Question Number 36747 by prof Abdo imad last updated on 05/Jun/18
![let f(x)= Σ_(n=1) ^∞ ((sin(nx))/n) x^n 1) prove that f is C^1 on ]−1,1[ 2)calculate f^′ (x) and prove that f(x)=arctan( ((xsinx)/(1−x cosx)))](Q36747.png)
$${let}\:{f}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{sin}\left({nx}\right)}{{n}}\:{x}^{{n}} \\ $$$$\left.\mathrm{1}\left.\right)\:{prove}\:{that}\:{f}\:{is}\:{C}^{\mathrm{1}} \:{on}\:\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$$\left.\mathrm{2}\right){calculate}\:{f}^{'} \left({x}\right)\:{and}\:{prove}\:{that} \\ $$$${f}\left({x}\right)={arctan}\left(\:\frac{{xsinx}}{\mathrm{1}−{x}\:{cosx}}\right) \\ $$
Commented by prof Abdo imad last updated on 06/Jun/18
![thefunction f_n (x) =((sin(nx))/n) x^n are C^1 on]−1,1[ and f_n ^′ (x)= sin(nx)x^(n−1) are continues on]−1,1[ also Σ f_n ^′ (x) converges unif. on]−1,1[ Σ f_n (x)conv.unif. f is C^1 on]−1,1[ 2) we have f^′ (x)= Σ_(n=1) ^∞ sin(nx) x^(n−1) =Im( Σ_(n=1) ^∞ e^(inx) x^(n−1) ) but Σ_(n=1) ^∞ e^(inx) x^(n−1) = Σ_(n=0) ^∞ e^(i(n+1)x) x^n = e^(ix) Σ_(n=0) ^∞ (x e^(ix) )^n =e^(ix) (1/(1−xe^(ix) )) = (1/(e^(−ix) −x)) = (1/(cosx −isinx −x)) =((cosx −x +isin(x))/((cosx −x)^2 +sin^2 x)) ⇒ f^′ (x) = ((sinx)/(1−2 x cosx +x^2 )) ⇒ f(x) = ∫ ((sinx)/(x^2 −2xcosx +1))dx +c....becontinued...](Q36824.png)
$$\left.{thefunction}\:{f}_{{n}} \left({x}\right)\:=\frac{{sin}\left({nx}\right)}{{n}}\:{x}^{{n}} \:{are}\:{C}^{\mathrm{1}} \:{on}\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$$\left.{and}\:{f}_{{n}} ^{'} \left({x}\right)=\:{sin}\left({nx}\right){x}^{{n}−\mathrm{1}} \:{are}\:{continues}\:{on}\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$$\left.{also}\:\Sigma\:{f}_{{n}} ^{'} \left({x}\right)\:{converges}\:{unif}.\:{on}\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$$\left.\Sigma\:{f}_{{n}} \left({x}\right){conv}.{unif}.\:{f}\:{is}\:{C}^{\mathrm{1}} \:{on}\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} {sin}\left({nx}\right)\:{x}^{{n}−\mathrm{1}} \\ $$$$={Im}\left(\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{e}^{{inx}} \:{x}^{{n}−\mathrm{1}} \right)\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{e}^{{inx}} \:{x}^{{n}−\mathrm{1}} \:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{e}^{{i}\left({n}+\mathrm{1}\right){x}} \:{x}^{{n}} \\ $$$$=\:{e}^{{ix}} \:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left({x}\:{e}^{{ix}} \right)^{{n}} \:={e}^{{ix}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{xe}^{{ix}} } \\ $$$$=\:\:\frac{\mathrm{1}}{{e}^{−{ix}} \:−{x}}\:=\:\frac{\mathrm{1}}{{cosx}\:−{isinx}\:−{x}} \\ $$$$=\frac{{cosx}\:−{x}\:+{isin}\left({x}\right)}{\left({cosx}\:−{x}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {x}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{{sinx}}{\mathrm{1}−\mathrm{2}\:{x}\:{cosx}\:+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\int\:\:\:\:\:\:\:\frac{{sinx}}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcosx}\:+\mathrm{1}}{dx}\:+{c}....{becontinued}... \\ $$