Question Number 127206 by MathSh last updated on 27/Dec/20

Answered by Olaf last updated on 28/Dec/20
![Ω = Π_(k=1) ^(k=m) sin(((kπ)/(2m+1))) Solve (z+1)^(2m+1) = 1 (1) z_k = e^((2ikπ)/(2m+1)) −1, k = 0,1,2,...,2m z_k = 2ie^((ikπ)/(2m+1)) sin(((kπ)/(2m+1))) Let P(z) = Σ_(k=0) ^(2m) (z+1)^k (2) ⇒ P(0) = 2m+1 (3) P(z) = 0 ⇔ Σ_(k=0) ^(2m) (z+1)^k = 0 ⇔ (((z+1)^(2m+1) −1)/((z+1)−1)) = 0, z≠0 ⇔ z = z_k , k = 1,2,...,2m (4) (2) and (4) : P(z) = Π_(k=1) ^(2m) (z−z_k ) ⇒ P(0) = Π_(k=1) ^(2m) (−z_k ) = Π_(k=1) ^(2m) z_k P(0) = Π_(k=1) ^(2m) [2ie^((ikπ)/(2m+1)) sin(((kπ)/(2m+1)))] P(0) = 2^(2m) i^(2m) e^(((2m(2m+1))/2).((iπ)/(2m+1))) Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) P(0) = 2^(2m) (−1)^m e^(imπ) Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) P(0) = 2^(2m) Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) (5) (3) and (5) : P(0) = 2m+1 = 2^(2m) Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) ⇒ Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) = ((2m+1)/2^(2m) ) (6) But sin(((kπ)/(2m+1))) = sin(π−((kπ)/(2m+1))) = sin((((2m+1−k)π)/(2m+1))) For example : sin((π/(2m+1))) = sin(((2mπ)/(2m+1))) ⇒ Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) = Π_(k=1) ^m sin^2 (((kπ)/(2m+1))) = (Π_(k=1) ^m sin(((kπ)/(2m+1))))^2 = Ω^2 (7) (6) and (7) : Ω = ((√(2m+1))/2^m )](https://www.tinkutara.com/question/Q127239.png)
Commented by MathSh last updated on 28/Dec/20

Answered by mnjuly1970 last updated on 28/Dec/20

Commented by MathSh last updated on 28/Dec/20

Commented by mnjuly1970 last updated on 28/Dec/20
