let f(x) =ln(1+sinx) developp f at fourier serie


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Question Number 96955 by mathmax by abdo last updated on 05/Jun/20

$let f (x) = \ln (1 + sinx)$ $developp f at fourier serie$
	Answered by mathmax by abdo last updated on 06/Jun/20
	$f(x) =ln(1+sinx) ⇒f^′ (x) =((cosx)/(1+sinx)) =(((e^(ix) +e^(−ix) )/2)/(1+((e^(ix) −e^(−ix) )/(2i)))) =i((e^(ix) +e^(−ix) )/(2i +e^(ix) −e^(−ix) )) =_(e^(ix) =z) i((z+z^(−1) )/(2i+z−z^(−1) )) =i×((z^2 +1)/(2iz +z^2 −1)) =i×((z^2 +1)/(z^2 +2iz −1)) =i ((z^2 +1)/((z+i)^2 )) =i ((z^2 +1)/(−(1−iz)^2 )) =−i((z^2 +1)/((1−iz)^2 )) we have (1/(1−iz)) =Σ_(n=0) ^∞ (iz)^n =Σ_(n=0) ^∞ i^n z^n ⇒(i/((1−iz)^2 )) =Σ_(n=1) ^∞ ni^n z^(n−1) ⇒ f^′ (x) =−i(z^2 +1)Σ_(n=1) ^∞ ni^n z^(n−1) =−(z^2 +1)Σ_(n=1) ^∞ n i^(n+1) z^(n−1) =−Σ_(n=1) ^∞ ni^(n+1) z^(n+1) −Σ_(n=1) ^∞ ni^(n+1) z^(n−1) =−Σ_(n=1) ^∞ ni^(n+1) e^(i(n+1)x) −Σ_(n=1) ^∞ ni^(n+1) e^(i(n−1)x) =−Σ_(n=1) ^∞ ni^(n+1) {cos(n+1)x +isin(n+1)x}−Σ_(n=1) ^∞ ni^(n+1) {cos(n−1)x+isin(n−1)x} =−Σ_(n=1) ^∞ n i^(n+1) ( cos(n+1)x+cos(n−1)x)+Σ_(n=1) ^∞ ni^n (sin(n+1)x+sin(n−1)x) =−Σ_(n=1) ^∞ (2n)(−1)^n i(cos(2n+1)x+cos(2n−1)x) +Σ_(n=0) ^∞ (2n+1)(−1)^n (cos(2n+2)x +cos(2nx))+Σ_(n=1) ^∞ (2n)(−1)^n (sin(2n+1)x+sin(2n−1)x) +Σ_(n=0) ^∞ (2n+1)(−1)^n i{sin(2n+2)x +sin(2n)x} f^′ (x) is tlreal ⇒f^′ (x)=Σ_(n=0) ^∞ (2n+1)(−1)^n {cos(2n+2)x +cos(2nx)} +Σ_(n=1) ^∞ (2n)(−1)^n {sin(2n+1)x) +sin(2n−1)x} ⇒ f(x) =Σ_(n=0) ^∞ (2n+1)(−1)^n {(1/(2n+2))sin(2n+2)x+(1/(2n))sin(2nx)} +Σ_(n=1) ^∞ (2n)(−1)^n {−(1/(2n+1)) cos(2n+1)x −(1/(2n−1)) cos(2n−1)x} +C rest to find C....be continued....$
	$f (x) = \ln (1 + sinx) \Rightarrow f^{^{'}} (x) = \frac{cosx}{1 + sinx} = \frac{\frac{e^{ix} + e^{- ix}}{2}}{1 + \frac{e^{ix} - e^{- ix}}{2 i}}$ $= i \frac{e^{ix} + e^{- ix}}{2 i + e^{ix} - e^{- ix}} =_{e^{ix} = z} i \frac{z + z^{- 1}}{2 i + z - z^{- 1}} = i \times \frac{z^{2} + 1}{2 iz + z^{2} - 1}$ $= i \times \frac{z^{2} + 1}{z^{2} + 2 iz - 1} = i \frac{z^{2} + 1}{{(z + i)}^{2}} = i \frac{z^{2} + 1}{- {(1 - iz)}^{2}} = - i \frac{z^{2} + 1}{{(1 - iz)}^{2}} we have$ $\frac{1}{1 - iz} = \sum_{n = 0}^{\infty} {(iz)}^{n} = \sum_{n = 0}^{\infty} i^{n} z^{n} \Rightarrow \frac{i}{{(1 - iz)}^{2}} = \sum_{n = 1}^{\infty} {ni}^{n} z^{n - 1} \Rightarrow$ $f^{^{'}} (x) = - i (z^{2} + 1) \sum_{n = 1}^{\infty} {ni}^{n} z^{n - 1}$ $= - (z^{2} + 1) \sum_{n = 1}^{\infty} n i^{n + 1} z^{n - 1} = - \sum_{n = 1}^{\infty} {ni}^{n + 1} z^{n + 1} - \sum_{n = 1}^{\infty} {ni}^{n + 1} z^{n - 1}$ $= - \sum_{n = 1}^{\infty} {ni}^{n + 1} e^{i (n + 1) x} - \sum_{n = 1}^{\infty} {ni}^{n + 1} e^{i (n - 1) x}$ $= - \sum_{n = 1}^{\infty} {ni}^{n + 1} {\cos (n + 1) x + isin (n + 1) x} - \sum_{n = 1}^{\infty} {ni}^{n + 1} {\cos (n - 1) x + isin (n - 1) x}$ $= - \sum_{n = 1}^{\infty} n i^{n + 1} (\cos (n + 1) x + \cos (n - 1) x) + \sum_{n = 1}^{\infty} {ni}^{n} (\sin (n + 1) x + \sin (n - 1) x)$ $= - \sum_{n = 1}^{\infty} (2 n) {(- 1)}^{n} i (\cos (2 n + 1) x + \cos (2 n - 1) x)$ $+ \sum_{n = 0}^{\infty} (2 n + 1) {(- 1)}^{n} (\cos (2 n + 2) x + \cos (2 nx)) + \sum_{n = 1}^{\infty} (2 n) {(- 1)}^{n} (\sin (2 n + 1) x + \sin (2 n - 1) x)$ $+ \sum_{n = 0}^{\infty} (2 n + 1) {(- 1)}^{n} i {\sin (2 n + 2) x + \sin (2 n) x}$ $f^{^{'}} (x) is tlreal \Rightarrow f^{^{'}} (x) = \sum_{n = 0}^{\infty} (2 n + 1) {(- 1)}^{n} {\cos (2 n + 2) x + \cos (2 nx)}$ $+ \sum_{n = 1}^{\infty} (2 n) {(- 1)}^{n} {\sin (2 n + 1) x) + \sin (2 n - 1) x} \Rightarrow$ $f (x) = \sum_{n = 0}^{\infty} (2 n + 1) {(- 1)}^{n} {\frac{1}{2 n + 2} \sin (2 n + 2) x + \frac{1}{2 n} \sin (2 nx)}$ $+ \sum_{n = 1}^{\infty} (2 n) {(- 1)}^{n} {- \frac{1}{2 n + 1} \cos (2 n + 1) x - \frac{1}{2 n - 1} \cos (2 n - 1) x} + C$ $rest to find C . . . . be continued . . . .$
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