Prove-that-C-n-k-1-2pi-pi-pi-2cos-2-n-cos-n-2-k-d-

Question Number 192688 by Erico last updated on 24/May/23

Prove that :

C_{n}^{k} = \frac{1}{2 π} {\int_{- π}}^{π} {(2 \cos \frac{θ}{2})}^{n} \cos [(\frac{n}{2} - k) θ] d θ

Answered by witcher3 last updated on 24/May/23

A = \frac{1}{π} Re \int_{0}^{π} {(e^{i \frac{x}{2}} + e^{- \frac{ix}{2}})}^{n} e^{i (\frac{n}{2} - k) x} dx, k \in [0, n]

{(e^{\frac{ix}{2}} + e^{- \frac{ix}{2}})}^{n} = \underset{m = 0}{\sum^{n}} C_{n}^{m} e^{\frac{imx}{2}} e^{- \frac{i}{2} (n - m) x}

= Σ C_{n}^{m} e^{ix (m - \frac{n}{2})}

A = \frac{1}{π} Σ C_{n}^{m} Re \int_{0}^{π} e^{ix (m - \frac{n}{2} + \frac{n}{2} - k)} dx

= \frac{1}{π} Σ C_{n}^{m} Re \int_{0}^{π} e^{ix (m - k)} dx

\int_{0}^{π} e^{i (m - k) x} dx = {\begin{cases} \frac{{(- 1)}^{m - k} - 1}{i (m - k)}, m \neq k \\ π, m = k \end{cases}

A = \frac{1}{π} Re (\sum_{m \neq k} \frac{{(- 1)}^{m - k} - 1}{i (m - k)} C_{n}^{m} + π C_{n}^{k})

= \frac{1}{π} . π C_{n}^{k} = C_{n}^{k}

\frac{1}{2 π} \int_{- π}^{π} {(2 \cos (\frac{x}{2}))}^{n} \cos ([\frac{n}{2} - k] x) dx = C_{n}^{k}

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