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How can calculate Laurent series for

$$f(z)=1/ \sin(z) $$ ??

I searched for it and found only the final result, is there a simple way to explain it ?

Gottfried Helms
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Foad Rezek
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4 Answers4

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Using the series for $\sin(z)$ and the formula for products of power series, we can get $$ \begin{align} \frac1{\sin(z)} &=\frac1z\frac{z}{\sin(z)}\\ &=\frac1z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)^{-1}\\ &=\frac1z\left(1+\frac{z^2}{6}+\frac{7z^4}{360}+\frac{31z^6}{15120}+\cdots\right)\\ &=\frac1z+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots \end{align} $$

Using the formula for products of power series

As given in the Wikipedia article linked above, $$ \left(\sum_{k=0}^\infty a_kz^k\right)\left(\sum_{k=0}^\infty b_kz^k\right) =\sum_{k=0}^\infty c_kz^k\tag{1} $$ where $$ c_k=\sum_{j=0}^ka_jb_{k-j}\tag{2} $$ Set $$ c_k=\left\{\begin{array}{} 1&\text{for }k=0\\ 0&\text{otherwise} \end{array}\right.\tag{3} $$ and $$ a_k=\left\{\begin{array}{} \frac{(-1)^j}{(2j+1)!}&\text{for }k=2j\\ 0&\text{for }k=2j+1 \end{array}\right.\tag{4} $$ Using $(2)$, $(3)$, and $(4)$, we can iteratively compute $b_k$.

For example, to compute the coefficient of $z^8$: $$ \begin{align} c_8=0 &=b_8-\frac16b_6+\frac1{120}b_4-\frac1{5040}b_2+\frac1{362880}b_0\\ &=b_8-\frac16\frac{31}{15120}+\frac1{120}\frac7{360}-\frac1{5040}\frac16+\frac1{362880}1\\ &=b_8-\frac{127}{604800} \end{align} $$ Thus, $b_8=\dfrac{127}{604800}$.

robjohn
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  • Can you please explain the 2nd equation transition ? how did you get rid of the pow(-1) ? – Foad Rezek Jan 23 '14 at 17:27
  • @FoadRezek: I have added a section on using the formula for products of power series. – robjohn Jan 23 '14 at 18:47
  • For future reference, to get rid of pow(-1), he treated the part O(z^2) as x and used $\frac{1}{1-x} = 1+x+x^2+...$. He then picked out the terms with z^2, z^4 etc. with the method in the link "product of power series". – XYSquared Jul 23 '20 at 07:17
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    @XYSquared: Actually, that is not what I did. This is why I mentioned $(1)$, because I use $(2)$, $(3)$, and $(4)$ to recursively compute the series for $\frac{x}{\sin(x)}$. – robjohn Jul 23 '20 at 19:19
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    For $k=0$, we get $\overbrace1^{c_0}=\overbrace1^{a_0}b_0$, which gives $b_0=1$. For $k>0$, $$0=b_k+\sum_{j=0}^{k-1}\frac{(-1)^{(k-j)/2}}{(k-j+1)!}b_j[k-j\text{ is even}]$$ and, if we notice that $b_k=0$ for all odd $k$, this can be rewritten as $$\begin{align} b_{2k}&=\sum_{j=0}^{k-1}\frac{(-1)^{k-j+1}}{(2k-2j+1)!}b_{2j}\ b_{2k+1}&=0\end{align}$$ – robjohn Jul 23 '20 at 19:43
  • Sorry! I didn't understand your answer fully before posting that. I think I get it now: $a_k$ are $1, 0, \frac{-1}{3!}, 0, \frac{1}{5!}, \dots$ while $b_k$ are the coefficients we want to solve and $c_k$ are just the coefficients to the "fake series" $1 + 0z^1 + 0z^2 + 0z^3 + \dots$. Never thought of this smart way to get the inverse! – XYSquared Jul 24 '20 at 06:38
  • Just curious: is there a closed-form solution to the recursive definition of coefficients $b_{2k}? – XYSquared Jul 24 '20 at 06:41
  • May I also ask which college/graduate courses deal with these series calculations? Are they courses about combinatorics? – XYSquared Jul 24 '20 at 06:43
  • There is no closed form for $b_{2k}$, unless you use Bernoulli numbers. – robjohn Jul 24 '20 at 07:01
  • What courses cover these calculations may vary. Some might be covered in a course on Discrete Math; some might be covered in an advanced course on series or special functions. It is being covered some here, and in other answers to questions tagged [tag:sequences-and-series]. – robjohn Jul 24 '20 at 07:12
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From the generating function of the Bernoulli polynomials: $${\frac {t{{\rm e}^{xt}}}{{{\rm e}^{t}}-1}}=\sum _{n=0}^{\infty }{ \frac {{\it B}_{n}\left(x\right) {t}^{n}}{n!}} \tag{1}$$ and from: $$ \frac{1}{\sin(z)}={\frac {-2i{{\rm e}^{-iz }}}{{{\rm e}^{-2iz}}-1}} \tag{2}$$ we set $$t=-2iz,\quad x=\frac{1}{2} \tag{3}$$ in $(1)$ to get: $${\frac {-2iz{{\rm e}^{-iz}}}{{{\rm e}^{-2iz}}-1}}=\sum _{n=0}^{ \infty }{\frac {{\it B_n} \left( \frac{1}{2} \right) \left( -2\,iz \right) ^{n}}{n!}} \tag{4}$$ and we then use the known value that relates the polynomial evaluated at $\frac{1}{2}$, to the Bernoulli number: $${\it B_n} \left( \frac{1}{2} \right) = \left( {2}^{1-n}-1 \right) { \it B_n} \tag{5}$$ together with the fact that: $${\it B}_{2n-1} = \cases{1/2&$n=1$\cr 0&otherwise\cr} \tag{6}$$ and divide both sides of $(4)$ by $z$ to get: $$\begin{align} \frac{1}{\sin(z)}&=\sum _{n=0}^{ \infty }{2{\it B}_{2n}\frac { (-1)^n\left( 1-2^{2n-1} \right) z^{2n-1}}{(2n)!}}\\ &=\frac{1}{z}+\frac{1}{6}\,z+{\frac {7 }{360}}\,{z}^{3}+{\frac {31}{15120}}\,{z}^{5}+... \tag{7} \end{align}$$

Mark Viola
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Graham Hesketh
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  • Sorry I did not see that Martin had already linked to this type of solution... – Graham Hesketh Jan 23 '14 at 18:23
  • A similar and independent answer is at https://math.stackexchange.com/a/4427590/945479 – qifeng618 Apr 15 '22 at 03:14
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    See also Section 5 in 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 18 '22 at 01:37
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Idea: $1=\sin z {1\over\sin z} =$ (known Taylor series)(unknown Laurent series).

EDIT: Solution here.

Community
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Martín-Blas Pérez Pinilla
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0

Are you looking for this kind of expansion Laurent series 1/sin(z)? For $z\ne k\pi\:(k\in\mathbb Z)$, the expansion takes \begin{align} \frac1{\sin(z)}&=\sum_{k=-\infty}^\infty\frac{z(-1)^k}{z^2-k^2\pi^2}. \end{align} Its proof can be found here Lobachevsky Integral Formula.

MathArt
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