Bourbaki

Exercise 1, § 2

Functions of a Real Variable, VI, 292

Establish the formulae

tan⁡z=∑n=1∞(−1)n−122n(22n−1)b2nz2n−1(2n)! \tan z = \sum_{n=1}^{\infty} (-1)^{n-1} 2^{2n} (2^{2n} - 1) b_{2n} \frac{z^{2n-1}}{(2n)!}
1sin⁡z=1z+∑n=1∞(−1)n−12(22n−1−1)b2nz2n−1(2n)! \frac{1}{\sin z} = \frac{1}{z} + \sum_{n=1}^{\infty} (-1)^{n-1} 2 (2^{2n-1} - 1) b_{2n} \frac{z^{2n-1}}{(2n)!}

here the series on the right-hand sides converge absolutely, the first for ∣z∣<π2|z| < \frac{\pi}{2} and the second for ∣z∣<π|z| < \pi (express tan⁡z\tan z and 1/sin⁡2z1/\sin 2z as linear combinations of cot⁡z\cot z and cot⁡2z\cot 2z). Deduce that the numbers 22n(22n−1)2nb2n\frac{2^{2n}(2^{2n} - 1)}{2n} b_{2n} are integers. (Use the following lemma: in two absolutely convergent series ∑n=0∞αnznn!\sum_{n=0}^{\infty} \alpha_n \frac{z^n}{n!}, ∑n=0∞βnznn!\sum_{n=0}^{\infty} \beta_n \frac{z^n}{n!} the coefficients αn\alpha_n and βn\beta_n are integers, then, in their product written in the form ∑n=0∞γnznn!\sum_{n=0}^{\infty} \gamma_n \frac{z^n}{n!}, the γn\gamma_n are integers.)

Solution

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Transcribed from the printed volume. Not translated.