Consider the well known continued fraction expansion $$ z \tan z = \frac{z^2}{1 - \cfrac{z^2}{3- \cfrac{z^2}{5 - \ldots}}} $$ of the tangent function going back to Euler and Lambert (Lambert used it for showing that $\tan z$ is irrational for rational nonzero values of $z$, which implies the irrationality of $\pi$; Legendre later observed that the same proof gives the irrationality of $\pi^2$). Wall. in his book on continued fractions, claims that the formula is valid "for all $z$".

Is there a nice way of determining the poles of $\tan z$ from looking at the right hand side of this expansion?


2 Answers 2


Just in case, there is a connection with Hankel determinants. Dividing both sides by $z^2$, one has $$\sum_k\mu_k z^k= \frac{\tan z}{z} = \frac{1}{1 - \cfrac{z^2}{3- \cfrac{z^2}{5 - \ldots}}} $$ Then one defines for all $m,n\geq 0$ the Hankel determinant $H^n_m$ is $$H^n_m=\det (\mu_{i+j})_{n\leq i,j\leq n+m-1}.$$ Let $z_1, z_2,\dots$ be the poles of $\frac{\tan z}{z}$, ordered by increase of the modulus. It's classically known (see Chapter 7 of Henrici's "Applied and computational complex analysis", Vol.1) that $$ \lim_{n\to\infty} \frac{H_m^{n+1}}{H_m^n}=\prod_{j=1}^m z_j^{-1},$$ whenever $|z_m|<|z_{m+1}|$. While I don't know $H_m^n$ for all $n$, for $n=0$ it can be read off directly from the continued fraction, using Theorem 11 of Krattenthaler's "Advanced determinantal calculus". In fact, it gives $H_m^0=1$. Perhaps $H_m^n$ are also directly related to the coefficients of the continued fraction.

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    $\begingroup$ Writing the $n$th partial quotient as $P_n(z)/Q_n(z)$, the $P_n$ and $Q_n$ (seem to) have only real roots and are probably well known orthogonal polynomials whose roots tend to even/odd multiples of $\pi/2$, just a guess. $\endgroup$ Dec 31, 2017 at 19:18

This verifies Henri Cohen's comment. Since the function is even, it is simpler to look at $\sqrt{x}tan(\sqrt{x})$, the $Q_n(x)$ is given by $[1,3-x,15-6x,x^2-45x+105,15x^2-420x+945,...]$. Since $deg(Q_n)=\lfloor n/2 \rfloor$, one should look at $p_n(x)=Q_{2n}(x)$, the roots are all real and distinct and interlace so that for each $k>0$ the $k$th smallest root decrease monotononically, the values of the roots divided by $(\pi/2)^2$ are given by $$1 [1.215854203708053257326553558]$$ $$2 [1.000556567684670043780569921, 17.23725648793612881611773345]$$ $$3 [1.000000264021497302735160628, 9.243071092446397955245651578, 74.86672290309583275487793689]$$ $$4 [1.000000000033926859936132129, 9.003785153434677150307469375, 27.94131378418484607770210908, 217.3842838410377339506305367]$$ $$5 [1.000000000000001649799381439, 9.000019418957686887495315571, 25.17287514526503834031255350, 62.70166691403289751607854631, 503.9732693572307379829582146]$$ $$6 [1.000000000000000000036873888, 9.000000041594544789352719498, 25.00432366956189670320729176, 50.59590302069961836345079636, 121.9182580673728203679597413, 1009.551573112532430359872689]$$ $$7 [1.000000000000000000000000430, 9.000000000044261664894057822, 25.00004695103961704597815305, 49.10575115766100355065900035, 88.08889135341595365433652728, 215.8656260646040985633268236, 1824.794335221891993855132913]$$ $$8 [1.000000000000000000000000000, 9.000000000000026169987166719, 25.00000026069484992294309458, 49.00327767012444219447267331, 81.88750001909035955812019217, 142.0164417098733597317941152, 356.4649305639281633307802602, 3056.141713122931766511156386]$$ which gives very fast convergence to $(2k-1)^2$. One can also find the zeros by looking at roots of $p_n(x)=P_{2n}(x)$ or $P_{2n+1}(x)$. In fact for any real $\alpha$, if $P_n/Q_n$ are the convergents to $x tan(x)$, the roots of $P_{2n}(x)-\alpha xQ_{2n}(x)$ will still be real and distinct and converge monotonically to the root of $tan(x)=\alpha$.


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