let-f-x-ln-2-cosx-developp-f-at-fourier-serie-

Question Number 96655 by mathmax by abdo last updated on 03/Jun/20

let f (x) = \ln (2 + cosx) developp f at fourier serie

Answered by mathmax by abdo last updated on 04/Jun/20

f (x) = \ln (2 + cosx) \Rightarrow f^{^{'}} (x) = \frac{- sinx}{2 + cosx} = - \frac{\frac{e^{ix} - e^{- ix}}{2 i}}{2 + \frac{e^{ix} + e^{- ix}}{2}}

= - \frac{1}{i} \times \frac{e^{ix} - e^{- ix}}{4 + e^{ix} + e^{- ix}} changement e^{ix} = z give

f^{^{'}} (x) = i \times \frac{z - z^{- 1}}{4 + z + z^{- 1}} = i \times \frac{z^{2} - 1}{4 z + z^{2} + 1} = i \times \frac{z^{2} - 1}{z^{2} + 4 z + 1} let decompose

u (z) = \frac{z^{2} - 1}{z^{2} + 4 z + 1}

z^{2} + 4 z + 1 = 0 \to Δ^{^{'}} = 4 - 1 = 3 \Rightarrow z_{1} = - 2 + \sqrt{3} and z_{2} = - 2 - \sqrt{3}

u (z) = \frac{z^{2} + 4 z + 1 - 4 z - 2}{z^{2} + 4 z + 1} = 1 - \frac{4 z + 2}{z^{2} + 4 z + 1} = 1 - \frac{4 z + 2}{(z - z_{1}) (z - z_{2})} = 1 - \frac{a}{z - z_{1}} - \frac{b}{z - z_{2}}

a = \frac{{4 z}_{1} + 2}{z_{1} - z_{2}} = \frac{- 8 + 4 \sqrt{3} + 2}{2 \sqrt{3}} = \frac{- 6 + 4 \sqrt{3}}{2 \sqrt{3}} = \frac{- 3 + 2 \sqrt{3}}{\sqrt{3}} = - \sqrt{3} + 2

b = \frac{{4 z}_{2} + 2}{- 2 \sqrt{3}} = \frac{- 8 - 4 \sqrt{3} + 2}{- 2 \sqrt{3}} = \frac{- 6 - 4 \sqrt{3}}{- 2 \sqrt{3}} = \frac{3 + 2 \sqrt{3}}{\sqrt{3}} = \sqrt{3} + 2 \Rightarrow

f^{^{'}} (x) = i {1 - \frac{2 - \sqrt{3}}{z - z_{1}} - \frac{2 + \sqrt{3}}{z - z_{2}}} = i - (2 - \sqrt{3}) i \times \frac{1}{z - z_{1}} - (2 + \sqrt{3}) i \times \frac{1}{z - z_{2}}

= i + \frac{(2 - \sqrt{3}) i}{z_{1} - z} + \frac{(2 + \sqrt{3}) i}{z_{2} - z} = i + \frac{(2 - \sqrt{3}) i}{z_{1}} \times \frac{1}{1 - \frac{z}{z_{1}}} + \frac{(2 + \sqrt{3}) i}{z_{2}} \times \frac{1}{1 - \frac{z}{z_{2}}}

= i - i \times \frac{1}{1 - \frac{z}{z_{1}}} - i \times \frac{1}{1 - \frac{z}{z_{2}}} so if ∣ z ∣< \inf (∣ z_{1} ∣, ∣ z_{2} ∣) we get

f^{^{'}} (x) = i - i \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}} - i \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{2}^{n}} = i - i \sum_{n = 0}^{\infty} (\frac{1}{z_{1}^{n}} + \frac{1}{z_{2}^{n}}) z^{n}

= i - i \sum_{n = 0}^{\infty} (\frac{z_{1}^{n} + z_{2}^{n}}{{(z_{1} z_{2})}^{n}}) z^{n} = i - i \sum_{n = 0}^{\infty} {\frac{{(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n}}{- 1}} z^{n}

= i + i \sum_{n = 0}^{\infty} {{(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n}} z^{n} \Rightarrow

= i + i \sum_{n = 0}^{\infty} {{(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n}} e^{inx} \Rightarrow

= i + i \sum_{n =}^{\infty} {{(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n}} (\cos (nx) + i \sin (nx)}

f^{^{'}} (x) real \Rightarrow f^{^{'}} (x) = - \sum_{n = 0}^{\infty} {{(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n}} \sin (nx) \Rightarrow

f (x) = \sum_{n = 0}^{\infty} \frac{({(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n}) \cos (nx)}{n} + c

f (π) = 0 = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n} {{(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n}} + c \Rightarrow

c = - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n} ({(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n} \Rightarrow

f (x) = \sum_{n = 0}^{\infty} \frac{{{(- 2 + \sqrt{3})}^{n} + {(- 2 - \sqrt{3})}^{n}}{n} (\cos (nx) - {(- 1)}^{n}

rest to study the case ∣ z_{1} ∣<∣ z ∣<∣ z_{2} ∣ .