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As the title says I would like to find the first three nonzero terms in the Maclaurin series $$y=\frac{x}{\sin(x)}$$

I have the first few terms for the expansion for $\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}....$

For the next step can I just say the next few terms are: $1-\frac{6}{x^2}+\frac{120}{x^4}....$ or should I do algebraic long division? If so what by and could anyone start me off?

Thank you!

2 Answers2

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The Bernoulli polynomials $B_k(x)$ for $k\ge0$ are defined by the exponential generating function \begin{equation*} \frac{z\textrm{e}^{xz}}{\textrm{e}^z-1}=\sum_{k=0}^\infty B_k(x)\frac{z^k}{k!}, \quad |z|<2\pi \end{equation*} for $x\in\mathbb{R}$. By the Euler formula $$ \textrm{e}^{\textrm{i}x}=\cos x+\textrm{i}\sin x, $$ we find the relation $$ \sin x=\frac{\textrm{e}^{\textrm{i}x}-\textrm{e}^{-\textrm{i}x}}{2\textrm{i}}. $$ Then \begin{align*} \frac{x}{\sin x}&=\frac{2\textrm{i}x} {\textrm{e}^{\textrm{i}x}-\textrm{e}^{-\textrm{i}x}}\\ &=\frac{(2\textrm{i}x)\textrm{e}^{(2\textrm{i}x)/2}} {\textrm{e}^{(2\textrm{i}x)}-1}\\ &=\sum_{k=0}^\infty B_k\biggl(\frac12\biggr)\frac{(2\textrm{i}x)^k}{k!}\\ &=\sum_{k=0}^\infty (2\textrm{i})^kB_k\biggl(\frac12\biggr)\frac{x^k}{k!}\\ &=\sum_{k=0}^\infty (2\textrm{i})^{2k}B_{2k}\biggl(\frac12\biggr)\frac{x^{2k}}{(2k)!}\\ &=\sum_{k=0}^\infty (-1)^k2^{2k}B_{2k}\biggl(\frac12\biggr)\frac{x^{2k}}{(2k)!}\\ &=\sum_{k=0}^\infty (-1)^{k+1}2^{2k}\biggl(1-\frac1{2^{2k-1}}\biggr)B_{2k}\frac{x^{2k}}{(2k)!}\\ &=2\sum_{k=0}^\infty (-1)^{k+1}\bigl(2^{2k-1}-1\bigr)B_{2k}\frac{x^{2k}}{(2k)!} \end{align*} for $|x|<\pi$, where $B_k(0)=B_k$ and \begin{align*} B_0&=1, & B_2&=\frac{1}{6},& B_4&=-\frac{1}{30},& B_6&=\frac{1}{42},& B_8&=-\frac{1}{30},\\ B_{10}&=\frac{5}{66},& B_{12}&=-\frac{691}{2730},& B_{14}&=\frac{7}{6},& B_{16}&=-\frac{3617}{510}, & B_{18}&=\frac{43867}{798}. \end{align*}

qifeng618
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  • A similar and independent proof is at https://math.stackexchange.com/a/649039. – qifeng618 Apr 15 '22 at 14:13
  • This answer has been included in Section 5 of the paper [Xue-Yan Chen, Lan Wu, Dongkyu Lim, and Feng Qi, Two identities and closed-form formulas for the Bernoulli numbers in terms of central factorial numbers of the second kind, Demonstratio Mathematica 55 (2022), no. 1, 822--830; available online at https://doi.org/10.1515/dema-2022-0166.] – qifeng618 Nov 17 '22 at 17:59
  • The series expansion $$\csc x=\frac1{\sin x}=\frac1x+\sum_{k=1}^\infty\frac{2(2^{2k-1}-1)|B_{2k}|}{(2k)!}x^{2k-1}, \quad x^2<\pi^2$$ is useful. See page 42 in the monograph [I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8 ]. – qifeng618 Apr 03 '23 at 22:14
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You certainly cannot so: it is not a Maclaurin series.

Hint:

First simplify the fraction: $$\frac x{\sin x}=\frac x{x-\cfrac{x^3}6+\cfrac{x^5}{120}+o(x^5)}=\frac 1{1-\cfrac{x^2}6+\cfrac{x^4}{120}+o(x^4)},$$ then perform the division by increasing powers of the numerator $1$ by $1-\frac{x^2}6+\frac{x^4}{120}$ up to degree $4$.

Bernard
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