I would like to calculate the Riemann sum of sin &#x2061;<!-- ⁡ --> ( x ) . Fun starts he

Kaeden Hoffman 2022-06-30 Answered
I would like to calculate the Riemann sum of sin ( x ). Fun starts here:
R = π n j = 1 n sin ( π n j )
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Answers (2)

owerswogsnz
Answered 2022-07-01 Author has 12 answers
There's a way to find an expression for the sum
j = 1 n sin ( j θ )
by considering instead the geometric sum
1 + z + z 2 + + z n = z n + 1 1 z 1 for  z 1
in combination with Euler's formula by taking z = e i θ = cos θ + i sin θ and also using De Moivre's formula. Then you can find that
j = 1 n sin ( j θ ) = cos ( θ 2 ) cos ( ( n + 1 2 ) θ ) 2 sin ( θ 2 )
This is a standard exercise in most complex analysis books or actually any book that introduces complex numbers. In your case you just have to take θ = π n

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Agostarawz
Answered 2022-07-02 Author has 3 answers
Use
2 sin ( π 2 n ) sin ( π n j ) = cos ( π 2 n ( 2 j 1 ) ) cos ( π 2 n ( 2 j + 1 ) )
Thus the sum telescopes j = 1 n ( g ( j ) g ( j + 1 ) ) = g ( 1 ) g ( n + 1 )
R n = π n j = 1 n sin ( π n j ) = π 2 n sin ( π 2 n ) ( cos ( π 2 n ) cos ( π 2 n ( 2 n + 1 ) ) ) = π n 1 tan ( π 2 n )
The large n limit is easy:
lim n R n = 2 lim n π 2 n 1 tan ( π 2 n ) = 2 lim x 0 x tan ( x ) = 2

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