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)

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)))

Commented byDwaipayan Shikari last updated on 19/Sep/20

Great sir!