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.

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