f(x)=arctan(2sinx) developp f at fourier serie


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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
	$we have f^′ (x)=((2cosx)/(1+4sin^2 x))=((2cosx)/(1+4.((1−cos(2x))/2)))=((2cosx)/(1+2−2cos(2x))) =((2cosx)/(3−2cos(2x))) =((2((e^(ix) +e^(−ix) )/2))/(3−2((e^(2ix) +e^(−2ix) )/2)))=((e^(ix) +e^(−ix) )/(3−e^(2ix) −e^(−2ix) )) =_(e^(ix) =z) ((z+z^(−1) )/(3−z^2 −z^(−2) ))=((z^2 (z+z^(−1) ))/(3z^2 −z^4 −1))=((z^3 +z)/(−z^4 +3z^2 −1))=−((z^3 +z)/(z^4 −3z^2 +1)) z^4 −3z^2 +1=0 ⇒u^2 −3u+1=0 (u=z^2 ) Δ=9−4=5 ⇒u_1 =((3+(√5))/2) and u_2 =((3−(√5))/2) ⇒ f^′ (x)=−((z^3 +z)/((z^2 −u_1 )(z^2 −u_2 )))=(z^3 +z)((1/(z^2 −u_1 ))−(1/(z^2 −u_2 )))×(1/( (√5))) =(1/( (√5)))(((z(z^2 −u_1 )+u_1 z+z)/(z^2 −u_1 ))−((z(z^2 −u_2 )+u_2 z+z)/(z^2 −u_2 ))) =(1/( (√5))){(((1+u_1 )z)/(z^2 −u_1 ))−(((1+u_2 )z)/(z^2 −u_2 ))} ∣(u_1 /z^2 )∣−1=((3+(√5))/2)−1>0 ⇒∣(u_1 /z_2 )∣>1 ∣(u_2 /z^2 )∣−1=((3−(√5))/2)−1<0 ⇒∣(u_2 /z^2 )∣<1 ⇒ f^′ (x)=(1/( (√5))){−((1+u_1 )/u_1 )×(z/(1−(z^2 /u_1 )))−((1+u_2 )/z)×(1/(1−(u_2 /z^2 )))} =−(((1+u_1 )z)/( (√5)u_1 ))Σ_(n=0) ^∞ (z^(2n) /u_1 ^n )−(((1+u_2 ))/(z(√5)))Σ_(n=0) ^∞ (u_2 ^n /z^(2n) ) =−((1+u_1 )/(u_1 (√5)))Σ_(n=0) ^∞ (z^(2n+1) /u_1 ^n )−(((1+u_2 ))/( (√5)))Σ_(n=0) ^∞ (u_2 ^n /z^(2n+1) ) =−((1+u_1 )/(u_1 (√5)))Σ_(n=0) ^∞ u_1 ^(−n) e^(i(2n+1)x) −((1+u_2 )/( (√5)))Σ_(n=0) ^∞ u_2 ^n e^(−i(2n+1)x) ⇒ f(x)=−((1+u_1 )/(u_1 (√(5(i(2n+1))))))Σ_(n=0) ^∞ u_1 ^(−n) e^(i(2n+1)x) +((1+u_2 )/( (√5)i(2n+1)))Σ_(n=0) ^∞ u_2 ^(−n) e^(−i(2n+1)x) +K ...be continued....$
	$we have f^{^{'}} (x) = \frac{2 cosx}{1 + {4 \sin}^{2} x} = \frac{2 cosx}{1 + 4 . \frac{1 - \cos (2 x)}{2}} = \frac{2 cosx}{1 + 2 - 2 \cos (2 x)}$ $= \frac{2 cosx}{3 - 2 \cos (2 x)} = \frac{2 \frac{e^{ix} + e^{- ix}}{2}}{3 - 2 \frac{e^{2 ix} + e^{- 2 ix}}{2}} = \frac{e^{ix} + e^{- ix}}{3 - e^{2 ix} - e^{- 2 ix}}$ $=_{e^{ix} = z} \frac{z + z^{- 1}}{3 - z^{2} - z^{- 2}} = \frac{z^{2} (z + z^{- 1})}{{3 z}^{2} - z^{4} - 1} = \frac{z^{3} + z}{- z^{4} + {3 z}^{2} - 1} = - \frac{z^{3} + z}{z^{4} - {3 z}^{2} + 1}$ $z^{4} - {3 z}^{2} + 1 = 0 \Rightarrow u^{2} - 3 u + 1 = 0 (u = z^{2})$ $Δ = 9 - 4 = 5 \Rightarrow u_{1} = \frac{3 + \sqrt{5}}{2} and u_{2} = \frac{3 - \sqrt{5}}{2} \Rightarrow$ $f^{^{'}} (x) = - \frac{z^{3} + z}{(z^{2} - u_{1}) (z^{2} - u_{2})} = (z^{3} + z) (\frac{1}{z^{2} - u_{1}} - \frac{1}{z^{2} - u_{2}}) \times \frac{1}{\sqrt{5}}$ $= \frac{1}{\sqrt{5}} (\frac{z (z^{2} - u_{1}) + u_{1} z + z}{z^{2} - u_{1}} - \frac{z (z^{2} - u_{2}) + u_{2} z + z}{z^{2} - u_{2}})$ $= \frac{1}{\sqrt{5}} {\frac{(1 + u_{1}) z}{z^{2} - u_{1}} - \frac{(1 + u_{2}) z}{z^{2} - u_{2}}}$ $∣ \frac{u_{1}}{z^{2}} ∣ - 1 = \frac{3 + \sqrt{5}}{2} - 1 > 0 \Rightarrow∣ \frac{u_{1}}{z_{2}} ∣> 1$ $∣ \frac{u_{2}}{z^{2}} ∣ - 1 = \frac{3 - \sqrt{5}}{2} - 1 < 0 \Rightarrow∣ \frac{u_{2}}{z^{2}} ∣< 1 \Rightarrow$ $f^{^{'}} (x) = \frac{1}{\sqrt{5}} {- \frac{1 + u_{1}}{u_{1}} \times \frac{z}{1 - \frac{z^{2}}{u_{1}}} - \frac{1 + u_{2}}{z} \times \frac{1}{1 - \frac{u_{2}}{z^{2}}}}$ $= - \frac{(1 + u_{1}) z}{\sqrt{5} u_{1}} \sum_{n = 0}^{\infty} \frac{z^{2 n}}{u_{1}^{n}} - \frac{(1 + u_{2})}{z \sqrt{5}} \sum_{n = 0}^{\infty} \frac{u_{2}^{n}}{z^{2 n}}$ $= - \frac{1 + u_{1}}{u_{1} \sqrt{5}} \sum_{n = 0}^{\infty} \frac{z^{2 n + 1}}{u_{1}^{n}} - \frac{(1 + u_{2})}{\sqrt{5}} \sum_{n = 0}^{\infty} \frac{u_{2}^{n}}{z^{2 n + 1}}$ $= - \frac{1 + u_{1}}{u_{1} \sqrt{5}} \sum_{n = 0}^{\infty} u_{1}^{- n} e^{i (2 n + 1) x} - \frac{1 + u_{2}}{\sqrt{5}} \sum_{n = 0}^{\infty} u_{2}^{n} e^{- i (2 n + 1) x} \Rightarrow$ $f (x) = - \frac{1 + u_{1}}{u_{1} \sqrt{5 (i (2 n + 1))}} \sum_{n = 0}^{\infty} u_{1}^{- n} e^{i (2 n + 1) x} + \frac{1 + u_{2}}{\sqrt{5} i (2 n + 1)} \sum_{n = 0}^{\infty} u_{2}^{- n} e^{- i (2 n + 1) x} + K$ $. . . be continued . . . .$
	Commented by mathmax by abdo last updated on 09/Jul/21

	$to simplify calculus we can use this decomposition$ $f^{^{'}} (x) = \frac{2 cosx}{3 - 2 \cos (2 x)} = \frac{2 cosx}{3 - 2 ({2 \cos}^{2} x - 1)} = \frac{2 cosx}{3 - {4 \cos}^{2} x + 2}$ $= - \frac{2 cosx}{{4 \cos}^{2} x - 5} = - \frac{2 cosx}{(2 cosx - \sqrt{5}) (2 cosx + \sqrt{5})}$ $= \frac{1}{2 \sqrt{5}} (\frac{1}{2 cosx - \sqrt{5}} - \frac{1}{2 cosx + \sqrt{5}}) = \frac{1}{2 \sqrt{5}} (u (x) - v (x))$ $u (x) = \frac{1}{2 cosx - \sqrt{5}} = \frac{1}{e^{ix} + e^{- ix} - \sqrt{5}} =_{e^{ix} = z} \frac{1}{z + z^{- 1} - \sqrt{5}}$ $= \frac{z}{z^{2} + 1 - \sqrt{5} z} = \frac{z}{z^{2} - \sqrt{5} z + 1}$ $Δ = 5 - 4 = 1 \Rightarrow z_{1} = \frac{\sqrt{5} + 1}{2} and z_{2} = \frac{\sqrt{5} - 1}{2}$ $u (x) = \frac{z}{(z - z_{1}) (z - z_{2})} = z (\frac{1}{z - z_{1}} - \frac{1}{z - z_{2}}) = \frac{z}{z - z_{1}} - \frac{z}{z - z_{2}}$ $∣ \frac{z}{z_{1}} ∣ - 1 = \frac{2}{\sqrt{5} + 1} - 1 = \frac{2 - \sqrt{5} - 1}{(. . .)} < 0 \Rightarrow∣ \frac{z}{z_{1}} ∣< 1$ $∣ \frac{z}{z_{2}} ∣ - 1 = \frac{2}{\sqrt{5} - 1} - 1 = \frac{2 - \sqrt{5} + 1}{\sqrt{5} - 1} = \frac{3 - \sqrt{5}}{(. .)} > 0 \Rightarrow∣ \frac{z}{z_{2}} ∣> 1 \Rightarrow$ $u (x) = - \frac{z}{z_{1} (1 - \frac{z}{z_{1}})} - \frac{1}{1 - \frac{z_{2}}{z}}$ $= - \frac{z}{z_{1}} \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}} - \sum_{n = 0}^{\infty} \frac{z_{2}^{n}}{z^{n}}$ $z_{1} z_{2} = 1 \Rightarrow z_{2}^{n} = \frac{1}{z_{1}^{n}} \Rightarrow u (x) = - \frac{z}{z_{1}} \sum_{n = 0}^{\infty} \frac{z^{n}}{z_{1}^{n}} - \sum_{n = 0}^{\infty} \frac{1}{z_{1}^{n} z^{n}}$ $= - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n - 1} e^{i (n + 1) x} - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} e^{- inx}$ $= - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n - 1} (\cos (n + 1) x + isin (n + 1) x)$ $- \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} (\cos (nx) - isin (nx))$ $u (x) real \Rightarrow u (x) = - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- (n + 1)} \cos (n + 1) x$ $- \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} \cos (nx)$ $= - \sum_{n = 1}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} \cos (nx) - \sum_{n = 0}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} \cos (nx)$ $= - 1 - 2 \sum_{n = 1}^{\infty} {(\frac{\sqrt{5} + 1}{2})}^{- n} \cos (nx)$ $. . . be continued . . . .$
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