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Question Number 144701 by mathmax by abdo last updated on 28/Jun/21

calculate Σ_(n=1) ^∞ ((cos(nθ))/n^2 )

$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{n}\theta\right)}{\mathrm{n}^{\mathrm{2}} } \\ $$

Answered by Dwaipayan Shikari last updated on 28/Jun/21

Σ_(n=1) ^∞ ((sin (nθ))/n)=(1/(2i))Σ_(n=1) ^∞ (((e^(iθ) )^n )/n)−(((e^(−iθ) )^n )/n) =−(1/(2i))log (((1−e^(iθ) )/(1−e^(−iθ) )))=((π−θ)/2) Σ_(n=1) ^∞ ∫_0 ^θ ((sin (nθ))/n)=((2πθ−θ^2 )/4) Σ_(n=1) ^∞ ((cos (0))/n^2 )−((cos (nθ))/n^2 )=((2πθ−θ^2 )/4) Σ_(n=1) ^∞ ((cos (nθ))/n^2 )=(π^2 /6)−((πθ)/2)+(θ^2 /4)

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\:\left({n}\theta\right)}{{n}}=\frac{\mathrm{1}}{\mathrm{2}{i}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({e}^{{i}\theta} \right)^{{n}} }{{n}}−\frac{\left({e}^{−{i}\theta} \right)^{{n}} }{{n}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{i}}\mathrm{log}\:\left(\frac{\mathrm{1}−{e}^{{i}\theta} }{\mathrm{1}−{e}^{−{i}\theta} }\right)=\frac{\pi−\theta}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\theta} \frac{\mathrm{sin}\:\left({n}\theta\right)}{{n}}=\frac{\mathrm{2}\pi\theta−\theta^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\:\left(\mathrm{0}\right)}{{n}^{\mathrm{2}} }−\frac{\mathrm{cos}\:\left({n}\theta\right)}{{n}^{\mathrm{2}} }=\frac{\mathrm{2}\pi\theta−\theta^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\:\left({n}\theta\right)}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\pi\theta}{\mathrm{2}}+\frac{\theta^{\mathrm{2}} }{\mathrm{4}} \\ $$

Answered by mathmax by abdo last updated on 28/Jun/21

ϕ(θ)=Σ_(n=1) ^∞ ((cos(nθ))/n^2 ) ⇒ϕ^′ (θ)=−Σ_(n=1) ^∞ ((sin(nθ))/n) =−Im(Σ_(n=1) ^∞ (e^(inθ) /n))=−Im(Σ_(n=1) ^∞ (((e^(iθ) )^n )/n))=−Im(log(1−e^(iθ) )) we have log(1−e^(iθ) )=log(1−cosθ−isinθ) =log(2sin^2 ((θ/2))−2isin((θ/2))cos((θ/2)))=log(−2isin((θ/2))e^((iθ)/2) ) =log2 +log(−i)+log(sin((θ/2)))+((iθ)/2) =log(2sin((θ/2)))−((iπ)/2)+((iθ)/2) ⇒ϕ^′ (θ)=((π−θ)/2) ⇒ ϕ(θ)=(π/2)θ−(θ^2 /4) +K but K=ϕ(0)=(π^2 /6) ⇒ ϕ(θ)=−(θ^2 /4)+((πθ)/2) +(π^2 /6)

$$\varphi\left(\theta\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{n}\theta\right)}{\mathrm{n}^{\mathrm{2}} }\:\Rightarrow\varphi^{'} \left(\theta\right)=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{n}\theta\right)}{\mathrm{n}} \\ $$$$=−\mathrm{Im}\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{e}^{\mathrm{in}\theta} }{\mathrm{n}}\right)=−\mathrm{Im}\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{n}} }{\mathrm{n}}\right)=−\mathrm{Im}\left(\mathrm{log}\left(\mathrm{1}−\mathrm{e}^{\mathrm{i}\theta} \right)\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{log}\left(\mathrm{1}−\mathrm{e}^{\mathrm{i}\theta} \right)=\mathrm{log}\left(\mathrm{1}−\mathrm{cos}\theta−\mathrm{isin}\theta\right) \\ $$$$=\mathrm{log}\left(\mathrm{2sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)\right)=\mathrm{log}\left(−\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{e}^{\frac{\mathrm{i}\theta}{\mathrm{2}}} \right) \\ $$$$=\mathrm{log2}\:+\mathrm{log}\left(−\mathrm{i}\right)+\mathrm{log}\left(\mathrm{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)+\frac{\mathrm{i}\theta}{\mathrm{2}} \\ $$$$=\mathrm{log}\left(\mathrm{2sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)−\frac{\mathrm{i}\pi}{\mathrm{2}}+\frac{\mathrm{i}\theta}{\mathrm{2}}\:\Rightarrow\varphi^{'} \left(\theta\right)=\frac{\pi−\theta}{\mathrm{2}}\:\Rightarrow \\ $$$$\varphi\left(\theta\right)=\frac{\pi}{\mathrm{2}}\theta−\frac{\theta^{\mathrm{2}} }{\mathrm{4}}\:+\mathrm{K}\:\:\mathrm{but}\:\mathrm{K}=\varphi\left(\mathrm{0}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow \\ $$$$\varphi\left(\theta\right)=−\frac{\theta^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi\theta}{\mathrm{2}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

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