Prove that : C_n ^k = (1/(2π)) ∫^( π) _( −π) (2cos(θ/2))^n cos[((n/2)−k)θ]dθ


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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
	$A=(1/π)Re∫_0 ^π (e^(i(x/2)) +e^(−((ix)/2)) )^n e^(i((n/2)−k)x) dx,k∈[0,n] (e^((ix)/2) +e^(−((ix)/2)) )^n =Σ_(m=0) ^n C_n ^m e^((imx)/2) e^(−(i/2)(n−m)x) =ΣC_n ^m e^(ix(m−(n/2))) A=(1/π)ΣC_n ^m Re∫_0 ^π e^(ix(m−(n/2)+(n/2)−k)) dx =(1/π)ΣC_n ^m Re∫_0 ^π e^(ix(m−k)) dx ∫_0 ^π e^(i(m−k)x) dx= { (((((−1)^(m−k) −1)/(i(m−k))),m≠k)),((π,m=k)) :} A=(1/π)Re(Σ_(m≠k) (((−1)^(m−k) −1)/(i(m−k)))C_n ^m +πC_n ^k ) =(1/π).πC_n ^k =C_n ^k (1/(2π))∫_(−π) ^π (2cos((x/2)))^n cos([(n/2)−k]x)dx=C_n ^k$
	$A = \frac{1}{π} Re \int_{0}^{π} {(e^{i \frac{x}{2}} + e^{- \frac{ix}{2}})}^{n} e^{i (\frac{n}{2} - k) x} dx, k \in [0, n]$ ${(e^{\frac{ix}{2}} + e^{- \frac{ix}{2}})}^{n} = \underset{m = 0}{\sum^{n}} C_{n}^{m} e^{\frac{imx}{2}} e^{- \frac{i}{2} (n - m) x}$ $= Σ C_{n}^{m} e^{ix (m - \frac{n}{2})}$ $A = \frac{1}{π} Σ C_{n}^{m} Re \int_{0}^{π} e^{ix (m - \frac{n}{2} + \frac{n}{2} - k)} dx$ $= \frac{1}{π} Σ C_{n}^{m} Re \int_{0}^{π} e^{ix (m - k)} dx$ $\int_{0}^{π} e^{i (m - k) x} dx = {\begin{cases} \frac{{(- 1)}^{m - k} - 1}{i (m - k)}, m \neq k \\ π, m = k \end{cases}$ $A = \frac{1}{π} Re (\sum_{m \neq k} \frac{{(- 1)}^{m - k} - 1}{i (m - k)} C_{n}^{m} + π C_{n}^{k})$ $= \frac{1}{π} . π C_{n}^{k} = C_{n}^{k}$ $\frac{1}{2 π} \int_{- π}^{π} {(2 \cos (\frac{x}{2}))}^{n} \cos ([\frac{n}{2} - k] x) dx = C_{n}^{k}$
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