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Question Number 115300 by mathdave last updated on 24/Sep/20
find from fourier series an expression for log(tanx)
$${find}\:{from}\:{fourier}\:{series}\:{an} \\ $$$${expression}\:{for} \\ $$$$\mathrm{log}\left(\mathrm{tan}{x}\right) \\ $$
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))
$${ln}\left({tanx}\right)={ln}\left(\frac{{sinx}}{{cosx}}\right)={ln}\left({sinx}\right) \\ $$$$−{ln}\left({cosx}\right)\:={u}\left({x}\right)−{v}\left({x}\right) \\ $$$${u}\left({x}\right)={ln}\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right) \\ $$$$={ln}\left({e}^{{ix}} \left(\mathrm{1}−{e}^{−\mathrm{2}{ix}} \right)\right)−{ln}\left(\mathrm{2}{i}\right) \\ $$$$={ix}+{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}{ix}} \right)−{ln}\left(\mathrm{2}\right)−\frac{{i}\pi}{\mathrm{2}} \\ $$$${we}\:{have}\:\frac{{d}}{{dt}}{ln}\left(\mathrm{1}−{t}\right)=−\frac{\mathrm{1}}{\mathrm{1}−{t}} \\ $$$$=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}−{t}\right)=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}} }{{n}}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{e}^{−\mathrm{2}{ix}} \right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{−\mathrm{2}{inx}} }{{n}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left({cos}\left(\mathrm{2}{nx}\right)−{isin}\left(\mathrm{2}{nx}\right)\right) \\ $$$$\Rightarrow{u}\left({x}\right)={i}\left({x}−\frac{\pi}{\mathrm{2}}\right)−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}} \\ $$$$+{i}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{sin}\left(\mathrm{2}{nx}\right)}{{n}}\:{but}\:{u}\left({x}\right)\in{R}\:\Rightarrow \\ $$$${ln}\left({sinx}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}} \\ $$$${and}\:{x}−\frac{\pi}{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{nx}\right)}{{n}} \\ $$$${v}\left({x}\right)={ln}\left({cosx}\right)\:={ln}\left({sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\right) \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{n}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\right)}{{n}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({n}\pi−\mathrm{2}{nx}\right)}{{n}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} {cos}\left(\mathrm{2}{nx}\right)}{{n}} \\ $$$${ln}\left({tanx}\right)={u}\left({x}\right)−{v}\left({x}\right) \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{cos}\left(\mathrm{2}{nx}\right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}}\:{cos}\left(\mathrm{2}{nx}\right) \\ $$$${we}\:{change}\:{n}\:{by}\:\mathrm{2}{n}+\mathrm{1}\:{we}\:{get} \\ $$$${ln}\left({tanx}\right)\:=−\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{cos}\left(\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right){x}\right) \\ $$$${ln}\left({tanx}\right)=−\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\left(\mathrm{4}{n}+\mathrm{2}\right){x}\right)}{\mathrm{2}{n}+\mathrm{1}} \\ $$

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