calculate Σ_(k=0) ^n C_n ^k cos(((kπ)/(2n)))


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Question Number 72020 by mathmax by abdo last updated on 23/Oct/19

$c a l c u l a t e \sum_{k = 0}^{n} C_{n}^{k} c o s (\frac{k π}{2 n})$
Commented by mathmax by abdo last updated on 24/Oct/19
$let A_n =Σ_(k=0) ^n C_n ^k cos(((kπ)/(2n))) ⇒A_n =Re(Σ_(k=0) ^n C_n ^k e^(i((kπ)/(2n))) ) but Σ_(k=0) ^n C_n ^k e^(i((kπ)/(2n))) =Σ_(k=0) ^n C_n ^k (e^((iπ)/(2n)) )^k =(1+e^((iπ)/(2n)) )^n =(1+cos((π/(2n)))+isin((π/(2n))))^n ={2cos^2 ((π/(4n)))+2isin((π/(4n)))cos((π/(4n)))}^n =2^n cos^n ((π/(4n))){e^((iπ)/(4n)) }^n =2^n e^((iπ)/4) cos^n ((π/(4n))) =2^n cos^n ((π/(4n))){((√2)/2) +i((√2)/2)} ⇒A_n =(√2)2^(n−1) cos^n ((π/(4n))) ⇒ A_n =2^(n−(1/2)) cos^n ((π/(4n)))$
$l e t A_{n} = \sum_{k = 0}^{n} C_{n}^{k} c o s (\frac{k π}{2 n}) \Rightarrow A_{n} = R e (\sum_{k = 0}^{n} C_{n}^{k} e^{i \frac{k π}{2 n}}) b u t$ $\sum_{k = 0}^{n} C_{n}^{k} e^{i \frac{k π}{2 n}} = \sum_{k = 0}^{n} C_{n}^{k} {(e^{\frac{i π}{2 n}})}^{k} = {(1 + e^{\frac{i π}{2 n}})}^{n}$ $= {(1 + c o s (\frac{π}{2 n}) + i s i n (\frac{π}{2 n}))}^{n} = {2 {c o s}^{2} (\frac{π}{4 n}) + 2 i s i n (\frac{π}{4 n}) c o s (\frac{π}{4 n})}^{n}$ $= 2^{n} {c o s}^{n} (\frac{π}{4 n}) {e^{\frac{i π}{4 n}}}^{n} = 2^{n} e^{\frac{i π}{4}} {c o s}^{n} (\frac{π}{4 n})$ $= 2^{n} {c o s}^{n} (\frac{π}{4 n}) {\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} \Rightarrow A_{n} = \sqrt{2} 2^{n - 1} {c o s}^{n} (\frac{π}{4 n}) \Rightarrow$ $A_{n} = 2^{n - \frac{1}{2}} {c o s}^{n} (\frac{π}{4 n})$
	Answered by mind is power last updated on 23/Oct/19

	$= Re \sum_{k = 0}^{n} C_{n}^{k} e^{\frac{i π k}{2 n}} = Re {(1 + e^{\frac{i π}{2 n}})}^{n}$ $1 + e^{\frac{i π}{2 n}} = e^{\frac{i π}{4 n}} (e^{\frac{i π}{4 n}} + e^{\frac{- i π}{4 n}}) = 2 \cos (\frac{π}{4 n}) e^{i \frac{π}{4 n}}$ ${(1 + e^{\frac{i π}{2 n}})}^{n} = 2^{n} \cos^{n} (\frac{π}{4 n}) e^{i \frac{π}{4}}$ $= Re \sum_{k = 0}^{n} C_{n}^{k} e^{\frac{i π k}{2 n}} = Re {(1 + e^{\frac{i π}{2 n}})}^{n} = 2^{n} \cos (\frac{π}{4 n}) . \frac{\sqrt{2}}{2}$
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