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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?

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1 Answer 1

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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