Question Number 62343 by maxmathsup by imad last updated on 20/Jun/19
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{{x}\prod_{{k}=\mathrm{1}} ^{\infty} \:{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\right\}{dx} \\ $$
Commented by maxmathsup by imad last updated on 20/Jun/19
![let A_n =Π_(k=1) ^n cos((x/2^k )) and B_n =Π_(k=1) ^n sin((x/2^k )) ⇒ A_n .B_n =Π_(k=1) ^n cos((x/2^k ))sin((x/2^k )) =(1/2^n ) Π_(k=1) ^n sin((x/2^(k−1) )) =(1/2^n ) Π_(k=0) ^(n−1) sin((x/2^k )) =(1/2^n ) sin(x) Π_(k=1) ^n sin((x/2^k )) .(1/(sin((x/2^n )))) ⇒A_n .B_n = B_n .((sinx)/(2^n sin((x/2^n )))) ⇒ A_n =((sinx)/(2^n sin((x/2^n )))) we have 2^n sin((x/2^n ))∼2^n (x/2^n ) (=x) ⇒lim_(n→+∞) A_n = ((sinx)/x) ⇒ ∫_0 ^(π/4) {x Π_(k=1) ^∞ cos((x/2^k ))}dx =∫_0 ^(π/4) {x ((sinx)/x)}dx =∫_0 ^(π/4) sinx dx =[−cosx]_0 ^(π/4) =1−((√2)/2) .](https://www.tinkutara.com/question/Q62377.png)
$${let}\:{A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:{and}\:{B}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:\Rightarrow \\ $$$${A}_{{n}} .{B}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right){sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:{sin}\left({x}\right)\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\:.\frac{\mathrm{1}}{{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}\:\Rightarrow{A}_{{n}} \:.{B}_{{n}} =\:{B}_{{n}} .\frac{{sinx}}{\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{{sinx}}{\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}\:\:\:{we}\:{have}\:\:\mathrm{2}^{{n}} \:{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)\sim\mathrm{2}^{{n}} \:\frac{{x}}{\mathrm{2}^{{n}} }\:\left(={x}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {A}_{{n}} =\:\frac{{sinx}}{{x}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{{x}\:\prod_{{k}=\mathrm{1}} ^{\infty} \:{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\right\}{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left\{{x}\:\frac{{sinx}}{{x}}\right\}{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sinx}\:{dx}\:=\left[−{cosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:. \\ $$
Answered by mr W last updated on 20/Jun/19
![P_n (x)=Π_(k=1) ^n cos ((x/2^k )) 2 sin ((x/2^n ))P_n (x)=cos ((x/2))cos ((x/2^2 ))...cos ((x/2^(n−1) ))sin ((x/2^(n−1) )) ...... 2^n sin ((x/2^n ))P_n (x)=sin (x) ⇒P_n (x)=((sin (x))/(2^n sin ((x/2^n ))))=((sin (x))/x)×((x/2^n )/(sin ((x/2^n )))) ⇒lim_(n→∞) P_n (x)=((sin (x))/x)×lim_(n→∞) ((x/2^n )/(sin ((x/2^n ))))=((sin (x))/x) ∫_0 ^(π/4) {xΠ_(k=1) ^∞ cos((x/2^k ))}dx =∫_0 ^(π/4) sin x dx =[−cos x]_0 ^(π/4) =1−((√2)/2)](https://www.tinkutara.com/question/Q62360.png)
$${P}_{{n}} \left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}^{{k}} }\right) \\ $$$$\mathrm{2}\:\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}^{{n}} }\right){P}_{{n}} \left({x}\right)=\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}^{\mathrm{2}} }\right)…\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }\right) \\ $$$$…… \\ $$$$\mathrm{2}^{{n}} \:\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}^{{n}} }\right){P}_{{n}} \left({x}\right)=\mathrm{sin}\:\:\left({x}\right) \\ $$$$\Rightarrow{P}_{{n}} \left({x}\right)=\frac{\mathrm{sin}\:\left({x}\right)}{\mathrm{2}^{{n}} \mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}=\frac{\mathrm{sin}\:\left({x}\right)}{{x}}×\frac{\frac{{x}}{\mathrm{2}^{{n}} }}{\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{P}_{{n}} \left({x}\right)=\frac{\mathrm{sin}\:\left({x}\right)}{{x}}×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{{x}}{\mathrm{2}^{{n}} }}{\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}=\frac{\mathrm{sin}\:\left({x}\right)}{{x}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{{x}\prod_{{k}=\mathrm{1}} ^{\infty} \:{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)\right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{sin}\:{x}\:{dx} \\ $$$$=\left[−\mathrm{cos}\:{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 20/Jun/19
$${thanks}\:{sir}\:{mrw}. \\ $$