Question: "I'm having trouble proving(sin⁡(nθ2)sin⁡(θ2))2=|∑k=1|n|eikθ|2where n∈Z and θ∈R."

Taylor Haines

Answered question

2022-03-01

I'm having trouble proving

{(\frac{\sin (\frac{n θ}{2})}{\sin (\frac{θ}{2})})}^{2} = {| \sum_{k = 1}^{| n |} e^{i k θ} |}^{2}

where

n \in Z

and

θ \in R

lucratifar1

Beginner2022-03-02Added 5 answers

Using Euler's Formula ,

e^{i x} = \cos x + i \sin x

So,

e^{- i x} = \cos (- x) + i \sin (- x) = \cos x - i \sin x \Rightarrow 2 i \sin x

= e^{i x} - e^{- i x}

n > 0, \sum_{k = 1}^{| n |} e^{i k θ} = \sum_{k = 1}^{n} e^{i k θ} = e^{i θ} (\frac{e^{\in θ} - 1}{e^{i θ} - 1})

= \frac{e^{i n \frac{θ}{2}}}{e^{i \frac{θ}{2}}} \frac{(e^{i n \frac{θ}{2} - e^{- i n \frac{θ}{2}}}} {(e^{i \frac{θ}{2}} - e^{- i \frac{θ}{2}})}}{}

= (\cos \frac{(n - 1) θ}{2} + i \sin \frac{(n - 1) θ}{2}) \frac{\sin \frac{n θ}{2}}{\sin \frac{θ}{2}}

Taking modulus

∣ \sum_{k = 1}^{| n |} e^{i k θ}

= ∣ (\cos \frac{(n - 1) θ}{2} + i \sin \frac{(n - 1) θ}{2}) \frac{\sin \frac{n θ}{2}}{\sin \frac{θ}{2}}

= | \frac{\sin \frac{n θ}{2}}{\sin \frac{θ}{2}} |

Similiarly, for n<0

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