Can someone prove that \(\displaystyle{\sum_{{{i}={1}}}^{{{89}}}}{{\sin}^{{{2}{n}}}{\left({\frac{{\pi}}{{{180}}}}{i}\right)}}\) is a

Riley Quinn

Riley Quinn

Answered question

2022-03-16

Can someone prove that i=189sin2n(π180i) is a dyadic rational for all positive integers n, or find counterexample?

Answer & Explanation

IdodaHekbed7mx

IdodaHekbed7mx

Beginner2022-03-17Added 4 answers

j=0k1sin2n(jπk)=j=0k1(ejiπkejiπk2i)2n
=1(2i)2nj=0k1m=02n(2nm)(ejiπk)m(ejiπk)2nm
=1(4)nm=02n(2nm)(1)mj=0k1ej(2m2n)iπk
That last sum (over j) is a sum of roots of unity - specifically, if d=GCD(mn,k), the last sum is the sum of the kd roots of unity d times. So the sum is equal to 0, except when d=k, in which case all the terms are 1 and the sum is k. This implies that the summand of the sum over m is an integer, so the entire expression is an integer divided by 4n.
In the case that n<k, we get the simple expression
1(4)n(2nn)(1)nk=14n(2nn)k
For the original problem I posed, we have
2j=189sin2n(πi180)+1
=sin2n(0)+j=189sin2n(πi180)+sin2n(90π180)+j=91179sin2n(πi180)
=j=0179sin2n(πi180)
and we have proven that the last sum is a dyadic ratioanl, so j=189sin2n(πi180) is a dyadic rational as well.

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