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Was reading some notes and it states that $f(z) = \frac{1 - e^{iz}}{z^2}$ can be written as $f(x) = \frac{-iz}{z^2} + E(z)$ where $E(z)$ is bounded as $z \rightarrow 0.$ I don't exactly see why. Help is appreciated.

green frog
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1 Answers1

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Taylors series of $e^{iz}$ about $z=0$ is $$1+iz+\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\cdots$$ so $$\frac{1-e^{iz}}{z^2}=\frac{1}{z^2}\left(1-1-iz-\frac{(iz)^2}{2!}-\frac{(iz)^3}{3!}-\cdots\right)=\frac{-iz}{z^2}+E(z)$$ with $E(z)=-\frac{1}{z^2}\left(\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\cdots\right)$.

Prajwal Kansakar
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  • Thank you for your help. Just a question about the meaning of bounded as $z \rightarrow 0$ just to make sure: Does that simply mean that for a sequence of $z_n$ that approaches 0, there exists an $N$ such that for $n > N, E(z_n) < M$ for some $M?$ – green frog May 06 '17 at 06:35
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    yes $|E(z_n)|<M$ for some $M$. or $\lim_{z\rightarrow 0} |E(z)|$ is some finite number $M$. – Prajwal Kansakar May 06 '17 at 06:45