## Lambert’s continued fractions for tanh(x) and tan(x) revisited

In earlier posts (since deleted) I discussed some of the properties of continued fractions and showed how Lambert’s continued fractions for tanh x and tan x could be derived and used to prove that these functions are always irrational numbers for non-zero, rational x and consequently that the exponential number, e, and pi are irrational.

In what follows, I simplify and generalise those discussions. The following is an outline, the detail can be found in the pdf here – revised in May 2016.

Irrationality criteria and continued fractions

Suppose f(x) is some function that we can represent in the following manner:

$f(x) = s_1(x)=\displaystyle\frac{a_1(x)}{b_1(x)+s_2(x)}$

$s_k(x)=\displaystyle\frac{a_k(x)}{b_k(x)+s_{k+1}(x)} (k\geq2)$

where the $a_k, b_k$ are polynomials in x with integer coefficients and $s_k$ has the property that it becomes rapidly smaller in absolute value for a given x as k becomes larger. We call this a continued fraction representation of f(x) and it is not too hard to derive from it the following expression

$q_n(x)f(x)-p_n(x)=\displaystyle\frac{(-1)^{n-1}(a_na_{n-1}...a_1)s_{n+1}(x)}{q_n(x)+q_{n-1}(x)s_{n+1}(x)}=R_n(x)$

where p, q are defined recursively by

$p_{-1}=1$ $p_0=0$

$q_{-1}=0$ $q_0=1$

and for $k\geq0$

$p_{k+1}(x)=a_{k+1}p_{k-1}(x)+b_{k+1}p_k(x)$

$q_{k+1}(x)=a_{k+1}q_{k-1}(x)+b_{k+1}q_k(x)$

Clearly $p_n$ and $q_n$ are integral polynomials in x. If d(n) is the degree of whichever of $p_n$ and $q_n$ has the higher degree, and if x = u/v is rational with u, v being integers having greatest common factor 1, then if we multiply the above expression by $v^{d(n)}$ then it takes the form

$kf(x)-h = v^{d(n)}R_n(x)$

where k, h are integers and we can make use of the following theorem:

f(x) is irrational if there exist integers h and k such that 0 < |kf(x) – h| < ε for any given 1 > ε > 0

Lambert’s continued fractions

Suppose $f_1(x)$ and $f_2(x)$ are functions that are can be expressed as even, absolutely convergent Maclaurin series. Then we can obtain a sequence of equations

$f_{n+1}(x)\displaystyle\frac{f_{n+1}(0)}{f_n(0)}f_n(x)=x^2f_{n+2}(x)$

If we take $f_1(x)=\cosh x$ and $f_2(x)=\sinh x/x$
then it is not too hard to show that $\tanh x/x$ can be expressed as a continued fraction of the form

$\tanh x/x=\displaystyle\frac{1}{1+s_2(x)}$

$s_k(x)=\displaystyle\frac{x^2}{(2k-1)+s_{k+1}(x)}$

Analogously

$\tan x/x=\displaystyle\frac{1}{1+s_2(x)}$

$s_k(x)=\displaystyle\frac{-x^2}{(2k-1)+s_{k+1}(x)}$

These are Lambert’s continued fractions. The process described above can be carried through for them to show that tanh x/x and tan x/x, and hence tanh x and tan x, are irrational for non-zero rational x.