This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
k = 1 ∏ n − 1 sin ( n k π ) = k = 1 ∏ n − 1 ( 2 i e n i π k − e − n i π k ) = k = 1 ∏ n − 1 ( 2 i 1 ) ⋅ ( − e − n i π k ) ⋅ ( − e n i 2 π k + 1 ) = ( 2 i − 1 ) n − 1 ⋅ e n − i π ∑ k = 1 n − 1 k ⋅ k = 1 ∏ n − 1 ( 1 − e n 2 π i k ) = 2 n − 1 ⋅ ( − i ) n − 1 1 ⋅ e − n i π ⋅ 2 ( n − 1 ) n ⋅ k = 1 ∏ n − 1 ( 1 − w k ) = 2 n − 1 ⋅ ( − i ) n − 1 ( − i ) n − 1 ⋅ x → 1 lim x − 1 x n − 1 = 2 n − 1 n x n − 1 = k = 0 ∏ n − 1 ( x − w k ) = ( x − 1 ) k = 1 ∏ n − 1 ( x − w k ) ⇒ k = 1 ∏ n − 1 ( x − w k ) = x − 1 x n − 1 sin ( θ ) = 2 i e i θ − e − i θ w is nth root of unity
Problem Loading...
Note Loading...
Set Loading...
An obscure (to me) identity from MathWorld simplifies things a bit.
k = 1 ∏ n − 1 sin ( n k π ) = 2 n − 1 n
By symmetry: sin ( n k π ) = sin ( n n − k π )
k = 1 ∏ 3 sin ( n k π ) = k = 4 ∏ 6 sin ( n k π )
k = 1 ∏ 3 sin ( n k π ) = 2 n − 1 n = 6 4 7 = 8 7 = 0 . 3 3 0 7