let-g-x-log-cosx-2sinx-developp-f-at-fourier-serie-

Question Number 144702 by mathmax by abdo last updated on 28/Jun/21

let g (x) = \log (cosx + 2 sinx)

developp f at fourier serie

Answered by mathmax by abdo last updated on 28/Jun/21

we have cosx + 2 sinx) = \sqrt{5} (\frac{1}{\sqrt{5}} cosx + \frac{2}{\sqrt{5}} sinx)

let \cos α = \frac{1}{\sqrt{5}} and \sin α = \frac{2}{\sqrt{5}} \Rightarrow \tan α = 2 \Rightarrow α = \arctan 2 \Rightarrow

cosx + 2 sinx = \sqrt{5} \cos (x - α) \Rightarrow g (x) = \frac{1}{2} \log 5 + \log (\cos (x - α))

the problem is turned to developp φ (t) = \log (cost)

we have φ (t) = \log (\frac{e^{it} + e^{- t}}{2}) = \log (e^{it} + e^{- it}) - \log (2)

= it + \log (1 + e^{- 2 it}) - \log 2

we [have \log^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} \Rightarrow

\log (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} u^{n + 1}}{n + 1} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{u^{n}}{n}

\Rightarrow \log (1 + e^{- 2 it}) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{e^{- 2 int}}{n}

\Rightarrow φ (t) = - \log 2 + it + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} (\cos (2 nt) + isin (2 nt))

= - \log 2 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \cos (2 nt) + i (t + Σ (\dots) sint)

φ (t) real \Rightarrow φ (t) = - \log 2 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \cos (2 nt) \Rightarrow

\log (cosx + 2 sinx) = \frac{1}{2} \log 5 + φ (x - α)

= \frac{1}{2} \log 5 - \log 2 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \cos (2 n (x - α))

= \frac{1}{2} \log 5 - \log 2 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \cos (2 nx - 2 narctan (2))