Question Number 148303 by mathmax by abdo last updated on 26/Jul/21

f(x)=((cos(2x))/(sin(x))) developp f at fourier serie

Answered by Olaf_Thorendsen last updated on 27/Jul/21

f(x) = ((cos(2x))/(sinx)) cos(2x) = Σ_(n=1) ^∞ (cos((2nx)−cos(2(n+1)x) (telescopic sum) cos(2x) = Σ_(n=1) ^∞ (cos((2n+1)x−x)−cos((2n+1)x+x) cos(2x) = 2Σ_(n=1) ^∞ (1/2)(cos((2n+1)x−x)−cos((2n+1)x+x) cos(2x) = 2Σ_(n=1) ^∞ sin((2n+1)x).sinx ((cos(2x))/(sinx)) = 2Σ_(n=1) ^∞ sin((2n+1)x) a_0 = 0 and ∀n≥1, a_n = 0 (because f is odd) ∀n∈N^∗ , b_(2n) = 0 and ∀n∈N, b_(2n+1) = 2

Answered by mathmax by abdo last updated on 27/Jul/21

f(x)=(((e^(2ix) +e^(−2ix) )/2)/((e^(ix) −e^(−ix) )/(2i)))=i×((e^(2ix) +e^(−2ix) )/(e^(ix) −e^(−ix) )) changement e^(ix) =z give f(x)=g(z)=i.((z^2 +z^(−2) )/(z−z^(−1) )) =i((z^3 +z^(−1) )/(z^2 −1))=i((z^4 +1)/(z^3 −z)) =i .((z(z^3 −z)+z^2 +1)/(z^3 −z))=iz +i((z^2 +1)/(z(z^2 −1))) ((z^2 +1)/(z(z−1)(z+1)))=(a/z)+(b/(z−1))+(c/(z+1)) a=−1,b=1 ,c=(2/2)=1 ⇒((z^2 +1)/(z(z−1)(z+1))) =−(1/z)+(1/(z−1))+(1/(z+1))=−(1/z)−(1/(1−z))+(1/(1+z)) ⇒g(z)=iz−(1/z)−Σ_(n=0) ^∞ z^n +Σ_(n=0) ^∞ (−1)^n z^n =ie^(ix) −e^(−ix) +Σ_(n=0) ^∞ {(−1)^n −1}e^(inx) =i(cosx+isinx)−(cosx−isinx)+Σ_(n=0) ^∞ {(−1)^n −1}(cos(nx)+isin(nx)) =−sinx−cosx+i(cosx+sinx)+Σ_(n=0) ^∞ {(−1)^n −1}cos(nx)+iΣ(...) f(x)is real ⇒ f(x)=−sinx−cosx+Σ_(n=0) ^∞ {(−1)^n −1}cos(nx) f(x)=−sinx−cosx−2Σ_(n=0) ^∞ cos(2n+1)x