For a natural number $n$, let $$ \begin{equation} \beta_n(z)=\frac{(1+z)^n+(1-z)^n}{(1+z)^n-(1-z)^n}. \end{equation} $$ Then the coefficients of the numerator and denominator of $\beta_n$ are binomial. For example: $$\begin{equation} \beta_4(z)=\frac{z^4+6z^2+1}{4z^3+4z}=\frac{1}{4} z+\cfrac{1}{\frac{4}{5}z+\cfrac{1}{\frac{25}{16}z+\cfrac{1}{\frac{16}{5} z}}}. \end{equation} $$ Is there a simple formula for the coefficients of the continued fraction for any $n$?


  • the fact that all the coefficients are positive numbers follows from complex analysis, since $\Re\beta_n(z)>0$ for $\Re z >0$

  • computer calculation shows that the factorizations of the coefficients consist of small primes, less than $2(n+1)$

  • the motivation for the problem is the rational approximation of the square root function, since $(z\beta_n)(z^2)\approx z$ for $\Re z >0$

  • a relevant discussion: Binary eigenvalues matrices and continued fractions


1 Answer 1


Recall by De Moivre's Theorem

\begin{equation} \tan n\theta = \frac{n \tan\theta - C(n,3) \tan^3 \theta + \cdots}{1 - C(n,2)\tan^2\theta+ C(n,4)\tan^4 \theta-\cdots} \end{equation}

Let $z = i\tan w$. Substituting and applying this result $$ \beta_n(z) = \beta_n(i\tan w) = -i \cot nw $$ By 1 or 2 which clearly hold for a complex domain

$\displaystyle \tan(nx) = \cfrac{n\tan x}{1 -\cfrac{(n^{2} - 1^{2})\tan^{2}x}{3 -\cfrac{(n^{2} - 2^{2})\tan^{2}x}{5 -\cfrac{(n^{2} - 3^{2})\tan^{2}x}{7 -\cdots}}}}$

with final term $\dfrac{(n^{2} - (n - 1)^{2})\tan^{2}x}{(2n - 1)}$. Simple operations show

\begin{align} \beta_n(z) = -i \cot nw &= {\frac{1}{ni\tan w} +\cfrac{n^{-1}(n^{2} - 1^{2})i \tan w}{3 -\cfrac{(n^{2} - 2^{2})\tan^{2}w}{5 -\cfrac{(n^{2} - 3^{2})\tan^{2}w}{7 -\cdots}}}} \\ &= {\frac{1}{nz} +\cfrac{n^{-1}(n^{2} - 1^{2})z}{3 +\cfrac{(n^{2} - 2^{2})z^2}{5 + \cfrac{(n^{2} - 3^{2})z^2}{7 +\cdots}}}} \\ &= {\frac{z^{-1}}{n} +\cfrac{n^{-1}(n^{2} - 1^{2})}{3z^{-1} +\cfrac{(n^{2} - 2^{2})}{5z^{-1} + \cfrac{(n^{2} - 3^{2})}{7z^{-1} +\cdots}}}} \\ &= {a_1z^{-1} +\cfrac{1}{a_2z^{-1}+\cfrac{1}{a_2z^{-1} + \cfrac{1}{a_3z^{-1} +\cdots}}}} \end{align}

Where $a_1 = \frac{1}{n}$, $a_2 = 3\frac{n}{n^2-1},a_3 = 5\frac{n^2-1}{n(n^2-2^2)},a_4 = 7\frac{n(n^2-2^2)}{(n^2-1)(n^2-3^2)},\cdots$ to $n$ terms. In general

\begin{equation} a_k = (2k-1)\Big(\frac{n}{n^2-1}\frac{n^2-2^2}{n^2-3^2}\frac{n^2-4^2}{n^2-5^2}\cdots\Big)^{(-1)^k} \end{equation}

to $[k/2]$ fractional products, $[x]$ being the ceiling function, where if $k$ is odd the last term has denominator $1$.

Substituting $1/z$ for $z$ now gives an explicit solution of the problem by using $\beta_{2m}(z) = \beta_{2m}(z^{-1})$ and $\beta_{2m+1}(z)= \beta_{2m+1}(z^{-1})^{-1}$.

To verify your case substituting $n=4$ returns $a_1 = 1/4, a_2 = 4/5$, $a_3 = 25/16$, and $a_4 = 16/5$.

Edit: The form of $a_k$ can be simplified. For even $n$ and even $k$ \begin{equation} a_k = (2k-1) \frac{(n-(k-2))(n-(k-4))\cdots n \cdots (n+k-4)(n+k-2)}{(n-(k-1))(n-(k-3))\cdots(n+k-3)(n+k-1)} = (2k-1)\Bigg(\frac{2^{k-1}((n+k)/2-1)!}{((n-k)/2)!}\Bigg)^2 \frac{(n-k)!}{(n+k-1)!} = \frac{2^{2(k-1)}(2k-1)}{n+k-1}{n-k \choose (n-k)/2}{n+k-2 \choose (n+k-2)/2}^{-1} \end{equation} Setting $k=n$, $a_n = 2^{2(n-1)}((n-1)!)^2/(2n-2)!$ giving a power of $2$ in the numerator. For even $n$ and odd $k$ take the reciprocal of $a_{k+1}$ and multiply by $(2k-1)(2k+1)/(n^2-k^2)$

\begin{equation} a_k = \frac{2k-1}{2^{2k}(n-k)}{n-k-1 \choose (n-k-1)/2}^{-1}{n+k-1 \choose (n+k-1)/2} \end{equation}

Treating odd $n$ and odd $k$ separately because of $n=k=1$ \begin{align} a_k &= (2k-1)\frac{(n-(k-2))(n-(k-4))\cdots(n+k-4)(n+k-2)}{(n-(k-1))(n-(k-3)\cdots n \cdots (n+k-3)(n+k-1)} \\ &= (2k-1)\Bigg(\frac{2^{k-1}((n+k-2)/2)!}{((n-k)/2)!}\Bigg)^2 \frac{(n-k)!}{(n+k-1)!}\\ &= \frac{2^{2(k-1)}(2k-1)}{n+k-1}{n-k \choose (n-k)/2}{n+k-2 \choose (n+k-2)/2}^{-1} \end{align} For odd $n$ even $k$ take the reciprocal of $a_{k+1}$ and multiply by $(2k-1)(2k+1)/(n^2-k^2)$ \begin{equation} a_k = \frac{2k-1}{2^{2k}(n-k)}{n-k-1 \choose (n-k-1)/2}^{-1}{n+k-1 \choose (n+k-1)/2} \end{equation}

  • $\begingroup$ Thank you, it's good that your continued fraction terminates... $\endgroup$
    – DVD
    Oct 12, 2018 at 19:37
  • $\begingroup$ Excellent work! The computer calculation shows that $a_n$ has a power of $2$ in the numerator. Can you explain that? $\endgroup$
    – DVD
    Oct 15, 2018 at 20:25
  • $\begingroup$ Please see the following question with my answer for the application of your result: math.stackexchange.com/questions/2941564/… $\endgroup$
    – DVD
    Oct 15, 2018 at 21:55
  • 1
    $\begingroup$ You may wish to verify the factorial coefficients for odd $n$. $\endgroup$
    – Jonny Boy
    Oct 16, 2018 at 3:16
  • 1
    $\begingroup$ Apologies, I corrected a subtle error in the odd coefficients. $\endgroup$
    – Jonny Boy
    Oct 18, 2018 at 21:02

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