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Question Number 3650 by prakash jain last updated on 18/Dec/15
If a_n is sum of the all primitive n^(th) root of unity. Does Σ_(n=1) ^∞ a_n converge? An n^(th) root of unity, say z, is primitive if it is not k^(th) root of unity where k<n. or z^n =1 and z^k ≠1 (k=1,2,3,..,n−1) a_1 =1 a_2 =−1 (2^(nd) root of unity are 1 and −1, 1 is not primitive)
$$\mathrm{If}\:{a}_{{n}} \:\mathrm{is}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{all}\:\mathrm{primitive}\:{n}^{{th}} \:\mathrm{root}\:\mathrm{of} \\ $$$$\mathrm{unity}. \\ $$$$\mathrm{Does}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \:\mathrm{converge}? \\ $$$$\mathrm{An}\:{n}^{{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity},\:\mathrm{say}\:{z},\:\mathrm{is}\:\mathrm{primitive}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not} \\ $$$${k}^{{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{where}\:{k}<{n}. \\ $$$${or}\:{z}^{{n}} =\mathrm{1}\:\mathrm{and}\:{z}^{{k}} \neq\mathrm{1}\:\left(\mathrm{k}=\mathrm{1},\mathrm{2},\mathrm{3},..,{n}−\mathrm{1}\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =−\mathrm{1}\:\:\left(\mathrm{2}^{{nd}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{are}\:\mathrm{1}\:\mathrm{and}\:−\mathrm{1},\:\mathrm{1}\:\mathrm{is}\:\mathrm{not}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{primitive}\right) \\ $$
Answered by Yozzii last updated on 18/Dec/15
Let p(n) be the primitive roots of the equation z^n =1 and a_n =Σp(n). n=1⇒z=1⇒p(1)=1⇒a_1 =1. n=2⇒z^2 =1⇒z=±1⇒p(2)=−1⇒a_2 =−1. n=3⇒z^3 =1=e^(2πir) (−1≤r≤1) ⇒z=e^((2πir)/3) ⇒p(3)=e^(±((2πi)/3)) ⇒a_3 =2cos((2π)/3)=−1 n=4⇒z^4 =1⇒z=±1,±i⇒p(4)=±i⇒a_4 =0 n=5⇒z^5 =e^(2πir) (−2≤r≤2) ⇒z=e^((2πir)/5) =1,e^(±((2πi)/5)) ,e^(±((4πi)/5)) ⇒p(5)=e^(±((2πi)/5)) ,e^(±((4πi)/5)) ⇒a_5 =2(cos((2π)/5)+cos((4π)/5)) a_5 =2{2cos(((4π+2π)/5)/2)cos(((4π−2π)/5)/2)} a_5 =4cos((3π)/5)cos(π/5)=−1 n=6⇒z^6 =e^(2πir) (−2≤r≤3) z=e^((πir)/3) ⇒z=±1,e^(±((πi)/3)) ,e^(±((2πi)/3)) ⇒p(6)=e^(±((πi)/3)) ⇒a_6 =2cos(π/3)=1 n=7⇒z^7 =e^(2πir) (−3≤r≤3) ⇒z=e^((2πir)/7) ⇒z=1,e^((±2πi)/7) ,e^((±4πi)/7) ,e^((±6πi)/7) ⇒a_7 =2(cos((2π)/7)+cos((4π)/7)+cos((6π)/7))=−1 n=8⇒z^8 =e^(2πir) ⇒z=e^((πir)/4) (−3≤r≤4) ⇒z=±1,e^(±((3πi)/4)) ,±i,e^(±((πi)/4)) ⇒a_8 =2(cos((3π)/4)+cos(π/4))=0. n=9⇒z=e^((2πir)/9) (−4≤r≤4) ⇒z=1,e^(±((2πi)/9)) ,e^(±((4πi)/9)) ,e^(±((2πi)/3)) ,e^(±((8πi)/9)) ⇒a_9 =2(cos((2π)/9)+cos((4π)/9)+cos((8π)/9))=0 n=10⇒z=e^((πir)/5) (−4≤r≤5) ⇒z=±1,e^(±((πi)/5)) ,e^(±((2πi)/5)) ,e^(±((3πi)/5)) ,e^(±((4πi)/5)) ⇒a_(10) =2(cos(π/5)+cos((3π)/5))=1 I conject that for n being prime, a_n ≠0, so that lim_((n∈P)→∞) a_n ≠0⇒Σa_n does not converge. Suppose that n∈P−{2} and z^n =e^(2πir) with r=0,1,2,...,n−1. Then, z=e^((2πir)/n) . n∈P−{2}⇒ (r/n) never reduces nor vanishes for all r≠0. If r=0 we obtain the non−primitive root z=1. Otherwise, all arguments ((2πr)/n) are unique. In adding all of the primitive roots however, the imaginary parts vanish for there are an even number of roots (n−1 ∵ n∈O^+ ) and half of these roots are complex conjugates of the remaining roots. ∴ a_n =2Σ_(r=1) ^((n−1)/2) cos((2πr)/n) (n∈P−{2}) ⇒lim_((n∈P−{2})→∞) a_n ≠0 If n=2, a_2 =−1≠0. So lim_((n∈P)→∞) a_n ≠0. There exist infinitely many primes with P⊂N, so a_n always exists for n∈P⊂N. Calculations project that a_n =−1 for all n∈P. It has been proven that a_n =−1 ∀n∈P indeed in the comments below.
$${Let}\:{p}\left({n}\right)\:{be}\:{the}\:{primitive}\:{roots}\:{of}\:{the} \\ $$$${equation}\:{z}^{{n}} =\mathrm{1}\:{and}\:{a}_{{n}} =\Sigma{p}\left({n}\right). \\ $$$${n}=\mathrm{1}\Rightarrow{z}=\mathrm{1}\Rightarrow{p}\left(\mathrm{1}\right)=\mathrm{1}\Rightarrow{a}_{\mathrm{1}} =\mathrm{1}. \\ $$$${n}=\mathrm{2}\Rightarrow{z}^{\mathrm{2}} =\mathrm{1}\Rightarrow{z}=\pm\mathrm{1}\Rightarrow{p}\left(\mathrm{2}\right)=−\mathrm{1}\Rightarrow{a}_{\mathrm{2}} =−\mathrm{1}. \\ $$$${n}=\mathrm{3}\Rightarrow{z}^{\mathrm{3}} =\mathrm{1}={e}^{\mathrm{2}\pi{ir}} \:\left(−\mathrm{1}\leqslant{r}\leqslant\mathrm{1}\right)\: \\ $$$$\Rightarrow{z}={e}^{\frac{\mathrm{2}\pi{ir}}{\mathrm{3}}} \Rightarrow{p}\left(\mathrm{3}\right)={e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} \Rightarrow{a}_{\mathrm{3}} =\mathrm{2}{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}=−\mathrm{1} \\ $$$${n}=\mathrm{4}\Rightarrow{z}^{\mathrm{4}} =\mathrm{1}\Rightarrow{z}=\pm\mathrm{1},\pm{i}\Rightarrow{p}\left(\mathrm{4}\right)=\pm{i}\Rightarrow{a}_{\mathrm{4}} =\mathrm{0} \\ $$$${n}=\mathrm{5}\Rightarrow{z}^{\mathrm{5}} ={e}^{\mathrm{2}\pi{ir}} \:\left(−\mathrm{2}\leqslant{r}\leqslant\mathrm{2}\right) \\ $$$$\Rightarrow{z}={e}^{\frac{\mathrm{2}\pi{ir}}{\mathrm{5}}} =\mathrm{1},{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{4}\pi{i}}{\mathrm{5}}} \Rightarrow{p}\left(\mathrm{5}\right)={e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{4}\pi{i}}{\mathrm{5}}} \\ $$$$\Rightarrow{a}_{\mathrm{5}} =\mathrm{2}\left({cos}\frac{\mathrm{2}\pi}{\mathrm{5}}+{cos}\frac{\mathrm{4}\pi}{\mathrm{5}}\right) \\ $$$${a}_{\mathrm{5}} =\mathrm{2}\left\{\mathrm{2}{cos}\frac{\frac{\mathrm{4}\pi+\mathrm{2}\pi}{\mathrm{5}}}{\mathrm{2}}{cos}\frac{\frac{\mathrm{4}\pi−\mathrm{2}\pi}{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$$${a}_{\mathrm{5}} =\mathrm{4}{cos}\frac{\mathrm{3}\pi}{\mathrm{5}}{cos}\frac{\pi}{\mathrm{5}}=−\mathrm{1} \\ $$$${n}=\mathrm{6}\Rightarrow{z}^{\mathrm{6}} ={e}^{\mathrm{2}\pi{ir}} \:\left(−\mathrm{2}\leqslant{r}\leqslant\mathrm{3}\right) \\ $$$${z}={e}^{\frac{\pi{ir}}{\mathrm{3}}} \Rightarrow{z}=\pm\mathrm{1},{e}^{\pm\frac{\pi{i}}{\mathrm{3}}} ,{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} \Rightarrow{p}\left(\mathrm{6}\right)={e}^{\pm\frac{\pi{i}}{\mathrm{3}}} \\ $$$$\Rightarrow{a}_{\mathrm{6}} =\mathrm{2}{cos}\frac{\pi}{\mathrm{3}}=\mathrm{1} \\ $$$${n}=\mathrm{7}\Rightarrow{z}^{\mathrm{7}} ={e}^{\mathrm{2}\pi{ir}} \:\left(−\mathrm{3}\leqslant{r}\leqslant\mathrm{3}\right) \\ $$$$\Rightarrow{z}={e}^{\frac{\mathrm{2}\pi{ir}}{\mathrm{7}}} \Rightarrow{z}=\mathrm{1},{e}^{\frac{\pm\mathrm{2}\pi{i}}{\mathrm{7}}} ,{e}^{\frac{\pm\mathrm{4}\pi{i}}{\mathrm{7}}} ,{e}^{\frac{\pm\mathrm{6}\pi{i}}{\mathrm{7}}} \\ $$$$\Rightarrow{a}_{\mathrm{7}} =\mathrm{2}\left({cos}\frac{\mathrm{2}\pi}{\mathrm{7}}+{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}+{cos}\frac{\mathrm{6}\pi}{\mathrm{7}}\right)=−\mathrm{1} \\ $$$${n}=\mathrm{8}\Rightarrow{z}^{\mathrm{8}} ={e}^{\mathrm{2}\pi{ir}} \Rightarrow{z}={e}^{\frac{\pi{ir}}{\mathrm{4}}} \left(−\mathrm{3}\leqslant{r}\leqslant\mathrm{4}\right) \\ $$$$\Rightarrow{z}=\pm\mathrm{1},{e}^{\pm\frac{\mathrm{3}\pi{i}}{\mathrm{4}}} ,\pm{i},{e}^{\pm\frac{\pi{i}}{\mathrm{4}}} \\ $$$$\Rightarrow{a}_{\mathrm{8}} =\mathrm{2}\left({cos}\frac{\mathrm{3}\pi}{\mathrm{4}}+{cos}\frac{\pi}{\mathrm{4}}\right)=\mathrm{0}. \\ $$$${n}=\mathrm{9}\Rightarrow{z}={e}^{\frac{\mathrm{2}\pi{ir}}{\mathrm{9}}} \:\left(−\mathrm{4}\leqslant{r}\leqslant\mathrm{4}\right) \\ $$$$\Rightarrow{z}=\mathrm{1},{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{9}}} ,{e}^{\pm\frac{\mathrm{4}\pi{i}}{\mathrm{9}}} ,{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} ,{e}^{\pm\frac{\mathrm{8}\pi{i}}{\mathrm{9}}} \\ $$$$\Rightarrow{a}_{\mathrm{9}} =\mathrm{2}\left({cos}\frac{\mathrm{2}\pi}{\mathrm{9}}+{cos}\frac{\mathrm{4}\pi}{\mathrm{9}}+{cos}\frac{\mathrm{8}\pi}{\mathrm{9}}\right)=\mathrm{0} \\ $$$${n}=\mathrm{10}\Rightarrow{z}={e}^{\frac{\pi{ir}}{\mathrm{5}}} \:\left(−\mathrm{4}\leqslant{r}\leqslant\mathrm{5}\right) \\ $$$$\Rightarrow{z}=\pm\mathrm{1},{e}^{\pm\frac{\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{2}\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{3}\pi{i}}{\mathrm{5}}} ,{e}^{\pm\frac{\mathrm{4}\pi{i}}{\mathrm{5}}} \\ $$$$\Rightarrow{a}_{\mathrm{10}} =\mathrm{2}\left({cos}\frac{\pi}{\mathrm{5}}+{cos}\frac{\mathrm{3}\pi}{\mathrm{5}}\right)=\mathrm{1} \\ $$$$ \\ $$$${I}\:{conject}\:{that}\:{for}\:{n}\:{being}\:{prime}, \\ $$$${a}_{{n}} \neq\mathrm{0},\:{so}\:{that}\:\underset{\left({n}\in\mathbb{P}\right)\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \neq\mathrm{0}\Rightarrow\Sigma{a}_{{n}} \:{does} \\ $$$${not}\:{converge}.\: \\ $$$$ \\ $$$${Suppose}\:{that}\:{n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\:{and}\:{z}^{{n}} ={e}^{\mathrm{2}\pi{ir}} \:{with} \\ $$$${r}=\mathrm{0},\mathrm{1},\mathrm{2},…,{n}−\mathrm{1}.\:{Then},\:{z}={e}^{\frac{\mathrm{2}\pi{ir}}{{n}}} . \\ $$$${n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\Rightarrow\:\frac{{r}}{{n}}\:{never}\:{reduces}\:{nor}\:{vanishes} \\ $$$${for}\:{all}\:{r}\neq\mathrm{0}.\:{If}\:{r}=\mathrm{0}\:{we}\:{obtain}\:{the}\:{non}−{primitive} \\ $$$${root}\:{z}=\mathrm{1}.\:{Otherwise},\:{all}\:{arguments}\: \\ $$$$\frac{\mathrm{2}\pi{r}}{{n}}\:{are}\:{unique}.\:{In}\:{adding}\:{all}\:{of}\:{the} \\ $$$${primitive}\:{roots}\:{however},\:{the}\:{imaginary} \\ $$$${parts}\:{vanish}\:{for}\:{there}\:{are}\:{an}\:{even}\: \\ $$$${number}\:{of}\:{roots}\:\left({n}−\mathrm{1}\:\because\:{n}\in\mathbb{O}^{+} \right)\:{and}\:{half}\:{of}\:{these} \\ $$$${roots}\:{are}\:{complex}\:{conjugates}\:{of}\:{the}\:{remaining} \\ $$$${roots}. \\ $$$$\therefore\:{a}_{{n}} =\mathrm{2}\underset{{r}=\mathrm{1}} {\overset{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {\sum}}{cos}\frac{\mathrm{2}\pi{r}}{{n}}\:\left({n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\right) \\ $$$$\Rightarrow\underset{\left({n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\right)\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \neq\mathrm{0} \\ $$$${If}\:{n}=\mathrm{2},\:{a}_{\mathrm{2}} =−\mathrm{1}\neq\mathrm{0}.\:{So}\:\underset{\left({n}\in\mathbb{P}\right)\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \neq\mathrm{0}. \\ $$$${There}\:{exist}\:{infinitely}\:{many}\:{primes}\: \\ $$$${with}\:\mathbb{P}\subset\mathbb{N},\:{so}\:{a}_{{n}} \:{always}\:{exists}\:{for} \\ $$$${n}\in\mathbb{P}\subset\mathbb{N}. \\ $$$${Calculations}\:{project}\:{that}\:{a}_{{n}} =−\mathrm{1} \\ $$$${for}\:{all}\:{n}\in\mathbb{P}.\:{It}\:{has}\:{been}\:{proven}\:{that} \\ $$$${a}_{{n}} =−\mathrm{1}\:\forall{n}\in\mathbb{P}\:{indeed}\:{in}\:{the}\:{comments}\: \\ $$$${below}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by prakash jain last updated on 18/Dec/15
Thanks for detailed answer.
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{detailed}\:\mathrm{answer}.\: \\ $$
Commented by Yozzii last updated on 18/Dec/15
S(n)=Σ_(r=1) ^n cosrθ S(n)sin0.5θ=Σ_(r=1) ^n cosrθsin0.5θ (θ≠2nπ) 2cosrθsin0.5θ=sin(r+0.5)θ−sin(r−0.5)θ ∴S(n)sin0.5θ=(1/2){sin1.5θ−sin0.5θ +sin2.5θ−sin1.5θ+sin3.5θ−sin2.5θ +...+sin(n−2.5)θ−sin(n−3.5)θ +sin(n−1.5)θ−sin(n−2.5)θ +sin(n−0.5)θ−sin(n−1.5)θ +sin(n+0.5)θ−sin(n−0.5)θ} S(n)=((sin(n+0.5)θ−sin0.5θ)/(2sin0.5θ)) S(n)=((2cos(((n+0.5)θ+0.5θ)/2)sin(((n+0.5)θ−0.5θ)/2))/(2sin0.5θ)) S(n)=((cos((n+1)/2)θsin((nθ)/2))/(sin0.5θ)) n∈P−{2}⇒ ((n−1)/2)∈Z^+ so ((n−1)/2) can be safely substituted for n in S(n). ∴ S(((n−1)/2))=((cos(((n/2)−(1/2)+1)/2)θsin((n−1)/4)θ)/(sin0.5θ)) S(((n−1)/2))=((cos((n+1)/4)θsin((n−1)/4)θ)/(2cos(θ/4)sin(θ/4))) Now let θ=((2π)/n). S(0.5(n−1))=((cos((n+1)/(2n))πsin((n−1)/(2n))π)/(2cos(π/(2n))sin(π/(2n)))) S((1/2)(n−1))=((cos((π/2)+(π/(2n)))sin((π/2)−(π/(2n))))/(2cos(π/(2n))sin(π/(2n)))) cos(0.5π+α)=0−sin0.5πsinα and sin(0.5π−α)=cosα ⇒S((1/2)(n−1))=−0.5 ⇒a_n =2×(−0.5)=−1 ∀n∈P−{2}. o
$${S}\left({n}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{cosr}\theta\:\:\: \\ $$$${S}\left({n}\right){sin}\mathrm{0}.\mathrm{5}\theta=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{cosr}\theta{sin}\mathrm{0}.\mathrm{5}\theta\:\:\:\left(\theta\neq\mathrm{2}{n}\pi\right) \\ $$$$ \\ $$$$\mathrm{2}{cosr}\theta{sin}\mathrm{0}.\mathrm{5}\theta={sin}\left({r}+\mathrm{0}.\mathrm{5}\right)\theta−{sin}\left({r}−\mathrm{0}.\mathrm{5}\right)\theta \\ $$$$\therefore{S}\left({n}\right){sin}\mathrm{0}.\mathrm{5}\theta=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\mathrm{1}.\mathrm{5}\theta−{sin}\mathrm{0}.\mathrm{5}\theta\right. \\ $$$$+{sin}\mathrm{2}.\mathrm{5}\theta−{sin}\mathrm{1}.\mathrm{5}\theta+{sin}\mathrm{3}.\mathrm{5}\theta−{sin}\mathrm{2}.\mathrm{5}\theta \\ $$$$+…+{sin}\left({n}−\mathrm{2}.\mathrm{5}\right)\theta−{sin}\left({n}−\mathrm{3}.\mathrm{5}\right)\theta \\ $$$$+{sin}\left({n}−\mathrm{1}.\mathrm{5}\right)\theta−{sin}\left({n}−\mathrm{2}.\mathrm{5}\right)\theta \\ $$$$+{sin}\left({n}−\mathrm{0}.\mathrm{5}\right)\theta−{sin}\left({n}−\mathrm{1}.\mathrm{5}\right)\theta \\ $$$$\left.+{sin}\left({n}+\mathrm{0}.\mathrm{5}\right)\theta−{sin}\left({n}−\mathrm{0}.\mathrm{5}\right)\theta\right\} \\ $$$${S}\left({n}\right)=\frac{{sin}\left({n}+\mathrm{0}.\mathrm{5}\right)\theta−{sin}\mathrm{0}.\mathrm{5}\theta}{\mathrm{2}{sin}\mathrm{0}.\mathrm{5}\theta} \\ $$$${S}\left({n}\right)=\frac{\mathrm{2}{cos}\frac{\left({n}+\mathrm{0}.\mathrm{5}\right)\theta+\mathrm{0}.\mathrm{5}\theta}{\mathrm{2}}{sin}\frac{\left({n}+\mathrm{0}.\mathrm{5}\right)\theta−\mathrm{0}.\mathrm{5}\theta}{\mathrm{2}}}{\mathrm{2}{sin}\mathrm{0}.\mathrm{5}\theta} \\ $$$${S}\left({n}\right)=\frac{{cos}\frac{{n}+\mathrm{1}}{\mathrm{2}}\theta{sin}\frac{{n}\theta}{\mathrm{2}}}{{sin}\mathrm{0}.\mathrm{5}\theta} \\ $$$${n}\in\mathbb{P}−\left\{\mathrm{2}\right\}\Rightarrow\:\frac{{n}−\mathrm{1}}{\mathrm{2}}\in\mathbb{Z}^{+} \\ $$$${so}\:\frac{{n}−\mathrm{1}}{\mathrm{2}}\:{can}\:{be}\:{safely}\:{substituted}\:{for}\:{n}\:{in}\:{S}\left({n}\right). \\ $$$$ \\ $$$$\therefore\:{S}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)=\frac{{cos}\frac{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}}{\mathrm{2}}\theta{sin}\frac{{n}−\mathrm{1}}{\mathrm{4}}\theta}{{sin}\mathrm{0}.\mathrm{5}\theta} \\ $$$${S}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)=\frac{{cos}\frac{{n}+\mathrm{1}}{\mathrm{4}}\theta{sin}\frac{{n}−\mathrm{1}}{\mathrm{4}}\theta}{\mathrm{2}{cos}\frac{\theta}{\mathrm{4}}{sin}\frac{\theta}{\mathrm{4}}} \\ $$$$ \\ $$$${Now}\:{let}\:\theta=\frac{\mathrm{2}\pi}{{n}}. \\ $$$${S}\left(\mathrm{0}.\mathrm{5}\left({n}−\mathrm{1}\right)\right)=\frac{{cos}\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\pi{sin}\frac{{n}−\mathrm{1}}{\mathrm{2}{n}}\pi}{\mathrm{2}{cos}\frac{\pi}{\mathrm{2}{n}}{sin}\frac{\pi}{\mathrm{2}{n}}} \\ $$$${S}\left(\frac{\mathrm{1}}{\mathrm{2}}\left({n}−\mathrm{1}\right)\right)=\frac{{cos}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}{n}}\right){sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}{n}}\right)}{\mathrm{2}{cos}\frac{\pi}{\mathrm{2}{n}}{sin}\frac{\pi}{\mathrm{2}{n}}} \\ $$$${cos}\left(\mathrm{0}.\mathrm{5}\pi+\alpha\right)=\mathrm{0}−{sin}\mathrm{0}.\mathrm{5}\pi{sin}\alpha \\ $$$${and}\:{sin}\left(\mathrm{0}.\mathrm{5}\pi−\alpha\right)={cos}\alpha \\ $$$$\Rightarrow{S}\left(\frac{\mathrm{1}}{\mathrm{2}}\left({n}−\mathrm{1}\right)\right)=−\mathrm{0}.\mathrm{5} \\ $$$$\Rightarrow{a}_{{n}} =\mathrm{2}×\left(−\mathrm{0}.\mathrm{5}\right)=−\mathrm{1}\:\forall{n}\in\mathbb{P}−\left\{\mathrm{2}\right\}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${o} \\ $$
Commented by prakash jain last updated on 18/Dec/15
p>2 prime−(so odd) x^p −1=0 1+x+x^2 +..+x^(p−1) =0 (1) All roots of (1) are primitive (x−r_1 )(x−r_2 )(x−r_3 )...(x−r_(p−1) )=0 x^(p−1) −(r_1 +r_2 +..+r_(p−1) )x^(p−2) +....+(−1)^(p−1) Π_(i=1) ^(p−1) r_i =0 so r_1 +r_2 +..+r_(p−1) =−1 Same as what you stated that sum of primitive roots =−1 for n∈P.
$${p}>\mathrm{2}\:{prime}−\left({so}\:{odd}\right) \\ $$$${x}^{{p}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +..+{x}^{{p}−\mathrm{1}} =\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{All}\:\mathrm{roots}\:\mathrm{of}\:\left(\mathrm{1}\right)\:\mathrm{are}\:\mathrm{primitive} \\ $$$$\left({x}−{r}_{\mathrm{1}} \right)\left({x}−{r}_{\mathrm{2}} \right)\left({x}−{r}_{\mathrm{3}} \right)…\left({x}−{r}_{{p}−\mathrm{1}} \right)=\mathrm{0} \\ $$$${x}^{{p}−\mathrm{1}} −\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +..+{r}_{{p}−\mathrm{1}} \right){x}^{{p}−\mathrm{2}} +….+\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \underset{{i}=\mathrm{1}} {\overset{{p}−\mathrm{1}} {\prod}}{r}_{{i}} \:=\mathrm{0} \\ $$$${so} \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} +..+{r}_{{p}−\mathrm{1}} =−\mathrm{1} \\ $$$$\mathrm{Same}\:\mathrm{as}\:\mathrm{what}\:\mathrm{you}\:\mathrm{stated}\:\mathrm{that}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{primitive} \\ $$$$\mathrm{roots}\:=−\mathrm{1}\:\mathrm{for}\:{n}\in\mathbb{P}. \\ $$
Commented by Yozzii last updated on 18/Dec/15
Ahh. Nicely .
$${Ahh}.\:{Nicely}\:. \\ $$

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