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

Question Number 145748 by mathmax by abdo last updated on 07/Jul/21

f (x) = \arctan (2 sinx)

developp f at fourier serie

Answered by mathmax by abdo last updated on 09/Jul/21

we have f^{^{'}} (x) = \frac{2 cosx}{1 + {4 \sin}^{2} x} = \frac{2 cosx}{1 + 4 . \frac{1 - \cos (2 x)}{2}} = \frac{2 cosx}{1 + 2 - 2 \cos (2 x)}

= \frac{2 cosx}{3 - 2 \cos (2 x)} = \frac{2 \frac{e^{ix} + e^{- ix}}{2}}{3 - 2 \frac{e^{2 ix} + e^{- 2 ix}}{2}} = \frac{e^{ix} + e^{- ix}}{3 - e^{2 ix} - e^{- 2 ix}}

=_{e^{ix} = z} \frac{z + z^{- 1}}{3 - z^{2} - z^{- 2}} = \frac{z^{2} (z + z^{- 1})}{{3 z}^{2} - z^{4} - 1} = \frac{z^{3} + z}{- z^{4} + {3 z}^{2} - 1} = - \frac{z^{3} + z}{z^{4} - {3 z}^{2} + 1}

z^{4} - {3 z}^{2} + 1 = 0 \Rightarrow u^{2} - 3 u + 1 = 0 (u = z^{2})

Δ = 9 - 4 = 5 \Rightarrow u_{1} = \frac{3 + \sqrt{5}}{2} and u_{2} = \frac{3 - \sqrt{5}}{2} \Rightarrow

f^{^{'}} (x) = - \frac{z^{3} + z}{(z^{2} - u_{1}) (z^{2} - u_{2})} = (z^{3} + z) (\frac{1}{z^{2} - u_{1}} - \frac{1}{z^{2} - u_{2}}) \times \frac{1}{\sqrt{5}}

= \frac{1}{\sqrt{5}} (\frac{z (z^{2} - u_{1}) + u_{1} z + z}{z^{2} - u_{1}} - \frac{z (z^{2} - u_{2}) + u_{2} z + z}{z^{2} - u_{2}})

= \frac{1}{\sqrt{5}} {\frac{(1 + u_{1}) z}{z^{2} - u_{1}} - \frac{(1 + u_{2}) z}{z^{2} - u_{2}}}

∣ \frac{u_{1}}{z^{2}} ∣ - 1 = \frac{3 + \sqrt{5}}{2} - 1 > 0 \Rightarrow∣ \frac{u_{1}}{z_{2}} ∣> 1

∣ \frac{u_{2}}{z^{2}} ∣ - 1 = \frac{3 - \sqrt{5}}{2} - 1 < 0 \Rightarrow∣ \frac{u_{2}}{z^{2}} ∣< 1 \Rightarrow

f^{^{'}} (x) = \frac{1}{\sqrt{5}} {- \frac{1 + u_{1}}{u_{1}} \times \frac{z}{1 - \frac{z^{2}}{u_{1}}} - \frac{1 + u_{2}}{z} \times \frac{1}{1 - \frac{u_{2}}{z^{2}}}}

= - \frac{(1 + u_{1}) z}{\sqrt{5} u_{1}} \sum_{n = 0}^{\infty} \frac{z^{2 n}}{u_{1}^{n}} - \frac{(1 + u_{2})}{z \sqrt{5}} \sum_{n = 0}^{\infty} \frac{u_{2}^{n}}{z^{2 n}}

= - \frac{1 + u_{1}}{u_{1} \sqrt{5}} \sum_{n = 0}^{\infty} \frac{z^{2 n + 1}}{u_{1}^{n}} - \frac{(1 + u_{2})}{\sqrt{5}} \sum_{n = 0}^{\infty} \frac{u_{2}^{n}}{z^{2 n + 1}}

= - \frac{1 + u_{1}}{u_{1} \sqrt{5}} \sum_{n = 0}^{\infty} u_{1}^{- n} e^{i (2 n + 1) x} - \frac{1 + u_{2}}{\sqrt{5}} \sum_{n = 0}^{\infty} u_{2}^{n} e^{- i (2 n + 1) x} \Rightarrow

f (x) = - \frac{1 + u_{1}}{u_{1} \sqrt{5 (i (2 n + 1))}} \sum_{n = 0}^{\infty} u_{1}^{- n} e^{i (2 n + 1) x} + \frac{1 + u_{2}}{\sqrt{5} i (2 n + 1)} \sum_{n = 0}^{\infty} u_{2}^{- n} e^{- i (2 n + 1) x} + K

\dots be continued \dots .

Commented by mathmax by abdo last updated on 09/Jul/21

to simplify calculus we can use this decomposition

f^{^{'}} (x) = \frac{2 cosx}{3 - 2 \cos (2 x)} = \frac{2 cosx}{3 - 2 ({2 \cos}^{2} x - 1)} = \frac{2 cosx}{3 - {4 \cos}^{2} x + 2}

= - \frac{2 cosx}{{4 \cos}^{2} x - 5} = - \frac{2 cosx}{(2 cosx - \sqrt{5}) (2 cosx + \sqrt{5})}

= \frac{1}{2 \sqrt{5}} (\frac{1}{2 cosx - \sqrt{5}} - \frac{1}{2 cosx + \sqrt{5}}) = \frac{1}{2 \sqrt{5}} (u (x) - v (x))

u (x) = \frac{1}{2 cosx - \sqrt{5}} = \frac{1}{e^{ix} + e^{- ix} - \sqrt{5}} =_{e^{ix} = z} \frac{1}{z + z^{- 1} - \sqrt{5}}

= \frac{z}{z^{2} + 1 - \sqrt{5} z} = \frac{z}{z^{2} - \sqrt{5} z + 1}

Δ = 5 - 4 = 1 \Rightarrow z_{1} = \frac{\sqrt{5} + 1}{2} and z_{2} = \frac{\sqrt{5} - 1}{2}

u (x) = \frac{z}{(z - z_{1}) (z - z_{2})} = z (\frac{1}{z - z_{1}} - \frac{1}{z - z_{2}}) = \frac{z}{z - z_{1}} - \frac{z}{z - z_{2}}

∣ \frac{z}{z_{1}} ∣ - 1 = \frac{2}{\sqrt{5} + 1} - 1 = \frac{2 - \sqrt{5} - 1}{(\dots)} < 0 \Rightarrow∣ \frac{z}{z_{1}} ∣< 1

∣ \frac{z}{z_{2}} ∣ - 1 = \frac{2}{\sqrt{5} - 1} - 1 = \frac{2 - \sqrt{5} + 1}{\sqrt{5} - 1} = \frac{3 - \sqrt{5}}{(. .)} > 0 \Rightarrow∣ \frac{z}{z_{2}} ∣> 1 \Rightarrow

u (x) = - \frac{z}{z_{1} (1 - \frac{z}{z_{1}})} - \frac{1}{1 - \frac{z_{2}}{z}}

= - \frac{z}{z_{1}} \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}} - \sum_{n = 0}^{\infty} \frac{z_{2}^{n}}{z^{n}}

z_{1} z_{2} = 1 \Rightarrow z_{2}^{n} = \frac{1}{z_{1}^{n}} \Rightarrow u (x) = - \frac{z}{z_{1}} \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}} - \sum_{n = 0}^{\infty} \frac{1}{z_{1}^{n} z^{n}}

= - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n - 1} e^{i (n + 1) x} - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} e^{- inx}

= - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n - 1} (\cos (n + 1) x + isin (n + 1) x)

- \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} (\cos (nx) - isin (nx))

u (x) real \Rightarrow u (x) = - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- (n + 1)} \cos (n + 1) x

- \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} \cos (nx)

= - \sum_{n = 1}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} \cos (nx) - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} \cos (nx)

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

\dots be continued \dots .