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Find the coefficient of $\dfrac{1}{z}$ in $\dfrac{1}{z^4\cdot\sin z}$

My attempt is as follows:-

$$y=\dfrac{1}{z^4\cdot\sin z}$$ $$y=\dfrac{1}{z^4\cdot\left(z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}\cdots\cdots\infty\right)}$$

$$y=\dfrac{\cdot\left(1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}\cdots\cdots\infty\right)^{-1}}{z^5}$$

$$y=\dfrac{\cdot\left(1-\left(\dfrac{z^2}{3!}-\dfrac{z^4}{5!}\right)-\left(\dfrac{z^6}{7!}-\dfrac{z^8}{9!}\right)\cdots\cdots\infty\right)^{-1}}{z^5}$$

$$y=\dfrac{\cdot\left(1-\dfrac{z^2}{3!}\cdot\left(1-\dfrac{z^2}{5\cdot4}\right)-\dfrac{z^6}{7!}\left(1-\dfrac{z^2}{9\cdot8}\right)\cdots\cdots\infty\right)^{-1}}{z^5}$$

I was stuck here and didn't get any pattern further.Please help me in this.

user3290550
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    This is very close to being a duplicate of https://math.stackexchange.com/questions/648923/calculate-laurent-series-for-1-sinz. Take a look at robjohn's excellent answer on how to invert a power series. – Sten Nov 17 '19 at 12:54
  • You could try a Laurent series expansion of $\csc{X}$ or split the sum below into two terms, with the first containing up to $z^4$ and the second containing the rest of the terms, and using binomial expansion. – David Nov 17 '19 at 12:57
  • @Sten, in robjohn's answer, how did he evaluate the inverse of $\left(1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}\right)$ – user3290550 Nov 17 '19 at 13:01
  • @user3290550 He used the fact that the $\frac{\sin z}{z} \cdot \frac{z}{\sin z} = 1$, and then the expression for how two power series are multiplied together – Sten Nov 17 '19 at 13:04
  • You can do long division of $1$ by $1-z^2/3!+z^4/5!-...$ up to the term that you need. – conditionalMethod Nov 17 '19 at 13:07

3 Answers3

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You need to find the expansion of $$\left(1-\frac{z^2}{6}+\frac{z^4}{120}-\cdots\right)^{-1}$$ up to the $z^4$ term. This expansion only has even powers of $z$ within so can be written as $c_0+c_2z^2+c_4z^4+\cdots$. Therefore $$1= (c_0+c_2z^2+c_4z^4+\cdots)\left(1-\frac{z^2}{6}+\frac{z^4}{120}-\cdots\right).$$ Comparing the coefficients of $z^0$, $z^2$, $z^4$ within this product gives $$1=c_0$$ $$0=c_2-\frac{c_0}6,$$ $$0=c_4-\frac{c_2}6+\frac{c_0}{120}$$ which you can solve successively for $c_0$, $c_2$, $c_4$.

Angina Seng
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The simplest method to find the expansion of $\;\smash{\dfrac 1{1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}+\cdots}}$ is division by increasing powers .

Here, we'll make it up to order $4$ in order to have the coefficient of $\frac1z$: $$\begin{array}{c} & &\phantom{-}1&{}+\dfrac{z^2}6 &{}-\dfrac{7z^4}{360}&{}+o(z^4) \\ \phantom{-}1\phantom{+\dfrac{z^2}6 -\dfrac{z^4}{120}}&\Bigl( &\phantom{-}1&{}-\dfrac{z^2}6 &{}+\dfrac{z^4}{120}&{}+o(z^4)\\ -1+\dfrac{z^2}6 -\dfrac{z^4}{120}\\ \phantom{-1+{}}\dfrac{z^2}6 -\dfrac{z^4}{120}\\ \phantom{-1}-\dfrac{z^2}6+\dfrac{z^4}{36}\\ \phantom{-1-\dfrac{z^2}6}-\dfrac{7z^4}{360} \end{array}$$ So Laurent's expansion starts with $-\dfrac{7}{360z}$.

Bernard
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Use $\frac1{1-x}=1+x+x^2+\>...$ and continue with

$$y=\dfrac{\left(1-\left(\dfrac{z^2}{3!}-\dfrac{z^4}{5!}\right)-\>...\right)^{-1}}{z^5}$$

$$=\frac1{z^5}\left(1 +\left(\dfrac{z^2}{3!}-\dfrac{z^4}{5!}\right) + \left(\dfrac{z^2}{3!}\right)^2 +\>...\right)$$

$$= \frac1{z^5} \left( 1+ \frac 1{6}z^2+ \left( \frac 1{36}-\frac1{120}\right)z^4+ \>...\right)$$

Thus, the coefficient is $\frac 1{36}-\frac1{120}=\frac 7{360}$.

Quanto
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