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Question Number 197323 by Erico last updated on 13/Sep/23
If f(x)=((sin(x))/x) and S_n (α)=Σ_(k=1) ^n [f(kπ+(π/α))+f(kπ−(π/α))] (α>1) Prove that lim_(n→+∞) S_n (α)=1−f((π/α))
$$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}}\:\:\:\mathrm{and}\:\mathrm{S}_{\mathrm{n}} \left(\alpha\right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left[\mathrm{f}\left(\mathrm{k}\pi+\frac{\pi}{\alpha}\right)+\mathrm{f}\left(\mathrm{k}\pi−\frac{\pi}{\alpha}\right)\right]\:\:\:\:\left(\alpha>\mathrm{1}\right) \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\underset{\mathrm{n}\rightarrow+\infty} {\:\mathrm{lim}}\:\mathrm{S}_{\mathrm{n}} \left(\alpha\right)=\mathrm{1}−\mathrm{f}\left(\frac{\pi}{\alpha}\right) \\ $$
Answered by witcher3 last updated on 14/Sep/23
S_n (a)=Σ_1 ^n (f(kπ+(π/a))−f(kπ−(π/a))) T(a)Σ_(k=1) ^n (((−1)^k sin((π/a)))/(kπ+(π/a))) −T(−a)+T(a)=S_n (a) T(a)=((sin((π/a)))/π)Σ_(k=1) ^n (((−1)^k )/(k+(1/a)))...cv Σ(−1)^k ≤2,..(1/(k+(1/a))) decrease cv →0 T(a)..cv Σ_(k=1) ^∞ (((−1)^k )/(k+(1/a)))=Σ_(k=1) ^∞ (1/(2k+(1/a)))−(1/(2k−1+(1/a))) =(1/2)Σ_(k=1) ^∞ (1/(k+(1/(2a))))−(1/(k+(1/2)((1/a)−1))) =(1/2)(Ψ((1/2)((1/a)+1))−Ψ((1/(2a))+1)) S_n (a)=Γ(a)−Γ(−a) =((sin((π/a)))/(2π))(Ψ((1/2)+(1/(2a)))−Ψ((1/2)−(1/(2a)))−Ψ(1+(1/(2a)))+Ψ(1−(1/(2a)))) =((sin((π/a)))/(2π))(Ψ(1−((1/2)−(1/(2a)))−Ψ((1/2)−(1/(2a)))−2a−Ψ((1/(2a)))+Ψ(1−(1/(2a)))) =((sin((π/a)))/(2π))(πcot(π((1/2)−(1/(2a)))−2a+πcot((π/(2a))) =((sin((π/a)))/2)(tg((1/(2a)))+cot((1/(2a))))−((sin((π/a)))/(π/a)) =((sin((π/a)))/2)(((cos^2 ((1/(2a)))+sin^2 ((1/(2a))))/(sin((π/(2a)))cos((π/(2a))))))−f((π/a)) =1−f((π/a))
$$\mathrm{S}_{\mathrm{n}} \left(\mathrm{a}\right)=\underset{\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{f}\left(\mathrm{k}\pi+\frac{\pi}{\mathrm{a}}\right)−\mathrm{f}\left(\mathrm{k}\pi−\frac{\pi}{\mathrm{a}}\right)\right) \\ $$$$\mathrm{T}\left(\mathrm{a}\right)\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{sin}\left(\frac{\pi}{\mathrm{a}}\right)}{\mathrm{k}\pi+\frac{\pi}{\mathrm{a}}} \\ $$$$−\mathrm{T}\left(−\mathrm{a}\right)+\mathrm{T}\left(\mathrm{a}\right)=\mathrm{S}_{\mathrm{n}} \left(\mathrm{a}\right) \\ $$$$\mathrm{T}\left(\mathrm{a}\right)=\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{a}}\right)}{\pi}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}+\frac{\mathrm{1}}{\mathrm{a}}}…\mathrm{cv} \\ $$$$\Sigma\left(−\mathrm{1}\right)^{\mathrm{k}} \leqslant\mathrm{2},..\frac{\mathrm{1}}{\mathrm{k}+\frac{\mathrm{1}}{\mathrm{a}}}\:\mathrm{decrease}\:\mathrm{cv}\:\rightarrow\mathrm{0} \\ $$$$\mathrm{T}\left(\mathrm{a}\right)..\mathrm{cv} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}+\frac{\mathrm{1}}{\mathrm{a}}}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2k}+\frac{\mathrm{1}}{\mathrm{a}}}−\frac{\mathrm{1}}{\mathrm{2k}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{a}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2a}}}−\frac{\mathrm{1}}{\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{a}}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{a}}+\mathrm{1}\right)\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2a}}+\mathrm{1}\right)\right) \\ $$$$\mathrm{S}_{\mathrm{n}} \left(\mathrm{a}\right)=\Gamma\left(\mathrm{a}\right)−\Gamma\left(−\mathrm{a}\right) \\ $$$$=\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{a}}\right)}{\mathrm{2}\pi}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2a}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2a}}\right)−\Psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2a}}\right)+\Psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2a}}\right)\right) \\ $$$$=\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{a}}\right)}{\mathrm{2}\pi}\left(\Psi\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2a}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2a}}\right)−\mathrm{2a}−\Psi\left(\frac{\mathrm{1}}{\mathrm{2a}}\right)+\Psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2a}}\right)\right)\right. \\ $$$$=\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{a}}\right)}{\mathrm{2}\pi}\left(\pi\mathrm{cot}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2a}}\right)−\mathrm{2a}+\pi\mathrm{cot}\left(\frac{\pi}{\mathrm{2a}}\right)\right.\right. \\ $$$$=\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{a}}\right)}{\mathrm{2}}\left(\mathrm{tg}\left(\frac{\mathrm{1}}{\mathrm{2a}}\right)+\mathrm{cot}\left(\frac{\mathrm{1}}{\mathrm{2a}}\right)\right)−\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{a}}\right)}{\frac{\pi}{\mathrm{a}}} \\ $$$$=\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{a}}\right)}{\mathrm{2}}\left(\frac{\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2a}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2a}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2a}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2a}}\right)}\right)−\mathrm{f}\left(\frac{\pi}{\mathrm{a}}\right) \\ $$$$=\mathrm{1}−\mathrm{f}\left(\frac{\pi}{\mathrm{a}}\right) \\ $$$$ \\ $$

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