it a transcendental number?

Attention: This is $\color{red}{\texttt{not}}$ Dottie Number!

But I think Dottie Number is helpful for this question: Explaining $\cos_\infty(x)$

Community
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Aster
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  • It is definitely a transcendental equation, so it's not possible to express it via traditional set of analytical functions... the number itself is probably transcendental too, but these things usually require very difficult proofs. Remember, even transcendentality of $\pi$ took a century. – orion Dec 01 '17 at 08:44
  • Dottie Number in that link has a series expression with Bessel functions, and Dottie Number already know as a transcendental number. – Aster Dec 01 '17 at 08:46
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    I think the proper way of approaching this is to look at $x=\cos k x$ for a general $k$. – orion Dec 01 '17 at 08:49
  • In that way, you again get to the Kepler equation and you can use the same procedure to get to the series representation. – orion Dec 01 '17 at 08:51
  • Niven's theorem implies at least that the number must be irrational. – Peter Dec 01 '17 at 08:53
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    For $x\approx 1$ and $k\approx 0$ (our case) because of $$x-\cos(kx)=1-\cos(k)+(x-1)(k\sin(k)+1)+O(k^2(x-1)^2)$$ the function $x-\cos(kx)$ can very well be approximated by $1-\cos(k)+(x-1)(k\sin(k)+1)$, hence the solution of $x=\cos(kx)$ is very close to $$\frac{\cos(k)-1}{k\sin(k)+1}+1$$ For $k=\frac{\pi}{180}$, the error is about $3.5 \cdot 10^{-12}$ – Peter Dec 01 '17 at 10:44

2 Answers2

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If we directly follow the solution given in your link, we need to get to the Kepler equation $M=E-e \sin E$ with a know solution.

We start at $$x=\cos kx$$ Substitute $$E=\frac{\pi}{2}-kx$$ multiply the above equation with $k$ and do the substitution: $$\frac{\pi}{2}-E=k \sin E$$

$$\frac{\pi}{2}=E+k \sin E$$ Comparing, we get $e=-k=-\pi/180$ and $M=\frac{\pi}{2}$.

The series from Kepler equation wiki page doesn't converge that well (you need a few terms to even get close to $1$), because $E\sim \frac{\pi}{2}-\frac{\pi}{180}$ isn't small in our case.

$$E={\displaystyle {\frac {1}{1-e}}M-{\frac {e}{(1-e)^{4}}}{\frac {M^{3}}{3!}}+{\frac {(9e^{2}+e)}{(1-e)^{7}}}{\frac {M^{5}}{5!}}-{\frac {(225e^{3}+54e^{2}+e)}{(1-e)^{10}}}{\frac {M^{7}}{7!}}+{\frac {(11025e^{4}+4131e^{3}+243e^{2}+e)}{(1-e)^{13}}}{\frac {M^{9}}{9!}}+\cdots }$$

Using all of the terms, it barely gets below $1$ properly.

It's better if we attack the original equation above, and express it around $1$. The fact that iteration is very quick, and looking at the graph, you see that it might be best to just iterate the series expansion. Let's say we want up to fourth power terms... just put the series into itself:

$$x=1-\frac{(kx)^2}{2!}+\frac{(kx)^4}{4!}-\frac{(kx)^6}{6!}+\cdots$$ $$x=1-\frac{(k(1-\frac{(k(1-\frac{(kx)^2}{2!}))^2}{2!}))^2}{2!}+\frac{(k(1-\frac{(kx)^2}{2!}))^4}{4!}-\frac{(kx)^6}{6!}+\cdots$$ $$x=1-\frac{1}{2}k^2+\frac{13}{24}k^4+\cdots$$ First correction gives you 0.99984769 already, correct to $7$ places. A longer series (I used Mathematica) is

$$1 - \frac{1}{2}k^2 + \frac{13}{24} k^4 - \frac{541}{720} k^6 + \frac{9509}{8064} k^8 - \frac{ 7231801}{3628800}k^{10} + \frac{1695106117}{479001600} k^{12} - \frac{567547087381}{87178291200} k^{14}+\cdots$$

Actually, getting the closed-form Taylor expansion is pretty straight-forward, because the inverse is explicit:

$$k=\frac{\arccos x}{x}$$

Therefore, must (formally) just use the Lagrange Inversion Theorem on the series of this function:

$$x(k)=1+\sum a_n k^n$$

where $a_n$ is $0$ for odd $n$ and for even $n$:

$$a_n=\frac{1}{n!}\lim_{x\to 1}\frac{d^{n-1}}{dx^{n-1}}\left(\frac{x(x-1)}{\arccos x}\right)^n$$ Checking with Mathematica, I get an extra factor of $1/2$ which I don't know where it comes from (I get $-1/4$ for the quadratic term). Help appreciated.

orion
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Using $t=\frac{\pi }{180}$, $c=\cos \left(t\right)$ and $s=\sin\left(t\right)$, the series expansion around $x=1$ is given by $$\cos \left(\frac{\pi x}{180}\right)-x=-(1-c)-(1- t s)(x-1)+$$ $$\sum_{n=2}^\infty \frac {t^n}{n!} \left(c \cos \left(\frac{\pi n}{2}\right)-s \sin \left(\frac{\pi n}{2}\right)\right) (x-1)^n$$ Truncating to some order and using power series reversion $$x=1+y+\frac{c t^2 }{2 (s t-1)}y^2+\frac{t^3 \left(3 c^2 t-s^2 t+s\right)}{6 (s t-1)^2}y^3+O\left(y^4\right)$$ where $y=\frac{1-c}{s t-1}$.

No problem since we know the value of $\sin (1{}^{\circ})$ (have a look here).

Using only the above terms, the estimate is $x=\color{red}{0.999847}648$ while the "exact" solution is $x=\color{red}{0.999847742}$;

Using twice more terms (that is to say to $O\left(y^7\right)$) $$x=\color{red}{0.999847741531088112959810768679799799181872}789$$ while the "exact" solution is $$x=\color{red}{0.999847741531088112959810768679799799181872586}$$

Claude Leibovici
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