Laurent series of the function at $z=0$

Question

Suppose the following function $f(z)$

$$ f(z)=\frac{\sin^{2}z}{z^{2}},\qquad z\in[-\pi,0)\land(0,\pi]. $$

The function has a removable singularity at $z=0$, since

$$ \lim f(z)_{z\rightarrow0}\equiv\lim f(z)_{z\rightarrow0}\frac{\sin2z}{2z}\equiv\lim f(z)_{z\rightarrow0}\frac{2\cos2z}{2}=1, $$

by L'Hospital rule, and $f(0)=1$. We want to expand $f(z)$ at $z=0$. The function has a simple pole of order $m=2$ at $0$. The Laurent series has the form of

$$ h(z)=\frac{b_{2}}{z^{2}}+\frac{b_{1}}{z}+\sum_{k=0}^{\infty}a_{k}z^{k}, $$

where \begin{align*} b_{2} & =\frac{1}{2\pi i}\oint_{C}\frac{f(\theta)}{\theta^{1}}\mathrm{d\theta},\\ b_{1} & =\frac{1}{2\pi i}\oint_{C}\frac{f(\theta)}{\theta^{2}}\mathrm{d\theta},\\ a_{1} & =\frac{1}{2\pi i}\oint_{C}\frac{f(\theta)}{\theta^{3}}\mathrm{d\theta},\\ a_{2} & =\frac{1}{2\pi i}\oint_{C}\frac{f(\theta)}{\theta^{4}}\mathrm{d\theta}, \end{align*} I can determine $b_{1}$ easily as \begin{align*} b_{1} & =\frac{1}{\left(m-1\right)!}\lim_{z\rightarrow z_{0}}\left\{ \frac{d^{m-1}}{dz^{m-1}}[(z-z_{0})^{m}f(z)]\right\} ,\\ & =\lim_{z\rightarrow0}\frac{d}{dz}\left[z^{2}\frac{\sin^{2}z}{z^{2}}\right]=\frac{d}{dz}\sin^{2}z=\left[\sin(2z)\right]_{z=}0=0. \end{align*} How to determine $b_{2}$, $a_{1}$, $a_{2}$? Is the following identity held \begin{align*} a_{1} & =\lim_{z\rightarrow0}(f^{\prime}(z)),\\ a_{2} & =\frac{\lim_{z\rightarrow0}(f^{(2)}(z))}{2!}, \end{align*} where $a_{1},a_{2}$ represent the limits (instead of function values) in the Taylor series? May I ask you for the explanation using the above mentioned formulas?

The solution based on the multiplication of $\sin^{2}z/z^{2}=(2z^{2}/2!-8z^{4}/4!+32z^{6}/6!+O(7))/z^{2}$ is known form me.

The Mathematica solution is

Normal[Series[Sin[z]^2/z^2, {z, 0, 4}]]

$$ h(z)=1-\frac{z^{2}}{3}+\frac{2z^{4}}{45}+O(4). $$

Thanks for your help.

Answer 1

I spent some time solving the integrals. For anyone interested in the solution:

Suppose the disc $C=\left\Vert z\right\Vert =1$ and the substitution

$$ z=e^{i\theta},\qquad\mathrm{dz}=ie^{i\theta}\mathrm{d\theta.} $$

Then \begin{align*} b_{2} & =\frac{1}{2\pi i}\int_{-\pi}^{\pi}\frac{\sin^{2}e^{i\theta}}{e^{i\theta}}ie^{i\theta}\mathrm{d\theta}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin^{2}e^{i\theta}d\theta,\\ b_{1} & =\frac{1}{2\pi i}\int_{-\pi}^{\pi}\frac{\sin^{2}e^{i\theta}}{e^{2i\theta}}ie^{i\theta}\mathrm{d\theta}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\sin^{2}e^{i\theta}}{e^{i\theta}}d\theta,\\ a_{1} & =\frac{1}{2\pi i}\int_{-\pi}^{\pi}\frac{\sin^{2}e^{i\theta}}{e^{2i\theta}}ie^{i\theta}\mathrm{d\theta}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\sin^{2}e^{i\theta}}{e^{2i\theta}}d\theta,\\ a_{2} & =\frac{1}{2\pi i}\int_{-\pi}^{\pi}\frac{\sin^{2}e^{i\theta}}{e^{2i\theta}}ie^{i\theta}\mathrm{d\theta}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\sin^{2}e^{i\theta}}{e^{3i\theta}}d\theta. \end{align*} Taking into account that \begin{align*} \frac{1}{2\pi}\int\sin^{2}e^{i\theta}d\theta & =\frac{1}{4\pi}\left(\theta-\int\cos(2e^{i\theta})\mathrm{d\theta}\right), \end{align*} the principal part is

\begin{align*} b_{2} & =\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin^{2}e^{i\theta}d\theta\\ & =\frac{1}{4\pi}\left[\theta\right]_{-\pi}^{\pi}-2\pi,\\ & =0.\\ b_{1} & =-\frac{1}{4\pi}ie^{-i\theta}\left[-1+\cos2e^{i\theta}-2e^{i\theta}\int_{-\pi}^{\pi}\sin(2e^{i\theta})\mathrm{d\theta}\right]_{-\pi}^{\pi},\\ & =\frac{1}{4\pi}\left[ie^{-i\theta}-ie^{-i\theta}\cos2e^{i\theta}\right]_{-\pi}^{\pi}+\frac{i}{2\pi}\int_{-\pi}^{\pi}\sin(2e^{i\theta})\mathrm{d\theta,}\\ & =\frac{1}{4\pi}\left(ie^{-i\pi}-ie^{-i\pi}\cos2e^{-i\pi}-ie^{i\pi}+ie^{i\pi}\cos2e^{i\pi}+0\right),\\ & =\frac{1}{4\pi}\left(-i+i\cos2+i-i\cos2+0\right),\\ & =0. \end{align*} Analogously, the remaining coefficients are \begin{align*} a_{0} & =-\frac{1}{8\pi}ie^{-2i\theta}\left[-1+\cos2e^{i\theta}-4e^{2i\theta}\int_{-\pi}^{\pi}\cos(2e^{i\theta})\mathrm{d\theta}-2e^{i\theta}\sin(2e^{i\theta})\right]_{-\pi}^{\pi}=1,\\ a_{1} & =\frac{1}{12\pi}ie^{-3i\theta}\left[1+\left(-1+2e^{i\theta}\right)\cos2e^{i\theta}+e^{i\theta}\sin2e^{i\theta}-4e^{3i\theta}\int_{-\pi}^{\pi}\sin(2e^{i\theta})\mathrm{d\theta}\right]_{-\pi}^{\pi}=0\\ a_{2} & =\frac{1}{12\pi}\left[-\frac{1}{24}ie^{-2i\theta}\left(-2+3e^{-2i\theta}\right)\cos2e^{i\theta}+\frac{1}{24}i\left(3e^{-4i\theta}-8\int_{-\pi}^{\pi}\cos(2e^{i\theta})\mathrm{d\theta}\right)+\right.\\ & =\left. +\frac{1}{12}ie^{-i\theta}\left(-2+e^{-2i\theta}\right)\sin2e^{i\theta}\right]_{-\pi}^{\pi},\\ & =-\frac{1}{3}. \end{align*}

Finally, the Laurent series lacks its principal part $$ h(z)=1-\frac{z^{3}}{3}+O(4). $$

Answer 2

This one is quite obvious , you don't even have to calculate b2 and b1 as they will surely be zero , since function has removable singularity it will not contain any bi's .

Now to find any ak it should be clear from the Laurent's series expansion that one must differentiate whole series k times to remove the (kth and lesser) degree monomial which is present with it and then divide by the factor k! Which has appeared after differentiating the series . Then finally tend the limit to zero so that all the terms after ak vanishes. Then finally we can calculate any coefficient ak

Answer 3

Using:

$$\cos(2z) = 1-2\sin^2(z)$$

we can write:

$$\frac{\sin^2(z)}{z^2}=\frac{1-\cos(2z)}{2 z^2} = 2\sum_{k=0}^{\infty}(-1)^k 2^{2k}\frac{z^{2k}}{(2k+2)!}$$