Question Number 3650 by prakash jain last updated on 18/Dec/15 | ||
$$\mathrm{If}\:{a}_{{n}} \:\mathrm{is}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{all}\:\mathrm{primitive}\:{n}^{{th}} \:\mathrm{root}\:\mathrm{of} \\ $$ $$\mathrm{unity}. \\ $$ $$\mathrm{Does}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \:\mathrm{converge}? \\ $$ $$\mathrm{An}\:{n}^{{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity},\:\mathrm{say}\:{z},\:\mathrm{is}\:\mathrm{primitive}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not} \\ $$ $${k}^{{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{where}\:{k}<{n}. \\ $$ $${or}\:{z}^{{n}} =\mathrm{1}\:\mathrm{and}\:{z}^{{k}} \neq\mathrm{1}\:\left(\mathrm{k}=\mathrm{1},\mathrm{2},\mathrm{3},..,{n}−\mathrm{1}\right) \\ $$ $${a}_{\mathrm{1}} =\mathrm{1} \\ $$ $${a}_{\mathrm{2}} =−\mathrm{1}\:\:\left(\mathrm{2}^{{nd}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{are}\:\mathrm{1}\:\mathrm{and}\:−\mathrm{1},\:\mathrm{1}\:\mathrm{is}\:\mathrm{not}\right. \\ $$ $$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{primitive}\right) \\ $$ | ||
Answered by Yozzii last updated on 18/Dec/15 | ||
$${Let}\:{p}\left({n}\right)\:{be}\:{the}\:{primitive}\:{roots}\:{of}\:{the} \\ $$ $${equation}\:{z}^{{n}} =\mathrm{1}\:{and}\:{a}_{{n}} =\Sigma{p}\left({n}\right). \\ $$ $${n}=\mathrm{1}\Rightarrow{z}=\mathrm{1}\Rightarrow{p}\left(\mathrm{1}\right)=\mathrm{1}\Rightarrow{a}_{\mathrm{1}} =\mathrm{1}. \\ $$ $${n}=\mathrm{2}\Rightarrow{z}^{\mathrm{2}} =\mathrm{1}\Rightarrow{z}=\pm\mathrm{1}\Rightarrow{p}\left(\mathrm{2}\right)=−\mathrm{1}\Rightarrow{a}_{\mathrm{2}} =−\mathrm{1}. \\ $$ $${n}=\mathrm{3}\Rightarrow{z}^{\mathrm{3}} =\mathrm{1}={e}^{\mathrm{2}\pi{ir}} \:\left(−\mathrm{1}\leqslant{r}\leqslant\mathrm{1}\right)\: \\ $$ $$\Rightarrow{z}={e}^{\frac{\mathrm{2}\pi{ir}}{\mathrm{3}}} \Rightarrow{p}\left(\mathrm{3}\right)={e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} \Rightarrow{a}_{\mathrm{3}} =\mathrm{2}{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}=−\mathrm{1} \\ $$ $${n}=\mathrm{4}\Rightarrow{z}^{\mathrm{4}} =\mathrm{1}\Rightarrow{z}=\pm\mathrm{1},\pm{i}\Rightarrow{p}\left(\mathrm{4}\right)=\pm{i}\Rightarrow{a}_{\mathrm{4}} =\mathrm{0} \\ $$ $${n}=\mathrm{5}\Rightarrow{z}^{\mathrm{5}} ={e}^{\mathrm{2}\pi{ir}} \:\left(−\mathrm{2}\leqslant{r}\leqslant\mathrm{2}\right) \\ $$ $$\Rightarrow{z}={e}^{\frac{\mathrm{2}\pi{ir}}{\mathrm{5}}} =\mathrm{1},{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{4}\pi{i}}{\mathrm{5}}} \Rightarrow{p}\left(\mathrm{5}\right)={e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{4}\pi{i}}{\mathrm{5}}} \\ $$ $$\Rightarrow{a}_{\mathrm{5}} =\mathrm{2}\left({cos}\frac{\mathrm{2}\pi}{\mathrm{5}}+{cos}\frac{\mathrm{4}\pi}{\mathrm{5}}\right) \\ $$ $${a}_{\mathrm{5}} =\mathrm{2}\left\{\mathrm{2}{cos}\frac{\frac{\mathrm{4}\pi+\mathrm{2}\pi}{\mathrm{5}}}{\mathrm{2}}{cos}\frac{\frac{\mathrm{4}\pi−\mathrm{2}\pi}{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$ $${a}_{\mathrm{5}} =\mathrm{4}{cos}\frac{\mathrm{3}\pi}{\mathrm{5}}{cos}\frac{\pi}{\mathrm{5}}=−\mathrm{1} \\ $$ $${n}=\mathrm{6}\Rightarrow{z}^{\mathrm{6}} ={e}^{\mathrm{2}\pi{ir}} \:\left(−\mathrm{2}\leqslant{r}\leqslant\mathrm{3}\right) \\ $$ $${z}={e}^{\frac{\pi{ir}}{\mathrm{3}}} \Rightarrow{z}=\pm\mathrm{1},{e}^{\pm\frac{\pi{i}}{\mathrm{3}}} ,{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} \Rightarrow{p}\left(\mathrm{6}\right)={e}^{\pm\frac{\pi{i}}{\mathrm{3}}} \\ $$ $$\Rightarrow{a}_{\mathrm{6}} =\mathrm{2}{cos}\frac{\pi}{\mathrm{3}}=\mathrm{1} \\ $$ $${n}=\mathrm{7}\Rightarrow{z}^{\mathrm{7}} ={e}^{\mathrm{2}\pi{ir}} \:\left(−\mathrm{3}\leqslant{r}\leqslant\mathrm{3}\right) \\ $$ $$\Rightarrow{z}={e}^{\frac{\mathrm{2}\pi{ir}}{\mathrm{7}}} \Rightarrow{z}=\mathrm{1},{e}^{\frac{\pm\mathrm{2}\pi{i}}{\mathrm{7}}} ,{e}^{\frac{\pm\mathrm{4}\pi{i}}{\mathrm{7}}} ,{e}^{\frac{\pm\mathrm{6}\pi{i}}{\mathrm{7}}} \\ $$ $$\Rightarrow{a}_{\mathrm{7}} =\mathrm{2}\left({cos}\frac{\mathrm{2}\pi}{\mathrm{7}}+{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}+{cos}\frac{\mathrm{6}\pi}{\mathrm{7}}\right)=−\mathrm{1} \\ $$ $${n}=\mathrm{8}\Rightarrow{z}^{\mathrm{8}} ={e}^{\mathrm{2}\pi{ir}} \Rightarrow{z}={e}^{\frac{\pi{ir}}{\mathrm{4}}} \left(−\mathrm{3}\leqslant{r}\leqslant\mathrm{4}\right) \\ $$ $$\Rightarrow{z}=\pm\mathrm{1},{e}^{\pm\frac{\mathrm{3}\pi{i}}{\mathrm{4}}} ,\pm{i},{e}^{\pm\frac{\pi{i}}{\mathrm{4}}} \\ $$ $$\Rightarrow{a}_{\mathrm{8}} =\mathrm{2}\left({cos}\frac{\mathrm{3}\pi}{\mathrm{4}}+{cos}\frac{\pi}{\mathrm{4}}\right)=\mathrm{0}. \\ $$ $${n}=\mathrm{9}\Rightarrow{z}={e}^{\frac{\mathrm{2}\pi{ir}}{\mathrm{9}}} \:\left(−\mathrm{4}\leqslant{r}\leqslant\mathrm{4}\right) \\ $$ $$\Rightarrow{z}=\mathrm{1},{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{9}}} ,{e}^{\pm\frac{\mathrm{4}\pi{i}}{\mathrm{9}}} ,{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} ,{e}^{\pm\frac{\mathrm{8}\pi{i}}{\mathrm{9}}} \\ $$ $$\Rightarrow{a}_{\mathrm{9}} =\mathrm{2}\left({cos}\frac{\mathrm{2}\pi}{\mathrm{9}}+{cos}\frac{\mathrm{4}\pi}{\mathrm{9}}+{cos}\frac{\mathrm{8}\pi}{\mathrm{9}}\right)=\mathrm{0} \\ $$ $${n}=\mathrm{10}\Rightarrow{z}={e}^{\frac{\pi{ir}}{\mathrm{5}}} \:\left(−\mathrm{4}\leqslant{r}\leqslant\mathrm{5}\right) \\ $$ $$\Rightarrow{z}=\pm\mathrm{1},{e}^{\pm\frac{\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{3}\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{4}\pi{i}}{\mathrm{5}}} \\ $$ $$\Rightarrow{a}_{\mathrm{10}} =\mathrm{2}\left({cos}\frac{\pi}{\mathrm{5}}+{cos}\frac{\mathrm{3}\pi}{\mathrm{5}}\right)=\mathrm{1} \\ $$ $$ \\ $$ $${I}\:{conject}\:{that}\:{for}\:{n}\:{being}\:{prime}, \\ $$ $${a}_{{n}} \neq\mathrm{0},\:{so}\:{that}\:\underset{\left({n}\in\mathbb{P}\right)\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \neq\mathrm{0}\Rightarrow\Sigma{a}_{{n}} \:{does} \\ $$ $${not}\:{converge}.\: \\ $$ $$ \\ $$ $${Suppose}\:{that}\:{n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\:{and}\:{z}^{{n}} ={e}^{\mathrm{2}\pi{ir}} \:{with} \\ $$ $${r}=\mathrm{0},\mathrm{1},\mathrm{2},...,{n}−\mathrm{1}.\:{Then},\:{z}={e}^{\frac{\mathrm{2}\pi{ir}}{{n}}} . \\ $$ $${n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\Rightarrow\:\frac{{r}}{{n}}\:{never}\:{reduces}\:{nor}\:{vanishes} \\ $$ $${for}\:{all}\:{r}\neq\mathrm{0}.\:{If}\:{r}=\mathrm{0}\:{we}\:{obtain}\:{the}\:{non}−{primitive} \\ $$ $${root}\:{z}=\mathrm{1}.\:{Otherwise},\:{all}\:{arguments}\: \\ $$ $$\frac{\mathrm{2}\pi{r}}{{n}}\:{are}\:{unique}.\:{In}\:{adding}\:{all}\:{of}\:{the} \\ $$ $${primitive}\:{roots}\:{however},\:{the}\:{imaginary} \\ $$ $${parts}\:{vanish}\:{for}\:{there}\:{are}\:{an}\:{even}\: \\ $$ $${number}\:{of}\:{roots}\:\left({n}−\mathrm{1}\:\because\:{n}\in\mathbb{O}^{+} \right)\:{and}\:{half}\:{of}\:{these} \\ $$ $${roots}\:{are}\:{complex}\:{conjugates}\:{of}\:{the}\:{remaining} \\ $$ $${roots}. \\ $$ $$\therefore\:{a}_{{n}} =\mathrm{2}\underset{{r}=\mathrm{1}} {\overset{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {\sum}}{cos}\frac{\mathrm{2}\pi{r}}{{n}}\:\left({n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\right) \\ $$ $$\Rightarrow\underset{\left({n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\right)\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \neq\mathrm{0} \\ $$ $${If}\:{n}=\mathrm{2},\:{a}_{\mathrm{2}} =−\mathrm{1}\neq\mathrm{0}.\:{So}\:\underset{\left({n}\in\mathbb{P}\right)\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \neq\mathrm{0}. \\ $$ $${There}\:{exist}\:{infinitely}\:{many}\:{primes}\: \\ $$ $${with}\:\mathbb{P}\subset\mathbb{N},\:{so}\:{a}_{{n}} \:{always}\:{exists}\:{for} \\ $$ $${n}\in\mathbb{P}\subset\mathbb{N}. \\ $$ $${Calculations}\:{project}\:{that}\:{a}_{{n}} =−\mathrm{1} \\ $$ $${for}\:{all}\:{n}\in\mathbb{P}.\:{It}\:{has}\:{been}\:{proven}\:{that} \\ $$ $${a}_{{n}} =−\mathrm{1}\:\forall{n}\in\mathbb{P}\:{indeed}\:{in}\:{the}\:{comments}\: \\ $$ $${below}. \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ | ||
Commented byprakash jain last updated on 18/Dec/15 | ||
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{detailed}\:\mathrm{answer}.\: \\ $$ | ||
Commented byYozzii last updated on 18/Dec/15 | ||
$${S}\left({n}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{cosr}\theta\:\:\: \\ $$ $${S}\left({n}\right){sin}\mathrm{0}.\mathrm{5}\theta=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{cosr}\theta{sin}\mathrm{0}.\mathrm{5}\theta\:\:\:\left(\theta\neq\mathrm{2}{n}\pi\right) \\ $$ $$ \\ $$ $$\mathrm{2}{cosr}\theta{sin}\mathrm{0}.\mathrm{5}\theta={sin}\left({r}+\mathrm{0}.\mathrm{5}\right)\theta−{sin}\left({r}−\mathrm{0}.\mathrm{5}\right)\theta \\ $$ $$\therefore{S}\left({n}\right){sin}\mathrm{0}.\mathrm{5}\theta=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\mathrm{1}.\mathrm{5}\theta−{sin}\mathrm{0}.\mathrm{5}\theta\right. \\ $$ $$+{sin}\mathrm{2}.\mathrm{5}\theta−{sin}\mathrm{1}.\mathrm{5}\theta+{sin}\mathrm{3}.\mathrm{5}\theta−{sin}\mathrm{2}.\mathrm{5}\theta \\ $$ $$+...+{sin}\left({n}−\mathrm{2}.\mathrm{5}\right)\theta−{sin}\left({n}−\mathrm{3}.\mathrm{5}\right)\theta \\ $$ $$+{sin}\left({n}−\mathrm{1}.\mathrm{5}\right)\theta−{sin}\left({n}−\mathrm{2}.\mathrm{5}\right)\theta \\ $$ $$+{sin}\left({n}−\mathrm{0}.\mathrm{5}\right)\theta−{sin}\left({n}−\mathrm{1}.\mathrm{5}\right)\theta \\ $$ $$\left.+{sin}\left({n}+\mathrm{0}.\mathrm{5}\right)\theta−{sin}\left({n}−\mathrm{0}.\mathrm{5}\right)\theta\right\} \\ $$ $${S}\left({n}\right)=\frac{{sin}\left({n}+\mathrm{0}.\mathrm{5}\right)\theta−{sin}\mathrm{0}.\mathrm{5}\theta}{\mathrm{2}{sin}\mathrm{0}.\mathrm{5}\theta} \\ $$ $${S}\left({n}\right)=\frac{\mathrm{2}{cos}\frac{\left({n}+\mathrm{0}.\mathrm{5}\right)\theta+\mathrm{0}.\mathrm{5}\theta}{\mathrm{2}}{sin}\frac{\left({n}+\mathrm{0}.\mathrm{5}\right)\theta−\mathrm{0}.\mathrm{5}\theta}{\mathrm{2}}}{\mathrm{2}{sin}\mathrm{0}.\mathrm{5}\theta} \\ $$ $${S}\left({n}\right)=\frac{{cos}\frac{{n}+\mathrm{1}}{\mathrm{2}}\theta{sin}\frac{{n}\theta}{\mathrm{2}}}{{sin}\mathrm{0}.\mathrm{5}\theta} \\ $$ $${n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\Rightarrow\:\frac{{n}−\mathrm{1}}{\mathrm{2}}\in\mathbb{Z}^{+} \\ $$ $${so}\:\frac{{n}−\mathrm{1}}{\mathrm{2}}\:{can}\:{be}\:{safely}\:{substituted}\:{for}\:{n}\:{in}\:{S}\left({n}\right). \\ $$ $$ \\ $$ $$\therefore\:{S}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)=\frac{{cos}\frac{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}}{\mathrm{2}}\theta{sin}\frac{{n}−\mathrm{1}}{\mathrm{4}}\theta}{{sin}\mathrm{0}.\mathrm{5}\theta} \\ $$ $${S}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)=\frac{{cos}\frac{{n}+\mathrm{1}}{\mathrm{4}}\theta{sin}\frac{{n}−\mathrm{1}}{\mathrm{4}}\theta}{\mathrm{2}{cos}\frac{\theta}{\mathrm{4}}{sin}\frac{\theta}{\mathrm{4}}} \\ $$ $$ \\ $$ $${Now}\:{let}\:\theta=\frac{\mathrm{2}\pi}{{n}}. \\ $$ $${S}\left(\mathrm{0}.\mathrm{5}\left({n}−\mathrm{1}\right)\right)=\frac{{cos}\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\pi{sin}\frac{{n}−\mathrm{1}}{\mathrm{2}{n}}\pi}{\mathrm{2}{cos}\frac{\pi}{\mathrm{2}{n}}{sin}\frac{\pi}{\mathrm{2}{n}}} \\ $$ $${S}\left(\frac{\mathrm{1}}{\mathrm{2}}\left({n}−\mathrm{1}\right)\right)=\frac{{cos}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}{n}}\right){sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}{n}}\right)}{\mathrm{2}{cos}\frac{\pi}{\mathrm{2}{n}}{sin}\frac{\pi}{\mathrm{2}{n}}} \\ $$ $${cos}\left(\mathrm{0}.\mathrm{5}\pi+\alpha\right)=\mathrm{0}−{sin}\mathrm{0}.\mathrm{5}\pi{sin}\alpha \\ $$ $${and}\:{sin}\left(\mathrm{0}.\mathrm{5}\pi−\alpha\right)={cos}\alpha \\ $$ $$\Rightarrow{S}\left(\frac{\mathrm{1}}{\mathrm{2}}\left({n}−\mathrm{1}\right)\right)=−\mathrm{0}.\mathrm{5} \\ $$ $$\Rightarrow{a}_{{n}} =\mathrm{2}×\left(−\mathrm{0}.\mathrm{5}\right)=−\mathrm{1}\:\forall{n}\in\mathbb{P}−\left\{\mathrm{2}\right\}. \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $${o} \\ $$ | ||
Commented byprakash jain last updated on 18/Dec/15 | ||
$${p}>\mathrm{2}\:{prime}−\left({so}\:{odd}\right) \\ $$ $${x}^{{p}} −\mathrm{1}=\mathrm{0} \\ $$ $$\mathrm{1}+{x}+{x}^{\mathrm{2}} +..+{x}^{{p}−\mathrm{1}} =\mathrm{0}\:\left(\mathrm{1}\right) \\ $$ $$\mathrm{All}\:\mathrm{roots}\:\mathrm{of}\:\left(\mathrm{1}\right)\:\mathrm{are}\:\mathrm{primitive} \\ $$ $$\left({x}−{r}_{\mathrm{1}} \right)\left({x}−{r}_{\mathrm{2}} \right)\left({x}−{r}_{\mathrm{3}} \right)...\left({x}−{r}_{{p}−\mathrm{1}} \right)=\mathrm{0} \\ $$ $${x}^{{p}−\mathrm{1}} −\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +..+{r}_{{p}−\mathrm{1}} \right){x}^{{p}−\mathrm{2}} +....+\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \underset{{i}=\mathrm{1}} {\overset{{p}−\mathrm{1}} {\prod}}{r}_{{i}} \:=\mathrm{0} \\ $$ $${so} \\ $$ $${r}_{\mathrm{1}} +{r}_{\mathrm{2}} +..+{r}_{{p}−\mathrm{1}} =−\mathrm{1} \\ $$ $$\mathrm{Same}\:\mathrm{as}\:\mathrm{what}\:\mathrm{you}\:\mathrm{stated}\:\mathrm{that}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{primitive} \\ $$ $$\mathrm{roots}\:=−\mathrm{1}\:\mathrm{for}\:{n}\in\mathbb{P}. \\ $$ | ||
Commented byYozzii last updated on 18/Dec/15 | ||
$${Ahh}.\:{Nicely}\:. \\ $$ | ||