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Question Number 192688 by Erico last updated on 24/May/23
Prove that :  C_n ^k  = (1/(2π)) ∫^(   π) _( −π) (2cos(θ/2))^n cos[((n/2)−k)θ]dθ
$$\mathrm{Prove}\:\mathrm{that}\:: \\ $$$$\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\underset{\:−\pi} {\int}^{\:\:\:\pi} \left(\mathrm{2cos}\frac{\theta}{\mathrm{2}}\right)^{\mathrm{n}} \mathrm{cos}\left[\left(\frac{\mathrm{n}}{\mathrm{2}}−\mathrm{k}\right)\theta\right]\mathrm{d}\theta \\ $$
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
$$\mathrm{A}=\frac{\mathrm{1}}{\pi}\mathrm{Re}\int_{\mathrm{0}} ^{\pi} \left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} +\mathrm{e}^{−\frac{\mathrm{ix}}{\mathrm{2}}} \right)^{\mathrm{n}} \mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{n}}{\mathrm{2}}−\mathrm{k}\right)\mathrm{x}} \mathrm{dx},\mathrm{k}\in\left[\mathrm{0},\mathrm{n}\right] \\ $$$$\left(\mathrm{e}^{\frac{\mathrm{ix}}{\mathrm{2}}} +\mathrm{e}^{−\frac{\mathrm{ix}}{\mathrm{2}}} \right)^{\mathrm{n}} =\underset{\mathrm{m}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{n}} ^{\mathrm{m}} \mathrm{e}^{\frac{\mathrm{imx}}{\mathrm{2}}} \mathrm{e}^{−\frac{\mathrm{i}}{\mathrm{2}}\left(\mathrm{n}−\mathrm{m}\right)\mathrm{x}} \\ $$$$=\Sigma\mathrm{C}_{\mathrm{n}} ^{\mathrm{m}} \mathrm{e}^{\mathrm{ix}\left(\mathrm{m}−\frac{\mathrm{n}}{\mathrm{2}}\right)} \\ $$$$\mathrm{A}=\frac{\mathrm{1}}{\pi}\Sigma\mathrm{C}_{\mathrm{n}} ^{\mathrm{m}} \mathrm{Re}\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{\mathrm{ix}\left(\mathrm{m}−\frac{\mathrm{n}}{\mathrm{2}}+\frac{\mathrm{n}}{\mathrm{2}}−\mathrm{k}\right)} \mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\pi}\Sigma\mathrm{C}_{\mathrm{n}} ^{\mathrm{m}} \mathrm{Re}\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{\mathrm{ix}\left(\mathrm{m}−\mathrm{k}\right)} \mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{\mathrm{i}\left(\mathrm{m}−\mathrm{k}\right)\mathrm{x}} \mathrm{dx}=\begin{cases}{\frac{\left(−\mathrm{1}\right)^{\mathrm{m}−\mathrm{k}} −\mathrm{1}}{\mathrm{i}\left(\mathrm{m}−\mathrm{k}\right)},\mathrm{m}\neq\mathrm{k}}\\{\pi,\mathrm{m}=\mathrm{k}}\end{cases} \\ $$$$\mathrm{A}=\frac{\mathrm{1}}{\pi}\mathrm{Re}\left(\underset{\mathrm{m}\neq\mathrm{k}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{m}−\mathrm{k}} −\mathrm{1}}{\mathrm{i}\left(\mathrm{m}−\mathrm{k}\right)}\mathrm{C}_{\mathrm{n}} ^{\mathrm{m}} +\pi\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \right) \\ $$$$=\frac{\mathrm{1}}{\pi}.\pi\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} =\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} \left(\mathrm{2cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)^{\mathrm{n}} \mathrm{cos}\left(\left[\frac{\mathrm{n}}{\mathrm{2}}−\mathrm{k}\right]\mathrm{x}\right)\mathrm{dx}=\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \\ $$$$ \\ $$

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