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Question Number 114566 by Akeyz last updated on 19/Sep/20

Answered by mathmax by abdo last updated on 19/Sep/20

$$\mathrm{we}\mathrm{have}\prod _{\mathrm{k}=1}^{\mathrm{\infty }}\cos \left(\frac{\mathrm{x}}{2^{\mathrm{k}}}\right)={\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{with}{\mathrm{A}}_{\mathrm{n}}\left(\mathrm{x}\right)=\prod _{\mathrm{k}=1}^{\mathrm{n}}\cos \left(\frac{\mathrm{x}}{2^{\mathrm{k}}}\right)$$$$\mathrm{let}{\mathrm{B}}_{\mathrm{n}}\left(\mathrm{x}\right)=\prod _{\mathrm{k}=1}^{\mathrm{n}}\sin \left(\frac{\mathrm{x}}{2^{\mathrm{k}}}\right)\Rightarrow {\mathrm{A}}_{\mathrm{n}}\left(\mathrm{x}\right).{\mathrm{B}}_{\mathrm{n}}\left(\mathrm{x}\right)=\prod _{\mathrm{k}=1}^{\mathrm{n}}\left(\cos \left(\frac{\mathrm{x}}{2^{\mathrm{k}}}\right)\sin \left(\frac{\mathrm{x}}{2^{\mathrm{k}}}\right)\right)$$$$=\prod _{\mathrm{k}=1}^{\mathrm{n}}\left(\frac{1}{2}\sin \left(\frac{\mathrm{x}}{2^{\mathrm{k}-1}}\right)\right)=\frac{1}{2^{\mathrm{n}}}\prod _{\mathrm{k}=0}^{\mathrm{n}-1}\sin \left(\frac{\mathrm{x}}{2^{\mathrm{k}}}\right)=\frac{\mathrm{sinx}}{2^{\mathrm{n}}}\times \frac{\prod _{\mathrm{k}=1}^{\mathrm{n}}\sin \left(\frac{\mathrm{x}}{2^{\mathrm{k}}}\right)}{\sin \left(\frac{\mathrm{x}}{2^{\mathrm{n}}}\right)}$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}}{2^{\mathrm{n}}\sin \left(\frac{\mathrm{x}}{2^{\mathrm{n}}}\right)}\sim \frac{\mathrm{sinx}}{2^{\mathrm{n}}\frac{\mathrm{x}}{2^{\mathrm{n}}}}=\frac{\mathrm{sinx}}{\mathrm{x}}\Rightarrow {\lim }_{\mathrm{n}\to +}{\mathrm{A}}_{\mathrm{n}}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}}{\mathrm{x}}$$$$\Rightarrow {\int }_0^{\frac{\pi \mathrm{u}}{4}}\mathrm{x}\prod _{\mathrm{k}=1}^{\mathrm{\infty }}\cos \left(\frac{\mathrm{x}}{2^{\mathrm{k}}}\right)\mathrm{dx}={\int }_0^{\frac{\pi }{4}}\mathrm{x}\times \frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}={[-\mathrm{cosx}]}_0^{\frac{\pi }{4}}$$$$=1-\frac{\sqrt{2}}{2}$$

Answered by Olaf last updated on 19/Sep/20

$$\sin 2\theta =2\sin \theta \cos \theta$$$$\cos \theta =\frac{1}{2}.\frac{\sin 2\theta }{\sin \theta }$$$$\mathrm{with}\theta =\frac{x}{2^k},\cos \left(\frac{x}{2^k}\right)=\frac{1}{2}.\frac{\sin \left(\frac{x}{2^{k-1}}\right)}{\sin \left(\frac{x}{2^k}\right)}$$$$\underset{k=1}{\prod ^n}\cos \left(\frac{x}{2^k}\right)=\frac{1}{2^n}.\frac{\sin x}{\sin \left(\frac{x}{2}\right)}.\frac{\sin \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{4}\right)}...\frac{\sin \left(\frac{x}{2^{n-1}}\right)}{\sin \left(\frac{x}{2^n}\right)}$$$$\underset{k=1}{\prod ^n}\cos \left(\frac{x}{2^k}\right)=\frac{1}{2^n}.\frac{\sin x}{\sin \left(\frac{x}{2^n}\right)}$$$$2^n\sin \left(\frac{x}{2^n}\right)\underset{\mathrm{\infty }}{\sim }{2}^n\frac{x}{2^n}=x$$$$\underset{k=1}{\prod ^{\mathrm{\infty }}}\cos \left(\frac{x}{2^k}\right)=\frac{\sin x}{x}$$$${\int }_0^{\frac{\pi }{4}}x\underset{k=1}{\prod ^{\mathrm{\infty }}}\cos \left(\frac{x}{2^k}\right)dx={\int }_0^{\frac{\pi }{4}}x.\frac{\sin x}{x}dx={[-\cos x]}_0^{\frac{\pi }{4}}$$$${\int }_0^{\frac{\pi }{4}}x\underset{k=1}{\prod ^{\mathrm{\infty }}}\cos \left(\frac{x}{2^k}\right)dx=1-\frac{1}{\sqrt{2}}$$$$$$$$$$

Commented by Dwaipayan Shikari last updated on 19/Sep/20

$$Greatsir!$$

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