Menu Close

calculate-n-1-cos-2nx-n-

Tinku Tara 0 Comments
Pin




Question Number 67034 by mathmax by abdo last updated on 22/Aug/19
calculate Σ_(n=1) ^∞ ((cos(2nx))/n)
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}} \\ $$
Commented by mathmax by abdo last updated on 23/Aug/19
let S =Σ_(n=1) ^∞ ((cos(2nx))/n) ⇒ S =Re(Σ_(n=1) ^∞ (e^(i2nx) /n)) let W(x) =Σ_(n=1) ^∞ (e^(i2nx) /n) ⇒W^′ (x) =Σ_(n=1) ^∞ 2ni (e^(i2nx) /n) =2iΣ_(n=1) ^∞ (e^(2ix) )^n =2i e^(2ix) Σ_(n=1) ^∞ (e^(2ix) )^(n−1) =2i e^(2ix) Σ_(n=0) ^∞ (e^(2ix) )^n =2i e^(2ix) ×(1/(1−e^(2ix) )) =((2i e^(2ix) )/(1−e^(2ix) )) ⇒ W(x) =−ln(1−e^(2ix) ) we have 1−e^(2ix) =1−cos(2x)−isin(2x) =2sin^2 (x)−2isin(x)cos(x) =−2i sin(x){cosx+i sinx} =−2isinx e^(ix) ⇒ln(1−e^(2ix) ) =ln(−2i)+ln(sinx)+ix =ln(2)+ln(−i) +ln(sinx)+ix =ln(2)+ln(e^(−((iπ)/2)) )+ln(sinx)+ix =ln(2)−((iπ)/2) +ix +ln(sinx) =ln(2sinx)+i(x−(π/2)) ⇒ Σ_(n=1) ^∞ ((cos(2nx))/n) =ln(2sinx) .
$${let}\:{S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}\:\Rightarrow\:{S}\:={Re}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{i}\mathrm{2}{nx}} }{{n}}\right) \\ $$$${let}\:{W}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{i}\mathrm{2}{nx}} }{{n}}\:\Rightarrow{W}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\mathrm{2}{ni}\:\:\frac{{e}^{{i}\mathrm{2}{nx}} }{{n}} \\ $$$$=\mathrm{2}{i}\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({e}^{\mathrm{2}{ix}} \right)^{{n}} \:=\mathrm{2}{i}\:{e}^{\mathrm{2}{ix}} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({e}^{\mathrm{2}{ix}} \right)^{{n}−\mathrm{1}} \:=\mathrm{2}{i}\:{e}^{\mathrm{2}{ix}} \sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{\mathrm{2}{ix}} \right)^{{n}} \\ $$$$=\mathrm{2}{i}\:{e}^{\mathrm{2}{ix}} ×\frac{\mathrm{1}}{\mathrm{1}−{e}^{\mathrm{2}{ix}} }\:=\frac{\mathrm{2}{i}\:{e}^{\mathrm{2}{ix}} }{\mathrm{1}−{e}^{\mathrm{2}{ix}} }\:\Rightarrow \\ $$$${W}\left({x}\right)\:=−{ln}\left(\mathrm{1}−{e}^{\mathrm{2}{ix}} \right)\:\:{we}\:{have}\:\mathrm{1}−{e}^{\mathrm{2}{ix}} =\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)−{isin}\left(\mathrm{2}{x}\right) \\ $$$$=\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right)−\mathrm{2}{isin}\left({x}\right){cos}\left({x}\right)\:=−\mathrm{2}{i}\:{sin}\left({x}\right)\left\{{cosx}+{i}\:{sinx}\right\} \\ $$$$=−\mathrm{2}{isinx}\:{e}^{{ix}} \:\Rightarrow{ln}\left(\mathrm{1}−{e}^{\mathrm{2}{ix}} \right)\:={ln}\left(−\mathrm{2}{i}\right)+{ln}\left({sinx}\right)+{ix} \\ $$$$={ln}\left(\mathrm{2}\right)+{ln}\left(−{i}\right)\:+{ln}\left({sinx}\right)+{ix}\:={ln}\left(\mathrm{2}\right)+{ln}\left({e}^{−\frac{{i}\pi}{\mathrm{2}}} \right)+{ln}\left({sinx}\right)+{ix} \\ $$$$={ln}\left(\mathrm{2}\right)−\frac{{i}\pi}{\mathrm{2}}\:+{ix}\:+{ln}\left({sinx}\right)\:={ln}\left(\mathrm{2}{sinx}\right)+{i}\left({x}−\frac{\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}\:={ln}\left(\mathrm{2}{sinx}\right)\:. \\ $$
Commented by mathmax by abdo last updated on 25/Aug/19
error at final line we have W(x)=−ln(1−e^(2ix) ) ⇒ W(x) =−ln(2sinx)−i(x−(π/2)) =−ln(2sinx)+i(((π−2x)/2)) ⇒ Σ_(n=1) ^∞ ((cos(2nx))/n) =−ln(2sinx)
$${error}\:{at}\:{final}\:{line}\:\:{we}\:{have}\:{W}\left({x}\right)=−{ln}\left(\mathrm{1}−{e}^{\mathrm{2}{ix}} \right)\:\Rightarrow \\ $$$${W}\left({x}\right)\:=−{ln}\left(\mathrm{2}{sinx}\right)−{i}\left({x}−\frac{\pi}{\mathrm{2}}\right)\:=−{ln}\left(\mathrm{2}{sinx}\right)+{i}\left(\frac{\pi−\mathrm{2}{x}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}\:=−{ln}\left(\mathrm{2}{sinx}\right) \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *