1) developp at fourier serie f(x)=ln(sinx) 2) developp at fourier serie g(x)=ln(cosx +sinx) 3)developp at fourier seri e h(x) =ln(cosx +2sinx)


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Question Number 97230 by mathmax by abdo last updated on 07/Jun/20

$1) developp at fourier serie f (x) = \ln (sinx)$ $2) developp at fourier serie g (x) = \ln (cosx + sinx)$ $3) developp at fourier seri e h (x) = \ln (cosx + 2 sinx)$
	Answered by mathmax by abdo last updated on 07/Jun/20
	$1)we have f^′ (x) =((cosx)/(sinx)) =((e^(ix) +e^(−ix) )/((e^(ix) −e^(−ix) )/i)) =i ((e^(ix) +e^(−ix) )/(e^x −e^(−ix) )) changement e^(ix) =z give f^′ (x) =i((z+z^(−1) )/(z−z^(−1) )) =i ((z^2 +1)/(z^2 −1)) =−i ((z^2 +1)/(1−z^2 )) =−i(z^2 +1)Σ_(n=0) ^∞ z^(2n) =−iΣ_(n=0) ^∞ (z^(2n+2) +z^(2n) ) =−i Σ_(n=0) ^∞ e^(i(2n+2)x) −i Σ_(n=0) ^∞ e^(i2nx) =−i{ Σ_(n=0) ^∞ (cos(2n+2)x +isin(2n+2)x) +Σ_(n=0) ^∞ ( cos(2nx)+isin(2nx))} =−i(Σ_(n=0) ^∞ cos(2n+2)x+Σ_(n=0) ^∞ cos(2nx)) +Σ_(n=0) ^∞ sin(2n+2)x +Σ_(n=0) ^∞ sin(2nx) but f^′ (x) is real ⇒f^′ (x) =Σ_(n=0) ^∞ sin(2n+2)x +Σ_(n=0) ^∞ sin(2nx) =Σ_(n=1) ^∞ sin(2nx) +Σ_(n=1) ^∞ sin(2nx) =2 Σ_(n=1) ^∞ sin(2nx) ⇒ f(x) =2Σ_(n=1) ^∞ (−(1/(2n)))cos(2nx) +C =−Σ_(n=1) ^∞ ((cos(2nx))/n) +C f((π/2)) =0 =−Σ_(n=1) ^∞ ((cos(nπ))/n) +C =−Σ_(n=1) ^∞ (((−1)^n )/n) +C =ln2 +C ⇒ C =−ln2 ⇒f(x) =−Σ_(n=1) ^∞ ((cos(2nx))/n) −ln(2)$
	$1) we have f^{^{'}} (x) = \frac{cosx}{sinx} = \frac{e^{ix} + e^{- ix}}{\frac{e^{ix} - e^{- ix}}{i}} = i \frac{e^{ix} + e^{- ix}}{e^{x} - e^{- ix}} changement e^{ix} = z give$ $f^{^{'}} (x) = i \frac{z + z^{- 1}}{z - z^{- 1}} = i \frac{z^{2} + 1}{z^{2} - 1} = - i \frac{z^{2} + 1}{1 - z^{2}} = - i (z^{2} + 1) \sum_{n = 0}^{\infty} z^{2 n}$ $= - i \sum_{n = 0}^{\infty} (z^{2 n + 2} + z^{2 n}) = - i \sum_{n = 0}^{\infty} e^{i (2 n + 2) x} - i \sum_{n = 0}^{\infty} e^{i 2 nx}$ $= - i {\sum_{n = 0}^{\infty} (\cos (2 n + 2) x + isin (2 n + 2) x) + \sum_{n = 0}^{\infty} (\cos (2 nx) + isin (2 nx))}$ $= - i (\sum_{n = 0}^{\infty} \cos (2 n + 2) x + \sum_{n = 0}^{\infty} \cos (2 nx)) + \sum_{n = 0}^{\infty} \sin (2 n + 2) x + \sum_{n = 0}^{\infty} \sin (2 nx)$ $but f^{^{'}} (x) is real \Rightarrow f^{^{'}} (x) = \sum_{n = 0}^{\infty} \sin (2 n + 2) x + \sum_{n = 0}^{\infty} \sin (2 nx)$ $= \sum_{n = 1}^{\infty} \sin (2 nx) + \sum_{n = 1}^{\infty} \sin (2 nx) = 2 \sum_{n = 1}^{\infty} \sin (2 nx) \Rightarrow$ $f (x) = 2 \sum_{n = 1}^{\infty} (- \frac{1}{2 n}) \cos (2 nx) + C = - \sum_{n = 1}^{\infty} \frac{\cos (2 nx)}{n} + C$ $f (\frac{π}{2}) = 0 = - \sum_{n = 1}^{\infty} \frac{\cos (n π)}{n} + C = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} + C = \ln 2 + C \Rightarrow$ $C = - \ln 2 \Rightarrow f (x) = - \sum_{n = 1}^{\infty} \frac{\cos (2 nx)}{n} - \ln (2)$
	Commented by mathmax by abdo last updated on 07/Jun/20

	$2) g (x) = \ln (cosx + sinx) \Rightarrow g (x) = \ln (\sqrt{2} \sin (x + \frac{π}{4}))$ $= \frac{1}{2} \ln (2) + \ln (x + \frac{π}{4}) = \frac{1}{2} \ln (2) - \sum_{n = 1}^{\infty} \frac{\cos (2 n (x + \frac{π}{4}))}{n} - \ln (2)$ $= - \frac{\ln (2)}{2} - \sum_{n = 1}^{\infty} \frac{\cos (2 nx + n \frac{π}{2})}{n}$
	Commented by mathmax by abdo last updated on 07/Jun/20

	$3) h (x) = \ln (cosx + 2 sinx) we have cosx + 2 sinx = \sqrt{5} (\frac{1}{\sqrt{5}} cosx + \frac{2}{\sqrt{5}} sinx)$ $let \sin α = \frac{1}{\sqrt{5}} and \cos α = \frac{2}{\sqrt{5}} \Rightarrow \tan α = \frac{1}{2} \Rightarrow α = \arctan (\frac{1}{2}) \Rightarrow$ $cosx + 2 sinx = \sqrt{5} \sin (x + α) \Rightarrow h (x) = \frac{1}{2} \ln 5 + \ln (\sin (x + α))$ $= \frac{\ln 5}{2} - \sum_{n = 1}^{\infty} \frac{\cos (2 n (x + α))}{n} - \ln (2)$ $= \frac{\ln 5}{2} - \ln (2) - \sum_{n = 1}^{\infty} \frac{\cos (2 nx + 2 n \arctan (\frac{1}{2}))}{n}$
	Answered by mathmax by abdo last updated on 07/Jun/20

	$another method for f (x) = \ln (sinx) we have$ $f (x) = \ln (\frac{e^{ix} - e^{- ix}}{2 i}) = \ln (e^{ix} - e^{- ix}) - \ln (2 i)$ $= \ln (e^{ix}) + \ln (1 - e^{- 2 ix}) - \ln (2 i) = ix - \ln 2 - lni + \ln (1 - e^{- 2 ix})$ $= ix - \ln (2) - \frac{i π}{2} - \sum_{n = 1}^{\infty} \frac{e^{- 2 inx}}{n}$ $= i (x - \frac{π}{2}) - \sum_{n = 1}^{\infty} \frac{\cos (2 nx)}{n} + i \sum_{n = 1}^{\infty} \frac{\sin (2 nx)}{n} - \ln (2)$ $f (x) \in R \Rightarrow f (x) = - \sum_{n = 1}^{\infty} \frac{\cos (2 nx)}{n} - \ln (2) also we get$ $\sum_{n = 1}^{\infty} \frac{\sin (2 nx)}{n} = \frac{π}{2} - x$
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