# Coefficients of binomial continued fractions

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

NOTES:

• 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

Recall by De Moivre's Theorem

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

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

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

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 $$k$$ $$$$a_k = (2k-1) \frac{(n-(k-2))(n-(k-4))\cdots n \cdots (n+k-4)(n+k-2)}{(n-(k-1))(n-(k-3))\cdots(n+k-3)(n+k-1)} = (2k-1)\Bigg(\frac{2^{k-1}((n+k)/2-1)!}{((n-k)/2)!}\Bigg)^2 \frac{(n-k)!}{(n+k-1)!} = \frac{2^{2(k-1)}(2k-1)}{n+k-1}{n-k \choose (n-k)/2}{n+k-2 \choose (n+k-2)/2}^{-1}$$$$ Setting $$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)$$

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

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

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