Maximum Sphere

Geometry Level 4

Find the value of dimension n n such that the surface area of unit n n -sphere { ( x 1 , x 2 , , x n + 1 ) R n + 1 x 1 2 + x 2 2 + + x n + 1 2 = 1 } \left\{ \left( x_1,x_2,\cdots,x_{n+1} \right) \in\mathbb{R}^{n+1}\, \mid \, x_1^2+x_2^2+\cdots+x_{n+1}^2=1 \right\} is a maximum.

Hint: The surface area of n n -sphere of radius R R is: S n ( R ) = 2 π n + 1 2 Γ ( n + 1 2 ) R n , S_n(R)=\frac{2\pi^{\frac {n+1}2}}{\Gamma (\frac {n+1}2)}R^n, where Γ ( ) \Gamma(\cdot) denotes the gamma function .


The answer is 6.

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1 solution

Guilherme Niedu
Aug 2, 2018

For an n-sphere , its surface area is given by:

S n ( R ) = 2 π n 2 Γ ( n 2 ) R n \large \displaystyle S_n(R) = \frac{ 2 \pi ^{\frac{n}{2}} }{ \Gamma(\frac{n}{2}) } R^n

Where Γ ( n ) \Gamma(n) is the Gamma function . Specifically, for an unit n-sphere ( R = 1 ) (R=1) and since in this problem we're looking for an (n+1)-sphere:

S ( n + 1 ) = 2 π n + 1 2 Γ ( n + 1 2 ) \large \displaystyle S(n+1) = \frac{ 2 \pi ^{\frac{n+1}{2}} }{ \Gamma(\frac{n+1}{2}) }

Differentiating it with respect to n n and making it equal to zero:

d S ( n + 1 ) d n = 2 π n + 1 2 ln ( π ) 1 2 Γ ( n + 1 2 ) 2 π n + 1 2 Γ ( n + 1 2 ) ψ ( n + 1 2 ) 1 2 [ Γ ( n + 1 2 ) ] 2 = 0 \large \displaystyle \frac{dS(n+1)}{dn} = \frac{2 \pi ^{\frac{n+1}{2}} \cdot \ln(\pi) \cdot \frac12 \cdot \Gamma(\frac{n+1}{2}) - 2 \pi ^{\frac{n+1}{2}} \cdot \Gamma(\frac{n+1}{2}) \cdot \psi (\frac{n+1}{2}) \cdot \frac12}{\left [ \Gamma(\frac{n+1}{2}) \right ] ^2 } = 0

ψ ( n + 1 2 ) = ln ( π ) \large \displaystyle \psi \left ( \frac{n+1}{2} \right )= \ln(\pi)

Where ψ ( n ) \psi(n) is the Digamma function . Solving numerically for positive n n , this gives:

n + 1 7.2569 \large \displaystyle n+1 \approx 7.2569

Since n + 1 n+1 is an integer, it should be 7 7 or 8 8 . Calculating both:

S ( 7 ) = 16 π 3 15 33.0734 \large \displaystyle S(7) = \frac{16 \pi^3}{15} \approx 33.0734

S ( 8 ) = π 4 3 32.4697 \large \displaystyle S(8) = \frac{\pi^4}{3} \approx 32.4697

So, the maximum occurs at n + 1 = 7 \large \displaystyle n + 1= 7 , or n = 6 \color{#3D99F6} \boxed{\large \displaystyle n = 6}

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