Question-82520

Question Number 82520 by peter frank last updated on 22/Feb/20

Commented by mathmax by abdo last updated on 22/Feb/20

l e t S_{n} = \sum_{k = 0}^{n} 2^{k} c o s (2 k θ) \Rightarrow S_{n} = R e (\sum_{k = 0}^{n} 2^{k} e^{i 2 k θ})

= R e (\sum_{k = 0}^{n} {(2 e^{i 2 θ})}^{k}) b u t w e h a v e

\sum_{k = 0}^{n} {(2 e^{2 i θ})}^{k} = \frac{1 - {(2 e^{2 i θ})}^{n + 1}}{1 - 2 e^{2 i θ}} = \frac{1 - 2^{n + 1} e^{2 (n + 1) i θ}}{1 - 2 c o s (2 θ) - 2 i s i n (2 θ)}

= \frac{(1 - 2 c o s (2 θ) + 2 i s i n (2 θ)) (1 - 2^{n + 1} e^{2 (n + 1) i θ})}{{(1 - 2 c o s (2 θ))}^{2} + 4 {s i n}^{2} (2 θ)}

= \frac{(1 - 2 c o s (2 θ) + 2 i s i n (2 θ)) {1 - 2^{n + 1} c o s 2 (n + 1) θ - 2^{n + 1} i s i n 2 (n + 1) θ}}{{(1 - 2 c o s (2 θ))}^{2} + 4 {s i n}^{2} (2 θ)}

= \frac{(1 - 2 c o s (2 θ)) (1 - 2^{n + 1} c o s (2 (n + 1) θ) - i (1 - 2 c o s (2 θ)) 2^{n + 1} s i n 2 (n + 1) θ + 2 i s i n (2 θ) (1 - 2^{n + 1} c o s 2 (n + 1) θ) + 2^{n + 2} s i n (2 θ) s i n 2 (n + 1) θ}{{(1 - 2 c o s (2 θ))}^{2} + 4 {s i n}^{2} (2 θ)}

\Rightarrow S_{n} = \frac{(1 - 2 c o s (2 θ)) (1 - 2^{n + 1} c o s (2 (n + 1) θ) + 2^{n + 2} s i n (2 θ) s i n 2 (n + 1) θ}{5 - 4 c o s (2 θ)}

Answered by mr W last updated on 22/Feb/20

z = 2 (\cos θ + i \sin θ) = 2 e^{i θ}

z^{k} = 2^{k} (\cos k θ + i \sin k θ) = {(2 e^{i θ})}^{k}

z^{0} + z^{1} + z^{2} + \dots + z^{n} = \underset{k = 0}{\sum^{n}} {(2 e^{i θ})}^{k} \leftarrow s u m o f G P

\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{{(2 e^{i θ})}^{n + 1} - 1}{2 e^{i θ} - 1}

\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{2^{n + 1} e^{i (n + 1) θ} - 1}{2 e^{i θ} - 1}

\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{2^{n + 1} [\cos (n + 1) θ + i \sin (n + 1) θ] - 1}{2 (\cos θ + i \sin θ) - 1}

\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{{[2^{n + 1} \cos (n + 1) θ - 1] + i [2^{n + 1} \sin (n + 1) θ]} [(2 \cos θ - 1) - i \sin θ]}{[(2 \cos θ - 1) + i \sin θ] [(2 \cos θ - 1) - i \sin θ]}

\underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) + i \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{{[2^{n + 1} \cos (n + 1) θ - 1] (2 \cos θ - 1) + 2^{n + 1} \sin (n + 1) θ \sin θ} + i {[2^{n + 1} \sin (n + 1) θ] (2 \cos θ - 1) - [2^{n + 1} \cos (n + 1) θ - 1] \sin θ}}{{(2 \cos θ - 1)}^{2} + \sin^{2} θ}

\Rightarrow \underset{k = 0}{\sum^{n}} 2^{k} \cos (k θ) = \frac{[2^{n + 1} \cos (n + 1) θ - 1] (2 \cos θ - 1) + 2^{n + 1} \sin θ \sin (n + 1) θ}{{(2 \cos θ - 1)}^{2} + \sin^{2} θ}

\Rightarrow \underset{k = 0}{\sum^{n}} 2^{k} \sin (k θ) = \frac{2^{n + 1} (2 \cos θ - 1) \sin (n + 1) θ - [2^{n + 1} \cos (n + 1) θ - 1] \sin θ}{{(2 \cos θ - 1)}^{2} + \sin^{2} θ}

Commented by peter frank last updated on 22/Feb/20

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