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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 byprakash jain last updated on 18/Dec/15

Thanks for detailed answer.

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{detailed}\:\mathrm{answer}.\: \\ $$

Commented byYozzii 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 byprakash 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 byYozzii last updated on 18/Dec/15

Ahh. Nicely .

$${Ahh}.\:{Nicely}\:. \\ $$

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