Question Number 114566 by Akeyz last updated on 19/Sep/20
Answered by mathmax by abdo last updated on 19/Sep/20
![we have Π_(k=1) ^∞ cos((x/2^k )) =lim_(n→+∞) A_n (x) with A_n (x)=Π_(k=1) ^n cos((x/2^k )) let B_n (x) =Π_(k=1) ^n sin((x/2^k )) ⇒A_n (x).B_n (x)=Π_(k=1) ^n (cos((x/2^k ))sin((x/2^k ))) =Π_(k=1) ^n ((1/2)sin((x/2^(k−1) ))) =(1/2^n ) Π_(k=0) ^(n−1) sin((x/2^k )) =((sinx)/2^n ) ×((Π_(k=1) ^n sin((x/2^k )))/(sin((x/2^n )))) ⇒ A_n (x) =((sinx)/(2^n sin((x/2^n ))))∼ ((sinx)/(2^n (x/2^n ))) =((sinx)/x) ⇒lim_(n→+) A_n (x) =((sinx)/x) ⇒∫_0 ^((πu)/4) x Π_(k=1) ^∞ cos((x/2^k ))dx =∫_0 ^(π/4) x×((sinx)/x)dx =[−cosx]_0 ^(π/4) =1−((√2)/2)](https://www.tinkutara.com/question/Q114583.png)
$$\mathrm{we}\:\mathrm{have}\:\prod_{\mathrm{k}=\mathrm{1}} ^{\infty} \:\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} \left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{A}_{\mathrm{n}} \left(\mathrm{x}\right)=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right) \\ $$$$\mathrm{let}\:\mathrm{B}_{\mathrm{n}} \left(\mathrm{x}\right)\:=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\:\Rightarrow\mathrm{A}_{\mathrm{n}} \left(\mathrm{x}\right).\mathrm{B}_{\mathrm{n}} \left(\mathrm{x}\right)=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\right) \\ $$$$=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}−\mathrm{1}} }\right)\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:\prod_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\:=\frac{\mathrm{sinx}}{\mathrm{2}^{\mathrm{n}} }\:×\frac{\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{n}} }\right)} \\ $$$$\Rightarrow\:\mathrm{A}_{\mathrm{n}} \left(\mathrm{x}\right)\:=\frac{\mathrm{sinx}}{\mathrm{2}^{\mathrm{n}} \:\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{n}} }\right)}\sim\:\frac{\mathrm{sinx}}{\mathrm{2}^{\mathrm{n}} \frac{\mathrm{x}}{\mathrm{2}^{\mathrm{n}} }}\:=\frac{\mathrm{sinx}}{\mathrm{x}}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+} \mathrm{A}_{\mathrm{n}} \left(\mathrm{x}\right)\:=\frac{\mathrm{sinx}}{\mathrm{x}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi\mathrm{u}}{\mathrm{4}}} \:\mathrm{x}\:\prod_{\mathrm{k}=\mathrm{1}} ^{\infty} \:\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{x}×\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}\:=\left[−\mathrm{cosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Answered by Olaf last updated on 19/Sep/20
![sin2θ = 2sinθcosθ cosθ =(1/2). ((sin2θ)/(sinθ)) with θ = (x/2^k ), cos((x/2^k )) = (1/2).((sin((x/2^(k−1) )))/(sin((x/2^k )))) Π_(k=1) ^n cos((x/2^k )) = (1/2^n ).((sinx)/(sin((x/2)))).((sin((x/2)))/(sin((x/4))))...((sin((x/2^(n−1) )))/(sin((x/2^n )))) Π_(k=1) ^n cos((x/2^k )) = (1/2^n ).((sinx)/(sin((x/2^n )))) 2^n sin((x/2^n )) ∼_∞ 2^n (x/2^n ) = x Π_(k=1) ^∞ cos((x/2^k )) = ((sinx)/x) ∫_0 ^(π/4) xΠ_(k=1) ^∞ cos((x/2^k ))dx = ∫_0 ^(π/4) x.((sinx)/x)dx = [−cosx]_0 ^(π/4) ∫_0 ^(π/4) xΠ_(k=1) ^∞ cos((x/2^k ))dx = 1−(1/( (√2)))](https://www.tinkutara.com/question/Q114581.png)
$$\mathrm{sin2}\theta\:=\:\mathrm{2sin}\theta\mathrm{cos}\theta \\ $$$$\mathrm{cos}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}.\:\frac{\mathrm{sin2}\theta}{\mathrm{sin}\theta} \\ $$$$\mathrm{with}\:\theta\:=\:\frac{{x}}{\mathrm{2}^{{k}} },\:\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }.\frac{\mathrm{sin}{x}}{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}}\right)}.\frac{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\frac{{x}}{\mathrm{4}}\right)}…\frac{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)}{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }.\frac{\mathrm{sin}{x}}{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)} \\ $$$$\mathrm{2}^{{n}} \mathrm{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)\:\underset{\infty} {\sim}\:\mathrm{2}^{{n}} \frac{{x}}{\mathrm{2}^{{n}} }\:=\:{x} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\:\frac{\mathrm{sin}{x}}{{x}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right){dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}.\frac{\mathrm{sin}{x}}{{x}}{dx}\:=\:\left[−\mathrm{cos}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right){dx}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 19/Sep/20
$${Great}\:{sir}! \\ $$