f-x-2-1-sinx-2-developp-f-at-fourier-serie-

Question Number 144464 by mathmax by abdo last updated on 25/Jun/21

f (x) = \frac{2}{{(1 + sinx)}^{2}}

developp f at fourier serie

Answered by mathmax by abdo last updated on 26/Jun/21

let f (a) = \frac{2}{a + sinx} \Rightarrow f^{^{'}} (a) = - \frac{2}{{(a + sinx)}^{2}} \Rightarrow - f^{^{'}} (1) = \frac{2}{{(1 + sinx)}^{2}}

we have f (a) = \frac{2}{a + \frac{e^{ix} - e^{- ix}}{2 i}} = \frac{4 i}{2 ia + e^{ix} - e^{- ix}}

=_{e^{ix} = z} \frac{4 i}{2 ia + z - z^{- 1}} = \frac{4 iz}{2 iaz + z^{2} - 1} = \frac{4 iz}{z^{2} + 2 iaz - 1}

Δ^{^{'}} = - a^{2} + 1 = 1 - a^{2} \Rightarrow z_{1} = - ia + \sqrt{1 - a^{2}}

z_{2} = - ia - \sqrt{1 - a^{2}} \Rightarrow f (a) = \frac{4 iz}{(z - z_{1}) (z - z_{2})}

= 4 iz (\frac{1}{z - z_{1}} - \frac{1}{z - z_{2}}) . \frac{1}{z_{1} - z_{2}} = \frac{4 i}{2 \sqrt{1 - a^{2}}} (\frac{z}{z - z_{1}} - \frac{z}{z - z_{2}})

∣ \frac{z}{z_{1}} ∣= 1 and ∣ \frac{z}{z_{2}} ∣= 1 \Rightarrow f (a) = \frac{2 i}{\sqrt{1 - a^{2}}} (\frac{z}{z_{1} (\frac{z}{z_{1}} - 1)} - \frac{z}{z_{2} (\frac{z}{z_{2}} - 1)})

= \frac{2 iz}{\sqrt{1 - a^{2}}} (\frac{1}{1 - \frac{z}{z_{2}}} - \frac{1}{1 - \frac{z}{z_{1}}}) = \frac{2 iz}{\sqrt{1 - a^{2}}} (\sum_{n = 0}^{\infty} \frac{z^{n}}{z_{2}^{n}} - \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}})

z_{1} = e^{iarctan (- \frac{a}{\sqrt{1 - a^{2}}})} and z_{2} = e^{iarctan (\frac{a}{\sqrt{1 - a^{2}}})} \Rightarrow

\Rightarrow \frac{1}{z_{1}^{n}} = e^{inarctan (\frac{a}{\sqrt{1 - a^{2}}})} + e^{- inarctan (\frac{a}{\sqrt{1 - a^{2}}})} = 2 Re (\dots .)

= 2 \cos (narctan (\frac{a}{\sqrt{1 - a^{2}}})) \Rightarrow

f (a) = \frac{2 iz}{\sqrt{1 - a^{2}}} \sum_{n = 0}^{\infty} 2 \cos (narctan (\frac{a}{\sqrt{1 - a^{2}}})) e^{inx}

= \frac{4 i}{\sqrt{1 - a^{2}}} \sum_{n = 0}^{\infty} \cos (narctan (\frac{a}{\sqrt{1 - a^{2}}})) (\cos (n + 1) x + isin (n + 1) x)

= 4 i (\dots .) - \frac{4}{\sqrt{1 - a^{2}}} \sum_{n = 0}^{\infty} \cos (narctan (\frac{a}{\sqrt{1 - a^{2}}})) \sin (n + 1) x

f (a) is reaal \Rightarrow

f (a) = \frac{- 4}{\sqrt{1 - a^{2}}} \sum_{n = 0}^{\infty} \cos (narctan (\frac{a}{\sqrt{1 - a^{2}}})) \sin (n + 1) x

rest to calculate f^{^{'}} (a) \dots . be continued