#2015_6 Do you like complex?

Geometry Level 4

For natural number n n state r = 1 n 1 sin ( r π n ) \displaystyle\prod _{ r=1 }^{ n-1 }{ \sin\left( \frac { r\pi }{ n } \right) } as an explicit formula of n n .

More Questions??

n . 2 n + 1 n.{ 2 }^{ n+1 } n 2 n 1 \frac { n }{ { 2 }^{ n-1 } } n . 2 n 1 n.{ 2 }^{ n-1 } n 2 n + 1 \frac { n }{ { 2 }^{ n+1 } }

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1 solution

Anandhu Raj
Feb 28, 2015

We know that,

z n 1 = ( z 1 ) r = 1 n 1 ( z α r ) { z }^{ n }-1=(z-1)\prod _{ r=1 }^{ n-1 }{ (z-{ \alpha }^{ r }) }

z n 1 ( z 1 ) = r = 1 n 1 ( z α r ) \frac { { z }^{ n }-1 }{ (z-1) } =\prod _{ r=1 }^{ n-1 }{ (z-{ \alpha }^{ r }) }

Applying limits on both sides,

l i m z 1 ( z n 1 ( z 1 ) ) = l i m z 1 ( r = 1 n 1 ( z α r ) ) \underset { z\rightarrow 1 }{ lim } \left( \frac { { z }^{ n }-1 }{ (z-1) } \right) =\underset { z\rightarrow 1 }{ lim } \left( \prod _{ r=1 }^{ n-1 }{ (z-{ \alpha }^{ r }) } \right)

n = ( r = 1 n 1 ( 1 α r ) ) \Rightarrow n=\left( \prod _{ r=1 }^{ n-1 }{ (1-{ \alpha }^{ r }) } \right)

= ( r = 1 n 1 ( 1 e i 2 π r n ) ) =\left( \prod _{ r=1 }^{ n-1 }{ (1-{ e }^{ \frac { i2\pi r }{ n } }) } \right)

= r = 1 n 1 { e i π r n ( e i π r n e i π r n ) } \prod _{ r=1 }^{ n-1 }{ \left\{ -{ e }^{ \frac { i\pi r }{ n } }\left( { e }^{ \frac { i\pi r }{ n } }-{ e }^{ \frac { -i\pi r }{ n } } \right) \right\} }

= r = 1 n 1 e i π r n . 2 i s i n ( π r n ) \prod _{ r=1 }^{ n-1 }{ -{ e }^{ \frac { i\pi r }{ n } }.2isin\left( \frac { \pi r }{ n } \right) }

n = r = 1 n 1 2 s i n ( π r n ) \left| n \right| =\prod _{ r=1 }^{ n-1 }{ 2sin\left( \frac { \pi r }{ n } \right) }

n = 2 n 1 r = 1 n 1 s i n ( π r n ) \Rightarrow \left| n \right| ={ 2 }^{ n-1 }\prod _{ r=1 }^{ n-1 }{ sin\left( \frac { \pi r }{ n } \right) }

r = 1 n 1 s i n ( π r n ) = n 2 n 1 \boxed{\therefore \prod _{ r=1 }^{ n-1 }{ sin\left( \frac { \pi r }{ n } \right) = } \frac { n }{ { 2 }^{ n-1 } } }

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