#33 Measure Your Calibre

Geometry Level 3

r = 0 n ( n r ) cos ( r θ ) = ? \large{\displaystyle{\sum^n_{r=0}} \binom{n}{r} \cos( r \theta) =\, ?}

Notation: ( M N ) = M ! N ! ( M N ) ! \dbinom MN = \dfrac {M!}{N! (M-N)!} denotes the binomial coefficient .


Bonus: r = 0 n ( n r ) ( tan r θ ) = ? \displaystyle \sum^n_{r=0} \binom{n}{r} (\tan r \theta) = \, ?

Inspired by Tapas Mazumdar , whose problem was about sines, mine is about cosines.
2 n cos n θ 2 cos n θ 2 2^n \cos^n\dfrac{θ}{2}\cos\dfrac{nθ}{2} 2 n 1 n cos n 1 θ 2 cos n θ 2 2^{n-1}n \cos^{n-1}\dfrac{θ}{2}\cos\dfrac{nθ}{2} 2 n sin n θ 2 sin n θ 2 2^n \sin^n\dfrac{θ}{2}\sin\dfrac{nθ}{2} 2 n cos n θ 2 sin n θ 2 2^n \cos^n\dfrac{θ}{2}\sin\dfrac{nθ}{2} 2 n cos n 1 n θ 2 cos n θ 2 2^n \cos^{n-1}n\dfrac{θ}{2}\cos\dfrac{nθ}{2}

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.

2 solutions

Chew-Seong Cheong
Apr 12, 2017

Relevant wiki: Euler's Formula

S = r = 0 n ( n r ) cos r θ = r = 0 n ( n r ) { e i r θ } By Euler’s formula: e i x = cos x + i sin x = { r = 0 n ( n r ) e i r θ } where { z } is the real part of complex number z . = { ( 1 + e i θ ) n } = { ( 1 + cos θ + i sin θ ) n } = { ( 1 + 2 cos 2 θ 2 1 + 2 i sin θ 2 cos θ 2 ) n } = { 2 n cos n θ 2 ( cos θ 2 + i sin θ 2 ) n } = { 2 n cos n θ 2 e i n θ 2 } = 2 n cos n θ 2 cos n θ 2 \begin{aligned} S & = \sum_{r=0}^n {n \choose r} \cos r\theta \\ & = \sum_{r=0}^n {n \choose r} \Re \left \{e^{ir\theta} \right \} & \small \color{#3D99F6} \text{By Euler's formula: } e^{ix} = \cos x + i \sin x \\ & = \Re \left \{\sum_{r=0}^n {n \choose r} e^{ir\theta} \right \} & \small \color{#3D99F6} \text{where }\Re\{z\} \text{ is the real part of complex number } z. \\ & = \Re \left \{\left(1+e^{i\theta} \right)^n \right \} \\ & = \Re \left \{\left(1+\cos \theta + i\sin \theta \right)^n \right \} \\ & = \Re \left \{\left(1+2\cos^2 \frac \theta 2 - 1 + 2 i \sin \frac \theta 2 \cos \frac \theta 2 \right)^n \right \} \\ & = \Re \left \{2^n \cos^n \frac \theta 2 \left( \cos \frac \theta 2 + i \sin \frac \theta 2 \right)^n \right \} \\ & = \Re \left \{2^n \cos^n \frac \theta 2 \cdot e^{i \frac {n\theta} 2} \right \} \\ & = \boxed {2^n \cos^n \dfrac \theta 2 \cos \dfrac {n\theta} 2} \end{aligned}

Sir, what about the bonus?

Md Zuhair - 4 years, 2 months ago

Log in to reply

It may not have a simple form.

Chew-Seong Cheong - 4 years, 2 months ago
Md Zuhair
Apr 11, 2017

r = 0 n ( n r ) cos r θ \large{\displaystyle{\sum^n_{r=0}} \binom{n}{r} \cos r \theta } \implies

LET ME THINK ABOUT THE BONUS!

Great solution...did the same.

A Former Brilliant Member - 4 years, 1 month ago

Log in to reply

Thank you brother, Do you have whatsapp?

If so, please share your number, we have whatsapp brilliant group , well add yoi

Md Zuhair - 4 years, 1 month ago

Log in to reply

No....I don't use whatsapp.

A Former Brilliant Member - 4 years, 1 month ago

Log in to reply

@A Former Brilliant Member Ohkay, Thanks.

Md Zuhair - 4 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...