Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 129825 by 0731619177 last updated on 19/Jan/21

	Answered by Olaf last updated on 20/Jan/21
	$Let u_0 = 0 and u_n = (√(2+(√(2+(√(2...)))))), n ≥ 1 u_(n+1) = (√(2+u_n )) By induction : 0 ≤ u_n ≤ 2 ⇒ 0 ≤ u_(n+1) ≤ 2 So let u_n = 2cosθ_n (u_0 = 0 ⇒ θ_0 = (π/2)) u_(n+1) = (√(2+2cosθ_n )) = 2(√((1+cosθ_n )/2)) u_(n+1) = 2(√(cos^2 (θ_n /2))) = 2cos(θ_n /2) = 2cosθ_(n+1) ⇒ θ_(n+1) = (θ_n /2) Finally, θ_n = (θ_0 /2^n ) = (π/2^(n+1) ), u_n = 2cos(π/2^(n+1) ) sin2α = 2sinαcosα cosα = ((sin2α)/(2sinα)) ⇒ u_n = 2cosθ_n = ((sin2θ_n )/(sinθ_n )) = ((sin(π/2^n ))/(sin(π/2^(n+1) ))) Π_(n=1) ^N (2/u_n ) = 2^N Π_(n=1) ^N ((sin(π/2^(n+1) ))/(sin(π/2^n ))) = 2^N ((sin(π/2^(N+1) ))/(sin(π/2))) (telescopic product) Π_(n=1) ^N (2/u_n ) = 2^N sin(π/2^(N+1) ) ∼_∞ 2^N (π/2^(N+1) ) = (π/2) (1) 3!! = 6! = 720 (2) (√(Γ(2))) = (√(1!)) = 1 (3) With (1), (2) and (3) the result is : {ln[−((π/2))]^(720) }^1 = 720.ln(π/2)$
	$Let u_{0} = 0 and u_{n} = \sqrt{2 + \sqrt{2 + \sqrt{2 . . .}}}, n ⩾ 1$ $u_{n + 1} = \sqrt{2 + u_{n}}$ $By induction :$ $0 ⩽ u_{n} ⩽ 2 \Rightarrow 0 ⩽ u_{n + 1} ⩽ 2$ $So let u_{n} = 2 \cos θ_{n} (u_{0} = 0 \Rightarrow θ_{0} = \frac{π}{2})$ $u_{n + 1} = \sqrt{2 + 2 \cos θ_{n}} = 2 \sqrt{\frac{1 + \cos θ_{n}}{2}}$ $u_{n + 1} = 2 \sqrt{\cos^{2} \frac{θ_{n}}{2}} = 2 \cos \frac{θ_{n}}{2} = 2 \cos θ_{n + 1}$ $\Rightarrow θ_{n + 1} = \frac{θ_{n}}{2}$ $Finally, θ_{n} = \frac{θ_{0}}{2^{n}} = \frac{π}{2^{n + 1}}, u_{n} = 2 \cos \frac{π}{2^{n + 1}}$ $\sin 2 α = 2 \sin α \cos α$ $\cos α = \frac{\sin 2 α}{2 \sin α}$ $\Rightarrow u_{n} = 2 \cos θ_{n} = \frac{\sin 2 θ_{n}}{\sin θ_{n}} = \frac{\sin \frac{π}{2^{n}}}{\sin \frac{π}{2^{n + 1}}}$ $\underset{n = 1}{\prod^{N}} \frac{2}{u_{n}} = 2^{N} \underset{n = 1}{\prod^{N}} \frac{\sin \frac{π}{2^{n + 1}}}{\sin \frac{π}{2^{n}}} = 2^{N} \frac{\sin \frac{π}{2^{N + 1}}}{\sin \frac{π}{2}}$ $(telescopic product)$ $\underset{n = 1}{\prod^{N}} \frac{2}{u_{n}} = 2^{N} \sin \frac{π}{2^{N + 1}} \underset{\infty}{\sim} 2^{N} \frac{π}{2^{N + 1}} = \frac{π}{2} (1)$ $3!! = 6! = 720 (2)$ $\sqrt{Γ (2)} = \sqrt{1!} = 1 (3)$ $With (1), (2) and (3) the result is :$ ${\ln {[- (\frac{π}{2})]}^{720}}^{1} = 720 . \ln \frac{π}{2}$
	Commented by mr W last updated on 20/Jan/21

	$3!! \neq (3!)!$ $3!! = 3 \times 1 = 3$
	Commented by Olaf last updated on 20/Jan/21

	$Sir, why 3!! = 3 \times 1 ?$ ${It}^{'} s the first time I see this notation .$
	Commented by Olaf last updated on 20/Jan/21

	$Dear mr W, my own calculator$ $gives : 3!! = (3!)! = 720$
	Commented by Olaf last updated on 20/Jan/21

	Commented by mr W last updated on 20/Jan/21

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum