# Is this seemingly novel irrational constant also transcendental?

I recently discovered a constant that is constructed as follows:

$\chi=\sum_{n=1}^\infty (\frac{\cos{n}}{2|\cos{n}|}+\frac{1}{2}) 2^{-n}$

Furthermore I can prove that it is an irrational number whose decimal approximation is .555609809015...

I conjecture that it is also a transcendental number but have not been able to prove this.

Anyone else want to take a crack at it? I am not a professional mathematician so if this number is not novel, then I appologize in advance.

One of the reasons I want to know this is that if this prototype number is proven transcendental than that would imply that an infinite set of similar transcendental numbers exists on the open interval (0,1) that are given by

$\chi_f=\sum_{n=1}^\infty (\frac{\cos{f(n)}}{2|\cos{f(n)}|}+\frac{1}{2}) 2^{-n}$

where f(n) is any algebraic function of n.

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Presumably f(n) = 0 is not allowed... –  Simon Lyons Jun 7 '12 at 15:47
Saying that $\chi$ is irrational is a way to saying that the sign of $\cos n$ is not eventually periodic, right? –  Pietro Majer Jun 7 '12 at 16:03
PS: I think the number's around 0.555610 –  Anthony Quas Jun 8 '12 at 1:47
Simon, the zeroes of f(n) are allowed if, as in the example, cos is used. However with other trig functions, f(n)=0 causes division by zero. –  Stuart LaForge Jun 8 '12 at 3:58
Pietro, yes, exactly. –  Stuart LaForge Jun 8 '12 at 3:59

Yes. It's known to be transcendental. The sequence of coefficients of your number is a variant of a Sturmian sequence. It has very low complexity. The definition of this: let the digit sequence be $a_1,a_2,a_3\ldots$ taking values in $\lbrace 0,1\ldots,d-1 \rbrace ^{\mathbb N}$. A subword of length $k$ is a string $a_ia_{i+1}a_{i+2}\ldots a_{i+k-1}$. The complexity, $p(k)$, is a function from $\mathbb N$ to $\mathbb N$ taking $k$ to the number of subwords of the sequence of length $k$.
In a 2007 paper in the Annals of Mathematics (vol 165, p547--565), Adamczewski and Bugeaud (On the complexity of algebraic numbers. I. Expansions in integer bases) showed that if a number is algebraic, then its digit sequence in base $b$ has complexity satisfying $p(k)/k\to\infty$.
In your case, the complexity of the sequence of base 2 digits satisfies $p(k)=2k$. How to see this? Define a map $f$ from $[0,1)$ to $\lbrace 0,1\rbrace$ by $f(x)=1$ if $x\in [0,1/2)$ and 0 otherwise. The $n$th term of your sequence is $f(\alpha n\bmod 1)$, where $\alpha=1/(2\pi)$. Write $T$ for the transformation from $[0,1)$ to itself given by $T(x)=x+\alpha\bmod 1$. Then the $n$th term is just $f(T^n0)$. The sub-block of the digit sequence of length $k$ starting at the $j$th term is $f(T^j0)\ldots f(T^{j+k-1}0)$. Since the $T^j0$ are dense in $[0,1)$, we need to ask how many blocks $f(x)\ldots f(T^{k-1}x)$ are possible.
Consider taking $x=0$ and moving it around the circle (=$[0,1)$) once. As you move it, the $T^ix$ also each move around the circle one time. The sequence changes each time one of the $(T^ix)_{0\le i< k}$ crosses 0 or 1/2. This is a total of $2k$ changes. Hence the sequence takes on $2k$ values as $x$ moves around the circle, hence the estimate for the complexity.
PS: This would be just fine for any linear function of $n$. I'm not sure how this would work for anything else... –  Anthony Quas Jun 7 '12 at 15:49