Find the value of dimension n such that the surface area of unit n -sphere { ( x 1 , x 2 , ⋯ , x n + 1 ) ∈ R n + 1 ∣ x 1 2 + x 2 2 + ⋯ + x n + 1 2 = 1 } is a maximum.
Hint: The surface area of n -sphere of radius R is: S n ( R ) = Γ ( 2 n + 1 ) 2 π 2 n + 1 R n , where Γ ( ⋅ ) denotes the gamma function .
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For an n-sphere , its surface area is given by:
S n ( R ) = Γ ( 2 n ) 2 π 2 n R n
Where Γ ( n ) is the Gamma function . Specifically, for an unit n-sphere ( R = 1 ) and since in this problem we're looking for an (n+1)-sphere:
S ( n + 1 ) = Γ ( 2 n + 1 ) 2 π 2 n + 1
Differentiating it with respect to n and making it equal to zero:
d n d S ( n + 1 ) = [ Γ ( 2 n + 1 ) ] 2 2 π 2 n + 1 ⋅ ln ( π ) ⋅ 2 1 ⋅ Γ ( 2 n + 1 ) − 2 π 2 n + 1 ⋅ Γ ( 2 n + 1 ) ⋅ ψ ( 2 n + 1 ) ⋅ 2 1 = 0
ψ ( 2 n + 1 ) = ln ( π )
Where ψ ( n ) is the Digamma function . Solving numerically for positive n , this gives:
n + 1 ≈ 7 . 2 5 6 9
Since n + 1 is an integer, it should be 7 or 8 . Calculating both:
S ( 7 ) = 1 5 1 6 π 3 ≈ 3 3 . 0 7 3 4
S ( 8 ) = 3 π 4 ≈ 3 2 . 4 6 9 7
So, the maximum occurs at n + 1 = 7 , or n = 6