If is the radius of a circle whose area is , find the radius of a circle whose area is .
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A = 2 k = 3 ∏ ∞ ( 1 − tan 4 2 n π ) = 2 k = 3 ∏ ∞ ( 1 − tan 2 2 k π ) ( 1 + tan 2 2 k π ) = 2 k = 3 ∏ ∞ 1 + tan 2 2 k π 1 − tan 2 2 k π ( 1 + tan 2 2 k π ) 2 = 2 k = 3 ∏ ∞ cos 2 n − 1 π sec 4 2 k π = 2 k = 3 ∏ ∞ cos 4 2 k π cos 2 k − 1 π = 2 cos 4 π k = 3 ∏ ∞ cos 3 2 k π 1 = k = 3 ∏ ∞ cos 3 2 k π 2 = n → ∞ lim 2 cos 3 4 π ( k = 2 ∏ n cos 2 k π 1 ) 3 = 2 1 ( n → ∞ lim 2 n − 1 sin 2 n π ) 3 = 2 1 ( n → ∞ lim 2 ⋅ 2 n π π sin 2 n π ) 3 = 1 6 π 3 See proof k = 2 ∏ n cos 2 k π = 2 n − 1 sin 2 n π 1
Therefore A = 1 6 π 2 = π R 2 , ⟹ R = 4 π and the radius of the circle with area R is r = 2 1 = lo g 2 ( 2 ) = lo g 2 sec 4 π = lo g 2 sec R .
Proof: Let us prove by induction the claim P ( n ) = k = 2 ∏ n cos 2 k π = 2 n − 1 sin 2 n π 1 for n ≥ 2 . For n = 2 , P ( 2 ) = cos 4 π = 2 1 = 2 sin 4 π 1 . The claim is true for n = 2 . Assuming that the claim is true for n , then:
P ( n + 1 ) = P ( n ) cos 2 n + 1 π = 2 n − 1 sin 2 n π cos 2 n + 1 π = 2 n sin 2 n + 1 π cos 2 n + 1 π cos 2 n + 1 π = 2 n sin 2 n + 1 π 1
The claim is also true of n + 1 and therefore is true for all n ≥ 2 .