# Laurent Series of $(z+1)^2\sin(z)$ centred at $0$

How am I to write the Laurent series of $$(z+1)^2\sin(z)$$ centred at $$0$$? I know $$(z+1)^2 \sin(z) = (z^2+2z+1) \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1)!} \\ \Rightarrow (z+1)^2 \sin(z) = \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+3}}{(2k+1)!} + \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+2}}{(2k+1)!} + \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1)!},$$ but how am I to appropriately re-index is summation such that I might have a (closed-form) expression, containing only a single sum, as in, say,
$$(z+1)^2 \sin(z) = \sum_{k = 0}^{\infty} a_k z^{k}?$$

• Some of the terms are z to an even power. So it cannot be put into the form of your final displayed formula. Nov 16, 2022 at 16:09
• Hey :) Now, there are some nonvanishing coefficents of $z^{k}$ for even $k$. So, you are looking for $a_k$s in $\sum_{k=0}^\infty a_kz^k$. For $k$ even, it's easy. then $a_k=\frac{(-1)^{k/2+1}}{(k-1)!}$. Nov 16, 2022 at 16:11
• @coffeemath Ah, yes! I have edited my question accordingly. Nov 16, 2022 at 16:16

We derive a series representation of $$(z+1)^2\sin(z)$$ evaluated at $$z=0$$ by calculating the coefficients $$a_n$$ in \begin{align*} \color{blue}{(z+1)^2\sin(z)}&=(z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\ &\,\,\color{blue}{=\sum_{n=0}^{\infty}a_nz^n}\tag{1}\\ &=\sum_{n=0}^{\infty}a_{2n}z^{2n}+\sum_{n=0}^{\infty}a_{2n+1}z^{2n+1}\tag{2}\\ \end{align*} At first we derive a representation (2) and then we merge the sums to obtain (1). We use the coefficient of operator $$[z^n]$$ to denote the coefficient of $$z^n$$ of a series. We obtain for $$n\geq 1$$: \begin{align*} \color{blue}{a_{2n}}&=[z^{2n}](z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\ &=\left([z^{2n-2}]+2[z^{2n-1}]+[z^{2n}]\right)\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\ &\,\,\color{blue}{=\frac{2(-1)^{n-1}}{(2n-1)!}}\tag{3}\\ \color{blue}{a_{2n+1}}&=[z^{2n+1}](z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\ &=\left([z^{2n-1}]+2[z^{2n}]+[z^{2n+1}]\right)\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\ &\,\,\color{blue}{=\frac{(-1)^{n-1}}{(2n-1)!}+\frac{(-1)^n}{(2n+1)!}}\tag{4} \end{align*}

We derive from (3) and (4) for $$n\geq 1$$: \begin{align*} a_n&=\frac{1+(-1)^n}{2}\,\color{blue}{\frac{2(-1)^{\frac{n}{2}-1}}{(n-1)!}} +\frac{1-(-1)^n}{2}\left(\color{blue}{\frac{(-1)^{\frac{n-1}{2}-1}}{(n-2)!}+\frac{(-1)^{\frac{n-1}{2}}}{n!}}\right)\\ &=\frac{\left(1+(-1)^n\right)(-1)^{\frac{n}{2}-1}}{(n-1)!} +\frac{\left(1-(-1)^n\right)(-1)^{\frac{n-1}{2}}}{2n!}\left(1+n-n^2\right)\tag{5}\\ \end{align*}

Comment:

• In (3) we select the coefficient of $$z^{2n-1}$$. The other coefficients with even power of $$z$$ are zero.

• In (4) we select the coefficients of $$z^{2n-1}$$ and $$z^{2n+1}$$. The coefficient with even power of $$z$$ is zero.

• In (5), we combine even and odd indexed coefficients $$a_n$$ by multiplying them by $$\frac{1+(-1)^n}{2}$$ and $$\frac{1-(-1)^n}{2}$$, respectively, and adding them.

We finally derive from (5) \begin{align*} &\color{blue}{(z+1)^2\sin(z)}\\ &\qquad=1+\sum_{n=1}^{\infty}\left(\left(1+(-1)^n\right)(-1)^{\frac{n}{2}-1}n\right.\\ &\qquad\qquad\qquad\qquad\left.+\left(1-(-1)^n\right)(-1)^{\frac{n-1}{2}}\frac{1+n-n^2}{2}\right)\frac{z^n}{n!}\\ &\qquad\color{blue}{=1+\sum_{n=1}^{\infty}\left(\left((-1)^{n-1}-1\right)n -i\left(1-(-1)^n\right)\frac{1+n-n^2}{2}\right)\frac{\left(iz\right)^n}{n!}}\\ \end{align*}