developp-at-fourier-serie-g-x-2-3-sin-2-x-

Question Number 98185 by abdomathmax last updated on 12/Jun/20

developp at fourier serie

g (x) = \frac{2}{3 + \sin^{2} x}

Answered by mathmax by abdo last updated on 12/Jun/20

g (x) = \frac{2}{3 + \sin^{2} x} = \frac{2}{3 + \frac{1 - \cos (2 x)}{2}} = \frac{4}{7 - \cos (2 x)} = \frac{4}{7 - \frac{e^{i 2 x} + e^{- i 2 x}}{2}}

= \frac{8}{14 - e^{2 ix} - e^{- 2 ix}} =_{e^{2 ix} = z} \frac{8}{14 - z - z^{- 1}} = \frac{8 z}{14 z - z^{2} - 1} = \frac{- 8 z}{z^{2} - 14 z + 1} = h (z)

z^{2} - 14 z + 1 = 0 \to Δ^{^{'}} = 7^{2} - 1 = 48 \Rightarrow z_{1} = 7 + \sqrt{48} = 7 + 4 \sqrt{3}

z_{2} = 7 - 4 \sqrt{3} so \frac{- 8 z}{z^{2} - 14 z + 1} = \frac{- 8 z}{(z - z_{1}) (z - z_{2})} = \frac{a}{z - z_{1}} + \frac{b}{z - z_{2}}

a = \frac{- {8 z}_{1}}{8 \sqrt{3}} = - \frac{7 + 4 \sqrt{3}}{\sqrt{3}} = - \frac{7}{\sqrt{3}} + 4

b = \frac{- {8 z}_{2}}{- 8 \sqrt{3}} = \frac{7 - 4 \sqrt{3}}{\sqrt{3}} = \frac{7}{\sqrt{3}} - 4 \Rightarrow h (z) = \frac{- a}{z_{1} - z} - \frac{b}{z_{2} - z} = \frac{- a}{z_{1} (1 - \frac{z}{z_{1}})} - \frac{b}{z_{2} (1 - \frac{z}{z_{2}})}

so for ∣ z ∣ < \inf (∣ z_{1} ∣, ∣ z_{2} ∣) h (z) = - \frac{a}{z_{1}} \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}} - \frac{b}{z_{2}} \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{2}^{n}}

= - \frac{a}{z_{1}} \sum_{n = 0}^{\infty} \frac{1}{z_{1}^{n}} e^{2 inx} - \frac{b}{z_{2}} \sum_{n = 0}^{\infty} \frac{1}{z_{2}^{n}} e^{2 inx}

= - \sum_{n = 0}^{\infty} (\frac{a}{z_{1}^{n + 1}} + \frac{b}{z_{2}^{n + 1}}) e^{2 inx} = g (x)