Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 196258 by York12 last updated on 21/Aug/23

$$If{(x_m+{iy}_m)}^{2n+1}=1,suchthat$$$$m\in \{1,2,3,....,2n\}\wedge x_m,y_m\in \mathbb{R}$$$$p=\underset{k=1}{\sum ^{2020}}\left[\frac{1-x_k+{iy}_k}{1+x_k+{iy}_k}\right],Find\left(\frac{p}{43}\right)$$

Commented by York12 last updated on 21/Aug/23

$$$$$$({\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}})={\mathrm{e}}^{\frac{2\mathrm{im}\pi }{2\mathrm{n}+1}},\mathrm{m}\in \{0,......2\mathrm{n}\}$$$${\mathrm{Z}}_{\mathrm{m}}={\mathrm{e}}^{2\mathrm{i}\frac{\mathrm{m}\pi }{2\mathrm{n}+1}},{\mathrm{ia}}_{\mathrm{k},\mathrm{n}}=\frac{2\mathrm{ik}\pi }{2\mathrm{n}+1}$$$$\mathrm{for}\mathrm{all}\mathrm{the}\mathrm{reste}\mathrm{n}=1010$$$$\underset{\mathrm{k}=1}{\sum ^{2020}}\frac{1-{\mathrm{x}}_{\mathrm{k}}+{\mathrm{iy}}_{\mathrm{k}}}{1+{\mathrm{x}}_{\mathrm{k}}+{\mathrm{iy}}_{\mathrm{k}}}=\underset{\mathrm{k}=0}{\sum ^{2020}}\frac{1-({\mathrm{x}}_{\mathrm{k}}-{\mathrm{iy}}_{\mathrm{k}})}{1+({\mathrm{x}}_{\mathrm{k}}+{\mathrm{iy}}_{\mathrm{k}})}=\mathrm{\Sigma }\frac{1-{\mathrm{e}}^{-{\mathrm{ia}}_{\mathrm{k}}}}{1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}$$$$=\mathrm{\Sigma }\frac{{\mathbf{e}}^{{\mathbf{ia}}_{\mathbf{k}}}-1}{{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}(1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}})}=\sum _{\mathrm{k}}\frac{2}{1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}-\frac{1}{{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}$$$$\mathrm{ler}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{2\mathrm{n}+1}-1$$$$\frac{{\mathrm{p}}^{\prime }\left(\mathrm{x}\right)}{\mathrm{p}\left(\mathrm{x}\right)}=\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{1}{\mathrm{X}-{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}\Rightarrow \frac{{\mathrm{p}}^{\prime }(-1)}{\mathrm{p}(-1)}=-\mathrm{\Sigma }\frac{1}{1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}$$$$\Rightarrow \underset{\mathrm{k}=0}{\sum ^{2020}}\frac{1}{1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}=-\frac{\left(2021\right)}{-2}=\frac{2021}{2}$$$$\underset{\mathrm{k}=0}{\sum ^{2020}}{\mathrm{e}}^{-{\mathrm{ia}}_{\mathrm{k}}}=\frac{1-{\left({\mathrm{e}}^{-\mathrm{i}\frac{2\pi }{2021}}\right)}^{2021}}{1-{\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{2021}}}=0$$$$\mathrm{P}=2.\frac{2021}{2}=2021=43.47$$$$\frac{\mathrm{p}}{43}=47$$$$$$

Commented by York12 last updated on 21/Aug/23

$$Cansomeoneexplain$$$$How\frac{p^{{}^{\prime }}\left(x\right)}{p\left(x\right)}=\underset{k=0}{\sum ^{2n}}\left(\frac{1}{x-e^{{ia}_k}}\right)$$$$please$$

Commented by sniper237 last updated on 22/Aug/23

$$Decompositioninsimpleelementsonk\left(X\right)$$

Commented by York12 last updated on 22/Aug/23

$$canyouwritedownpleaseIamreallylost$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com