数列an定义如下a1=根号2,an+1=根号(2-根号(4-an^2))

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数列an定义如下a1=根号2,an+1=根号(2-根号(4-an^2))
求an的通项

令bn=an/2,则an=2bn,带入得
4b[n+1]²=2-√(4-4bn²)=2(1-√(1-bn²))
=>2b[n+1]²=1-√(1-bn²) 显然0≤bn≤1
∴可设bn=sinβn,0≤βn≤π/2∴得
2sin²β[n+1]=1-cosβn=2sin²(βn/2)
=>sin²β[n+1]=sin²(βn/2)
=>sinβ[n+1]=sin(βn/2)
=>β[n+1]=βn/2
而b1=a1/2=√2/2=sinπ/4,∴β1=π/4
∴βn=β1/2^(n-1)=π/2^(n+1),
∴an=2bn=2sinβn=2sin(π/2^(n+1))

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