Question Number 147417 by tabata last updated on 20/Jul/21
$${how}\:{can}\:{find}\:{taylor}\:{series}\:{of}\:{f}\left({z}\right)={cot}\left({z}\right)\:{when}\:{z}=\mathrm{5}\pi \\ $$
Answered by mathmax by abdo last updated on 21/Jul/21
$$\mathrm{f}\left(\mathrm{z}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{5}\pi\right)}{\mathrm{n}!}\left(\mathrm{z}−\mathrm{5}\pi\right)^{\mathrm{n}} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\mathrm{tanz}}\:\Rightarrow\mathrm{f}\left(\mathrm{5}\pi\right)\mathrm{is}\:\mathrm{not}\:\mathrm{defined}..! \\ $$