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Question Number 115300 by mathdave last updated on 24/Sep/20
find from fourier series an expression for log(tanx)
findfromfourierseriesanexpressionforlog(tanx)
Answered by Bird last updated on 25/Sep/20
ln(tanx)=ln(((sinx)/(cosx)))=ln(sinx) −ln(cosx) =u(x)−v(x) u(x)=ln(((e^(ix) −e^(−ix) )/(2i))) =ln(e^(ix) (1−e^(−2ix) ))−ln(2i) =ix+ln(1−e^(−2ix) )−ln(2)−((iπ)/2) we have (d/dt)ln(1−t)=−(1/(1−t)) =−Σ_(n=0) ^∞ t^n ⇒ln(1−t)=−Σ_(n=0) ^∞ (t^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (t^n /n) ⇒ ln(1−e^(−2ix) ) =−Σ_(n=1) ^∞ (e^(−2inx) /n) =−Σ_(n=1) ^∞ (1/n)(cos(2nx)−isin(2nx)) ⇒u(x)=i(x−(π/2))−Σ_(n=1) ^∞ ((cos(2nx))/n) +iΣ_(n=1) ^∞ ((sin(2nx))/n) but u(x)∈R ⇒ ln(sinx) =−Σ_(n=1) ^∞ ((cos(2nx))/n) and x−(π/2) +Σ_(n=1) ^∞ ((sin(2nx))/n) v(x)=ln(cosx) =ln(sin((π/2)−x)) =−Σ_(n=1) ^∞ ((cos(2n((π/2)−x)))/n) =−Σ_(n=1) ^∞ ((cos(nπ−2nx))/n) =−Σ_(n=1) ^∞ (((−1)^n cos(2nx))/n) ln(tanx)=u(x)−v(x) =−Σ_(n=1) ^∞ ((cos(2nx))/n) +Σ_(n=1) ^∞ (((−1)^n )/n)cos(2nx) =Σ_(n=1) ^∞ (((−1)^n −1)/n) cos(2nx) we change n by 2n+1 we get ln(tanx) =−2Σ_(n=0) ^∞ (1/(2n+1))cos(2(2n+1)x) ln(tanx)=−2Σ_(n=0) ^∞ ((cos((4n+2)x))/(2n+1))
ln(tanx)=ln(sinxcosx)=ln(sinx)−ln(cosx)=u(x)−v(x)u(x)=ln(eix−e−ix2i)=ln(eix(1−e−2ix))−ln(2i)=ix+ln(1−e−2ix)−ln(2)−iπ2wehaveddtln(1−t)=−11−t=−∑n=0∞tn⇒ln(1−t)=−∑n=0∞tn+1n+1=−∑n=1∞tnn⇒ln(1−e−2ix)=−∑n=1∞e−2inxn=−∑n=1∞1n(cos(2nx)−isin(2nx))⇒u(x)=i(x−π2)−∑n=1∞cos(2nx)n+i∑n=1∞sin(2nx)nbutu(x)∈R⇒ln(sinx)=−∑n=1∞cos(2nx)nandx−π2+∑n=1∞sin(2nx)nv(x)=ln(cosx)=ln(sin(π2−x))=−∑n=1∞cos(2n(π2−x))n=−∑n=1∞cos(nπ−2nx)n=−∑n=1∞(−1)ncos(2nx)nln(tanx)=u(x)−v(x)=−∑n=1∞cos(2nx)n+∑n=1∞(−1)nncos(2nx)=∑n=1∞(−1)n−1ncos(2nx)wechangenby2n+1wegetln(tanx)=−2∑n=0∞12n+1cos(2(2n+1)x)ln(tanx)=−2∑n=0∞cos((4n+2)x)2n+1

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