Best solutions to - "I am being asked to prove that∑k=0ncos⁡(kx)=12+sin⁡(2n+12x)2sin⁡(x2)"

Nyah Conrad

Answered question

2022-03-01

I am being asked to prove that

\sum_{k = 0}^{n} \cos (k x) = \frac{1}{2} + \frac{\sin (\frac{2 n + 1}{2} x)}{2 \sin (\frac{x}{2})}

benehmenshgf

Beginner2022-03-02Added 7 answers

\sum_{0 \leq r \leq n} e^{i k x} = \frac{e^{i (n + 1) x} - 1}{e^{i x} - 1}

= \frac{e^{\frac{i (n + 1) x}{2}}}{e^{\frac{i x}{2}}} \frac{(e^{\frac{i (n + 1) x}{2}} - e^{\frac{i (n + 1) x}{2}})}{(e^{\frac{i x}{2}} - e^{0 \frac{i x}{2}})}

= e^{\frac{\in x}{2}} \frac{2 i \sin \frac{(n + 1) x}{2}}{2 i \sin \frac{x}{2}}

e^{i y} - e^{- i y} = 2 i \sin y

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

using Euler's identity.
Its real part is

\cos \frac{n x}{2} \frac{\sin \frac{(n + 1) x}{2}}{\sin \frac{x}{2}} = \frac{2 \cos \frac{n x}{2} \sin \frac{(n + 1) x}{2}}{2 \sin \frac{x}{2}}

= \frac{\sin \frac{(2 n + 1) x}{2} + \sin \frac{x}{2}}{2 \sin \frac{x}{2}}

using

2 \sin A \cos B = \sin (A + B) + \sin (A - B)

Josef Beil

Beginner2022-03-03Added 12 answers

\sum_{k = 0}^{n} 2 \cos k θ = \sum_{k = 0}^{n} (e^{i θ k} + e^{- i θ k})

= \frac{e^{i θ (n + 1)} - 1}{e^{i θ} - 1} + \frac{e^{- i θ (n + 1)} - 1}{e^{- i θ} - 1}

= \frac{e^{\in θ} + e^{- \in θ} - e^{i (n + 1) θ} - e^{- i (n + 1) θ} - e^{i θ} - e^{- i θ} + 2}{2 - e^{i θ} - e^{- i θ}}

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

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

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

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