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Question Number 144053 by mnjuly1970 last updated on 21/Jun/21
........ Calculus........ Ω:=lim(1/π)∫_0 ^( 2π) (Σ_(k=1) ^n ((sin(kx))/( (√2^k ))))^2 dx=?
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……..\:{Calculus}…….. \\ $$$$\:\:\:\:\:\:\:\Omega:={lim}\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{sin}\left({kx}\right)}{\:\sqrt{\mathrm{2}^{{k}} }}\right)^{\mathrm{2}} {dx}=? \\ $$$$ \\ $$
Answered by mindispower last updated on 21/Jun/21
let n,m∈N ∫_0 ^(2π) sin(nx)sin(mx)dx=(1/2)∫_0 ^(2π) (cos(m−n)x−cos(n+m)x)dx =(1/2)∫_0 ^(2π) cos(m−n)xdx=0,n≠m =π,n=m.......(1) Ω=lim_(n→∞) .(1/π)∫_0 ^(2π) (Σ_(k=1) ^n ((sin(kx))/2^(k/2) ))^2 dx =lim_(n→∞) .(1/π)∫_0 ^(2π) Σ_(k=1) ^n .Σ_(m=1) ^n ((sin(kx)sin(mx))/2^((k+m)/2) )dx =lim_(n→∞) .(1/π)Σ_(k=1) ^n Σ_(m=1) ^n (1/2^((m+k)/2) )∫_0 ^(2π) sin(kx)sin(mx)dx withe(1) Ω=lim_(n→∞) (1/π_ )Σ_(k=1) ^n Σ_(m=k) (1/2^k ).π =lim_(n→∞) .(1/π)Σ_(k=1) ^n (π/2^k )=Σ_(k=1) ^n (1/2^k )=1 Ω=1
$${let}\:{n},{m}\in\mathbb{N} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sin}\left({nx}\right){sin}\left({mx}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left({cos}\left({m}−{n}\right){x}−{cos}\left({n}+{m}\right){x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} {cos}\left({m}−{n}\right){xdx}=\mathrm{0},{n}\neq{m} \\ $$$$=\pi,{n}={m}…….\left(\mathrm{1}\right) \\ $$$$\Omega=\underset{{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{sin}\left({kx}\right)}{\mathrm{2}^{\frac{{k}}{\mathrm{2}}} }\right)^{\mathrm{2}} {dx} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}.\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{sin}\left({kx}\right){sin}\left({mx}\right)}{\mathrm{2}^{\frac{{k}+{m}}{\mathrm{2}}} }{dx} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:.\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\frac{{m}+{k}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sin}\left({kx}\right){sin}\left({mx}\right){dx} \\ $$$${withe}\left(\mathrm{1}\right) \\ $$$$\Omega=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\pi_{} }\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{m}={k}} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{{k}} }.\pi \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\pi}{\mathrm{2}^{{k}} }=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{k}} }=\mathrm{1} \\ $$$$\Omega=\mathrm{1} \\ $$
Commented by mnjuly1970 last updated on 21/Jun/21
......thanks alot mr power.....
$$\:\:\:\:\:\:\:……{thanks}\:{alot}\:{mr}\:{power}….. \\ $$
Commented by mindispower last updated on 22/Jun/21
pleasur
$${pleasur} \\ $$

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