Sammy's question at Yahoo Answers (Laurent expansion)

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SUMMARY

The discussion focuses on finding the Laurent series for the function \(\frac{\cos z}{z}\) centered at \(z=0\). The Maclaurin series for \(\cos z\) is given as \(\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n}}{(2n)!}\). Consequently, the Laurent series expansion for \(\frac{\cos z}{z}\) is expressed as \(\frac{1}{z} + \sum_{n=1}^{\infty}\frac{(-1)^n z^{2n-1}}{(2n)!}\), valid for \(0 < |z| < +\infty\). This provides a clear representation of the function's behavior near the singularity at \(z=0\.

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Fernando Revilla
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Hello Sammy,

The Maclaurin expansion of $\cos z$ is: $$\cos z=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}\qquad (\forall z\in\mathbb{C})$$ so, the Laurent series expansion for $\cos z/z$ centered at $z=0$ is $$\frac{\cos z}{z}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n-1}}{(2n)!}=\frac{1}{z}+\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n-1}}{(2n)!}\quad (0<|z|<+\infty)$$ If you have further questions, you can post them in the http://www.mathhelpboards.com/f50/ section.
 

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