Jul 8, 2018 · See Richard P. Brent and David Harvey, Fast computation of Bernoulli, Tangent and Secant numbers; see also A multimodular algorithm for computing Bernoulli numbers, Math. Comp. 79 (2010), no. 272, 2361–2370. MR 2684369 (2011h:11019)

Apr 13, 2016 · These up down numbers are produced by the generating function $$\sec{(x)}+\tan{(x)}$$ and the even indexed coefficients are the Euler Numbers and the odd indexed coefficients are the Tangent Numbers. I know that the Bernoulli numbers are related to the Cotangent function but I'm not sure how to start or approach getting the Bernoulli numbers …

The statement defining the Bernoulli numbers says that the power series expansion of $\frac{z}{e^z-1}$ is ...

Mar 29, 2017 · I was trying to understand Bernoulli numbers. When I googled, I found this link.. It starts by saying that, The Bernoulli numbers are defined via the coefficients of the power series expansion of $\frac{t}{e^{t}-1}$,

Aug 22, 2022 · HINT: Use the definition of the Bernoulli numbers and the following fact: the function $\tan x$ has a Taylor expansion at $0$ with all coefficients $\ge 0$. This can be shown by induction using the equality

Dec 16, 2017 · Bai-Ni Guo and Feng Qi, An explicit formula for Bernoulli numbers in terms of Stirling numbers of the second kind, Journal of Analysis & Number Theory 3 (2015), no. 1, 27--30. There existed at least seven alternative proofs of the formulas \eqref{Higgins-Gould-B} and \eqref{Bernoulli-Stirling-eq} in the paper [1, 2] above and in the following ...

[7-28] I have decided to define the Bernoulli numbers as the inverse of the Harmonic numbers, since this puts in evidence their importance when proving Faulhaber's formula. I will keep looking for a proof of the closed form formula. DEF Denote by $\bf H$ the Harmonic numbers $\left(1,\frac 1 2,\frac 13,\frac 14 \ldots\right)$. We define the ...

Oct 19, 2020 · Find the first five Bernoulli numbers. I understand why the power series expansion of $\frac{z}{2} \cot (\frac{z}{2})$ contains only terms with even powers (it has to do with the fact that the function is even, and so all the derivatives of odd order will vanish at $0$ , so they odd order terms will vanish too).

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I am given in a previous exercise that the Bernoulli numbers are defined by $$ \frac{z}{e^z - 1} = \sum_{n = 0}^{\infty} \frac{B_n}{n!} z^n, $$ where $ B_n $ is the $ n^{th} $ Bernoulli number. I've been looking for a clever way to write $ \;\large\frac{1}{e^{iz} - e^{-iz}} \;$ in the form of a linear combination involving terms of the form ...