Question-7009

Question Number 7009 by Tawakalitu. last updated on 05/Aug/16

Commented by FilupSmith last updated on 06/Aug/16

do you mean (−1)^∞ if so: −1=e^(iπ) ∴let z=(−1)^x =e^(iπx) z=cos(πx)+isin(πx) x→∞ Re(z)=cos(πx) −1≤Re(x)≤1 −1≤cos(πx)≤1 Im(z)=sin(xπ) −1≤Im(z)≤1 −i≤isin(πx)≤i Therefore: −(i+1)≤z≤i+1 −(i+1)≤(−1)^∞ ≤i+1 If (−1)^∞ is non−complex then: lim_(n→∞) (−1)^(2n) =1 If complex: lim_(n→∞) (−1)^n =z

$$\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:\left(−\mathrm{1}\right)^{\infty} \\ $$$$\mathrm{if}\:\mathrm{so}: \\ $$$$−\mathrm{1}={e}^{{i}\pi} \\ $$$$\therefore\mathrm{let}\:{z}=\left(−\mathrm{1}\right)^{{x}} ={e}^{{i}\pi{x}} \\ $$$${z}=\mathrm{cos}\left(\pi{x}\right)+{i}\mathrm{sin}\left(\pi{x}\right) \\ $$$${x}\rightarrow\infty \\ $$$$\mathrm{Re}\left({z}\right)=\mathrm{cos}\left(\pi{x}\right) \\ $$$$−\mathrm{1}\leqslant\mathrm{Re}\left({x}\right)\leqslant\mathrm{1} \\ $$$$−\mathrm{1}\leqslant\mathrm{cos}\left(\pi{x}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{Im}\left({z}\right)=\mathrm{sin}\left({x}\pi\right) \\ $$$$−\mathrm{1}\leqslant\mathrm{Im}\left({z}\right)\leqslant\mathrm{1} \\ $$$$−{i}\leqslant{i}\mathrm{sin}\left(\pi{x}\right)\leqslant{i} \\ $$$$ \\ $$$$\mathrm{Therefore}: \\ $$$$−\left({i}+\mathrm{1}\right)\leqslant{z}\leqslant{i}+\mathrm{1} \\ $$$$−\left({i}+\mathrm{1}\right)\leqslant\left(−\mathrm{1}\right)^{\infty} \leqslant{i}+\mathrm{1} \\ $$$$ \\ $$$$\mathrm{If}\:\left(−\mathrm{1}\right)^{\infty} \:\mathrm{is}\:\mathrm{non}−\mathrm{complex}\:\mathrm{then}: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(−\mathrm{1}\right)^{\mathrm{2}{n}} =\mathrm{1} \\ $$$$\mathrm{If}\:\mathrm{complex}: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(−\mathrm{1}\right)^{{n}} ={z} \\ $$

Commented by Tawakalitu. last updated on 06/Aug/16

$${Tbanks}\:{for}\:{your}\:{help} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply