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Question-157706




Question Number 157706 by Oberon last updated on 26/Oct/21
Answered by mr W last updated on 26/Oct/21
2a_n =(√((2a_(n−1) +1)/2))  let b_n =2a_n   b_n =(√((b_n +1)/2))  2b_n ^2 −1=b_(n−1)   say b_n =cos (θ/2^n )  2b_n −1=2 cos^2  (θ/2^n )−1=cos (θ/2^(n−1) )=b_(n−1)  ✓  b_1 =cos (θ/2)=2a_1 =2×(1/2)(√(1/2))  cos (θ/2)=(√(1/2))  ⇒(θ/2)=(π/4) ⇒θ=(π/2)  ⇒b_n =cos (π/2^(n+1) )  ⇒a_n =(b_n /2)  ⇒a_n =(1/2)cos (π/2^(n+1) )
$$\mathrm{2}{a}_{{n}} =\sqrt{\frac{\mathrm{2}{a}_{{n}−\mathrm{1}} +\mathrm{1}}{\mathrm{2}}} \\ $$$${let}\:{b}_{{n}} =\mathrm{2}{a}_{{n}} \\ $$$${b}_{{n}} =\sqrt{\frac{{b}_{{n}} +\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{2}{b}_{{n}} ^{\mathrm{2}} −\mathrm{1}={b}_{{n}−\mathrm{1}} \\ $$$${say}\:{b}_{{n}} =\mathrm{cos}\:\frac{\theta}{\mathrm{2}^{{n}} } \\ $$$$\mathrm{2}{b}_{{n}} −\mathrm{1}=\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}^{{n}} }−\mathrm{1}=\mathrm{cos}\:\frac{\theta}{\mathrm{2}^{{n}−\mathrm{1}} }={b}_{{n}−\mathrm{1}} \:\checkmark \\ $$$${b}_{\mathrm{1}} =\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\mathrm{2}{a}_{\mathrm{1}} =\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\theta}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}\:\Rightarrow\theta=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{b}_{{n}} =\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$$\Rightarrow{a}_{{n}} =\frac{{b}_{{n}} }{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$

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