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Question Number 134328 by BHOOPENDRA last updated on 02/Mar/21

express f(x)=x as a sine series   in 0<x<π?

$${express}\:{f}\left({x}\right)={x}\:{as}\:{a}\:{sine}\:{series}\: \\ $$ $${in}\:\mathrm{0}<{x}<\pi? \\ $$

Answered by Dwaipayan Shikari last updated on 02/Mar/21

log(1+e^(ix) )=log((√(2+2sinx)))+itan^(−1) ((sinx)/(cosx+1))  (cosx−((cos2x)/2)+((cos3x)/3)−((cos4x)/4)+...)+i(sinx−((sin2x)/2)+((sin3x)/3)−..)  =log((√(2+2sinx)))+itan^(−1) ((2sin(x/2)cos(x/2))/(2cos^2 (x/2)))  sinx−((sin2x)/2)+((sin3x)/3)−((sin4x)/4)+...=tan^(−1) (tan(x/2))  (x/2)=sinx−((sin2x)/2)+((sin3x)/3)−((sin4x)/4)+...

$${log}\left(\mathrm{1}+{e}^{{ix}} \right)={log}\left(\sqrt{\mathrm{2}+\mathrm{2}{sinx}}\right)+{itan}^{−\mathrm{1}} \frac{{sinx}}{{cosx}+\mathrm{1}} \\ $$ $$\left({cosx}−\frac{{cos}\mathrm{2}{x}}{\mathrm{2}}+\frac{{cos}\mathrm{3}{x}}{\mathrm{3}}−\frac{{cos}\mathrm{4}{x}}{\mathrm{4}}+...\right)+{i}\left({sinx}−\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}+\frac{{sin}\mathrm{3}{x}}{\mathrm{3}}−..\right) \\ $$ $$={log}\left(\sqrt{\mathrm{2}+\mathrm{2}{sinx}}\right)+{itan}^{−\mathrm{1}} \frac{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$ $${sinx}−\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}+\frac{{sin}\mathrm{3}{x}}{\mathrm{3}}−\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}+...={tan}^{−\mathrm{1}} \left({tan}\frac{{x}}{\mathrm{2}}\right) \\ $$ $$\frac{{x}}{\mathrm{2}}={sinx}−\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}+\frac{{sin}\mathrm{3}{x}}{\mathrm{3}}−\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}+... \\ $$

Commented byDwaipayan Shikari last updated on 02/Mar/21

But for x=π it doesn′t hold true. 0<x<π  for example  ((((π/2)))/2)=sin((π/2))−((sin(π))/2)+((sin(((3π)/2)))/3)−((sin(2π))/4)+..  (π/4)=1−(1/3)+(1/5)−(1/7)+...

$${But}\:{for}\:{x}=\pi\:{it}\:{doesn}'{t}\:{hold}\:{true}.\:\mathrm{0}<{x}<\pi \\ $$ $${for}\:{example} \\ $$ $$\frac{\left(\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}}={sin}\left(\frac{\pi}{\mathrm{2}}\right)−\frac{{sin}\left(\pi\right)}{\mathrm{2}}+\frac{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)}{\mathrm{3}}−\frac{{sin}\left(\mathrm{2}\pi\right)}{\mathrm{4}}+.. \\ $$ $$\frac{\pi}{\mathrm{4}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+... \\ $$

Commented byBHOOPENDRA last updated on 02/Mar/21

thanks sir

$${thanks}\:{sir} \\ $$

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