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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^{t h} root of

unity .

Does \underset{n = 1}{\sum^{\infty}} a_{n} converge ?

An n^{t h} root of unity, say z, is primitive if it is not

k^{t h} root of unity where k < n .

o r z^{n} = 1 and z^{k} \neq 1 (k = 1, 2, 3, . ., n - 1)

a_{1} = 1

a_{2} = - 1 (2^{n d} root of unity are 1 and - 1, 1 is not

primitive)

Answered by Yozzii last updated on 18/Dec/15

L e t p (n) b e t h e p r i m i t i v e r o o t s o f t h e

e q u a t i o n z^{n} = 1 a n d a_{n} = Σ p (n) .

n = 1 \Rightarrow z = 1 \Rightarrow p (1) = 1 \Rightarrow a_{1} = 1 .

n = 2 \Rightarrow z^{2} = 1 \Rightarrow z = \pm 1 \Rightarrow p (2) = - 1 \Rightarrow a_{2} = - 1 .

n = 3 \Rightarrow z^{3} = 1 = e^{2 π i r} (- 1 ⩽ r ⩽ 1)

\Rightarrow z = e^{\frac{2 π i r}{3}} \Rightarrow p (3) = e^{\pm \frac{2 π i}{3}} \Rightarrow a_{3} = 2 c o s \frac{2 π}{3} = - 1

n = 4 \Rightarrow z^{4} = 1 \Rightarrow z = \pm 1, \pm i \Rightarrow p (4) = \pm i \Rightarrow a_{4} = 0

n = 5 \Rightarrow z^{5} = e^{2 π i r} (- 2 ⩽ r ⩽ 2)

\Rightarrow z = e^{\frac{2 π i r}{5}} = 1, e^{\pm \frac{2 π i}{5}}, e^{\pm \frac{4 π i}{5}} \Rightarrow p (5) = e^{\pm \frac{2 π i}{5}}, e^{\pm \frac{4 π i}{5}}

\Rightarrow a_{5} = 2 (c o s \frac{2 π}{5} + c o s \frac{4 π}{5})

a_{5} = 2 {2 c o s \frac{\frac{4 π + 2 π}{5}}{2} c o s \frac{\frac{4 π - 2 π}{5}}{2}}

a_{5} = 4 c o s \frac{3 π}{5} c o s \frac{π}{5} = - 1

n = 6 \Rightarrow z^{6} = e^{2 π i r} (- 2 ⩽ r ⩽ 3)

z = e^{\frac{π i r}{3}} \Rightarrow z = \pm 1, e^{\pm \frac{π i}{3}}, e^{\pm \frac{2 π i}{3}} \Rightarrow p (6) = e^{\pm \frac{π i}{3}}

\Rightarrow a_{6} = 2 c o s \frac{π}{3} = 1

n = 7 \Rightarrow z^{7} = e^{2 π i r} (- 3 ⩽ r ⩽ 3)

\Rightarrow z = e^{\frac{2 π i r}{7}} \Rightarrow z = 1, e^{\frac{\pm 2 π i}{7}}, e^{\frac{\pm 4 π i}{7}}, e^{\frac{\pm 6 π i}{7}}

\Rightarrow a_{7} = 2 (c o s \frac{2 π}{7} + c o s \frac{4 π}{7} + c o s \frac{6 π}{7}) = - 1

n = 8 \Rightarrow z^{8} = e^{2 π i r} \Rightarrow z = e^{\frac{π i r}{4}} (- 3 ⩽ r ⩽ 4)

\Rightarrow z = \pm 1, e^{\pm \frac{3 π i}{4}}, \pm i, e^{\pm \frac{π i}{4}}

\Rightarrow a_{8} = 2 (c o s \frac{3 π}{4} + c o s \frac{π}{4}) = 0 .

n = 9 \Rightarrow z = e^{\frac{2 π i r}{9}} (- 4 ⩽ r ⩽ 4)

\Rightarrow z = 1, e^{\pm \frac{2 π i}{9}}, e^{\pm \frac{4 π i}{9}}, e^{\pm \frac{2 π i}{3}}, e^{\pm \frac{8 π i}{9}}

\Rightarrow a_{9} = 2 (c o s \frac{2 π}{9} + c o s \frac{4 π}{9} + c o s \frac{8 π}{9}) = 0

n = 10 \Rightarrow z = e^{\frac{π i r}{5}} (- 4 ⩽ r ⩽ 5)

\Rightarrow z = \pm 1, e^{\pm \frac{π i}{5}}, e^{\pm \frac{2 π i}{5}}, e^{\pm \frac{3 π i}{5}}, e^{\pm \frac{4 π i}{5}}

\Rightarrow a_{10} = 2 (c o s \frac{π}{5} + c o s \frac{3 π}{5}) = 1

I c o n j e c t t h a t f o r n b e i n g p r i m e,

a_{n} \neq 0, s o t h a t \lim_{(n \in P) \to \infty} a_{n} \neq 0 \Rightarrow Σ a_{n} d o e s

n o t c o n v e r g e .

S u p p o s e t h a t n \in P - {2} a n d z^{n} = e^{2 π i r} w i t h

r = 0, 1, 2, \dots, n - 1 . T h e n, z = e^{\frac{2 π i r}{n}} .

n \in P - {2} \Rightarrow \frac{r}{n} n e v e r r e d u c e s n o r v a n i s h e s

f o r a l l r \neq 0 . I f r = 0 w e o b t a i n t h e n o n - p r i m i t i v e

r o o t z = 1 . O t h e r w i s e, a l l a r g u m e n t s

\frac{2 π r}{n} a r e u n i q u e . I n a d d i n g a l l o f t h e

p r i m i t i v e r o o t s h o w e v e r, t h e i m a g i n a r y

p a r t s v a n i s h f o r t h e r e a r e a n e v e n

n u m b e r o f r o o t s (n - 1 ∵ n \in O^{+}) a n d h a l f o f t h e s e

r o o t s a r e c o m p l e x c o n j u g a t e s o f t h e r e m a i n i n g

r o o t s .

∴ a_{n} = 2 \underset{r = 1}{\sum^{\frac{n - 1}{2}}} c o s \frac{2 π r}{n} (n \in P - {2})

\Rightarrow \lim_{(n \in P - {2}) \to \infty} a_{n} \neq 0

I f n = 2, a_{2} = - 1 \neq 0 . S o \lim_{(n \in P) \to \infty} a_{n} \neq 0 .

T h e r e e x i s t i n f i n i t e l y m a n y p r i m e s

w i t h P \subset N, s o a_{n} a l w a y s e x i s t s f o r

n \in P \subset N .

C a l c u l a t i o n s p r o j e c t t h a t a_{n} = - 1

f o r a l l n \in P . I t h a s b e e n p r o v e n t h a t

a_{n} = - 1 \forall n \in P i n d e e d i n t h e c o m m e n t s

b e l o w .

Commented by prakash jain last updated on 18/Dec/15

Thanks for detailed answer .

Commented by Yozzii last updated on 18/Dec/15

S (n) = \underset{r = 1}{\sum^{n}} c o s r θ

S (n) s i n 0.5 θ = \underset{r = 1}{\sum^{n}} c o s r θ s i n 0.5 θ (θ \neq 2 n π)

2 c o s r θ s i n 0.5 θ = s i n (r + 0.5) θ - s i n (r - 0.5) θ

∴ S (n) s i n 0.5 θ = \frac{1}{2} {s i n 1.5 θ - s i n 0.5 θ

+ s i n 2.5 θ - s i n 1.5 θ + s i n 3.5 θ - s i n 2.5 θ

+ \dots + s i n (n - 2.5) θ - s i n (n - 3.5) θ

+ s i n (n - 1.5) θ - s i n (n - 2.5) θ

+ s i n (n - 0.5) θ - s i n (n - 1.5) θ

+ s i n (n + 0.5) θ - s i n (n - 0.5) θ}

S (n) = \frac{s i n (n + 0.5) θ - s i n 0.5 θ}{2 s i n 0.5 θ}

S (n) = \frac{2 c o s \frac{(n + 0.5) θ + 0.5 θ}{2} s i n \frac{(n + 0.5) θ - 0.5 θ}{2}}{2 s i n 0.5 θ}

S (n) = \frac{c o s \frac{n + 1}{2} θ s i n \frac{n θ}{2}}{s i n 0.5 θ}

n \in P - {2} \Rightarrow \frac{n - 1}{2} \in Z^{+}

s o \frac{n - 1}{2} c a n b e s a f e l y s u b s t i t u t e d f o r n i n S (n) .

∴ S (\frac{n - 1}{2}) = \frac{c o s \frac{\frac{n}{2} - \frac{1}{2} + 1}{2} θ s i n \frac{n - 1}{4} θ}{s i n 0.5 θ}

S (\frac{n - 1}{2}) = \frac{c o s \frac{n + 1}{4} θ s i n \frac{n - 1}{4} θ}{2 c o s \frac{θ}{4} s i n \frac{θ}{4}}

N o w l e t θ = \frac{2 π}{n} .

S (0.5 (n - 1)) = \frac{c o s \frac{n + 1}{2 n} π s i n \frac{n - 1}{2 n} π}{2 c o s \frac{π}{2 n} s i n \frac{π}{2 n}}

S (\frac{1}{2} (n - 1)) = \frac{c o s (\frac{π}{2} + \frac{π}{2 n}) s i n (\frac{π}{2} - \frac{π}{2 n})}{2 c o s \frac{π}{2 n} s i n \frac{π}{2 n}}

c o s (0.5 π + α) = 0 - s i n 0.5 π s i n α

a n d s i n (0.5 π - α) = c o s α

\Rightarrow S (\frac{1}{2} (n - 1)) = - 0.5

\Rightarrow a_{n} = 2 \times (- 0.5) = - 1 \forall n \in P - {2} .

o

Commented by prakash jain last updated on 18/Dec/15

p > 2 p r i m e - (s o o d d)

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}) \dots (x - r_{p - 1}) = 0

x^{p - 1} - (r_{1} + r_{2} + . . + r_{p - 1}) x^{p - 2} + \dots . + {(- 1)}^{p - 1} \underset{i = 1}{\prod^{p - 1}} r_{i} = 0

s o

r_{1} + r_{2} + . . + r_{p - 1} = - 1

Same as what you stated that sum of primitive

roots = - 1 for n \in P .

Commented by Yozzii last updated on 18/Dec/15

A h h . N i c e l y .

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