Is this a Trigonometry Problem or a Number Theory Problem?

Number Theory Level 5

S = ∑ a ∈ P cos ⁡ ( 2 π a 1001 ) \large \mathcal{S} = \sum_{a \in \mathcal{P}} \cos \left( 2 \pi \dfrac{a}{1001} \right)

Let us a define a set P \large \mathcal{P} which contains all the natural numbers less than or equal to 1001 and are coprime to 1001.

Evaluate the value of S \mathcal{S} .


The answer is -1.

Surya Prakash by Surya Prakash , 22, India

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2 solutions

Aareyan Manzoor
Dec 5, 2015

the sum is equal to R e ( ∑ a ∈ P ( e x p ( 2 π a i 1001 ) ) Re(\sum_{a\in\mathcal{P}} (exp(\dfrac{2\pi ai}{1001})) this is just the sum of all the primitive 1001th roots of unity.1001= 7 ∗ 11 ∗ 13 7*11*13 . so μ ( 1001 ) = ( − 1 ) 3 = − 1 \mu(1001)=(-1)^3=-1 the real part is also − 1 \boxed{-1} . here μ ( n ) \mu(n) is the mobius function

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Lu Chee Ket
Dec 5, 2015

1001 = 7 × \times 11 × \times 13

Two numbers are coprime if and only if their greatest common divisor is 1.

With Excel: 

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A1 = ROW() {Natural number}

B1 = IF(AND(MOD(A1,7)<>0,MOD(A1,11)<>0,MOD(A1,13)<>0),A1,0)

C1 = IF(B1<>0,COS(2*PI()*B1/1001),0)

For A1 to A1001, we sum up C1 to C1001 and found equaled to -1.

Maximum cumulative sum = 114.264457263257 at row 251, row 252 and row 253;

Minimum cumulative sum = -115.269164927167 at row 750.

Answer: − 1 \boxed{-1}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

∑ k = 0 n − 1 e 2 π i k n = 0 = ∑ d ∣ n ∑ 0 ≤ k ≤ d − 1 , gcd ⁡ ( k , d ) = 1 e 2 π i k d \sum _{ k=0 }^{ n-1 }{ { e }^{ 2\pi i\frac { k }{ n } } } =0=\sum _{ d|n }{ \sum _{ 0\le k\le d-1, \text{ }\gcd(k,d)=1 }{ { e }^{ 2\pi i\frac { k }{ d } } } }

Then you can use mobius inversion to get ∑ 0 ≤ k ≤ n − 1 , ( k , n ) = 1 e 2 π i k n = μ ( n ) \sum _{ 0\le k\le n-1,(k,n)=1 }{ { e }^{ 2\pi i\frac { k }{ n } } } =\mu (n)

Julian Poon - 5 years, 6 months ago

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