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a(k) = a(k-1)² − 2 for k=1,2, … … (i)
Write (i) as a(k)+1 = (a(k-1)−1)(a(k-1)+1) ⟹ a(k-1)−1 = (a(k)+1)/(a(k-1)+1)
Shift index for a(k)−1 = (a(k+1)+1)/(a(k)+1)
∏ (k=0,n) (a(k)−1)/a(k) = ∏ (k=0,n) (a(k+1)+1)/(a(k)+1) * ∏ (k=0,n) 1/a(k)
∏ (k=0,n) (a(k)−1)/a(k) = (a(n+1)+1)/(a(0)+1) / ∏ (k=0,n) a(k) = (2/7)(a(n+1)+1) / ∏ (k=0,n) a(k) … (ii)
Make the sub a(k) = x(k)+1/x(k) in (i) … (iii)
x(k)+1/x(k) = x(k-1)²+1/x( k-1)² ⟹ x(k-1)²−x(k) = (x(k-1)²−x(k)) / (x(k)x(k-1)²)
Solutions are x(k)=x(k-1)² or1/x(k-1)². ( either will do, they result in the same a(k) ).
Use in (iii) for a(k) = x(k-1)²+1/x(k-1)² ⟹ a(k) = x(k-2)⁴+1/x(k-2)⁴ and so on giving
a(k) = x(0)^(2ᵏ) + 1/x(0)^(2ᵏ) = 2^(2ᵏ) + 2^(−2ᵏ) since x(0)=2 from a(0) = x(0)+1/x(0)
So ∏ (k=0,n) a(k) = ∏ (k=0,n) ( 2^(2ᵏ) + 2^(−2ᵏ) ) = (2/3)( 2^(2ⁿ⁺¹)−2^(−2ⁿ⁺¹) ) (see end)
Hence RHS of (ii) becomes (2/7)(3/2)( 2^(2ⁿ⁺¹)+2^(−2ⁿ⁺¹ )+1 ) / ( 2^(2ⁿ⁺¹)−2^(−2ⁿ⁺¹) )
As n→∞ this approaches (3/7)(1) = 3/7
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Consider P = (a+b)(a²+b²)(a⁴+b⁴)(a⁸+b⁸) … (a^(2ⁿ)+b^(2ⁿ))
P(a−b) = (a²−b²)(a²+b²)(a⁴+b⁴)(a⁸+b⁸) … (a^(2ⁿ)+b^(2ⁿ))
= (a⁴−b⁴)(a⁴+b⁴)(a⁸+b⁸) … (a^(2ⁿ)+b^(2ⁿ))
= (a⁸−b⁸)(a⁸+b⁸) … (a^(2ⁿ)+b^(2ⁿ))
= (a¹⁶−b¹⁶) … (a^(2ⁿ)+b^(2ⁿ))
= (a^(2ⁿ⁺¹)−b^(2ⁿ⁺¹))
∴ P = ( a^(2ⁿ⁺¹)−b^(2ⁿ⁺¹) ) / (a−b)
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