Hypersphere!

Algebra Level 5

Let V n ( r ) V_n (r) be the volume of the n n -dimensional ball of radius r r . So, V 2 ( r ) = π r 2 , V 3 ( r ) = 4 3 π r 3 , V 4 ( r ) = π 2 r 4 2 , . . . . V_2(r)=\pi r^2, V_3(r)=\dfrac {4}{3}\pi r^3, V_4(r)= \dfrac{\pi^2 r^4}{2},.... \ We also set V 0 ( r ) = 1 V_0(r)=1 .

Let S = V 0 ( 1 ) + V 2 ( 1 ) + V 4 ( 1 ) + . . . . S=V_0(1)+V_2(1)+V_4(1)+. . . . \ . Find the value of ⌊ S ⌋ . \lfloor S \rfloor.

Notation: ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 23.

Sathvik Acharya by Sathvik Acharya , 17, India

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1 solution

The volume of n n -hypersphere is given by V n ( r ) = S n r 2 n V_n(r) = \dfrac {S_nr^2}n for n ≥ 2 n \ge 2 , where S n = 2 π n / 2 Γ ( n 2 ) S_n = \dfrac {2\pi^{n/2}}{\Gamma \left(\frac n2\right)} and Γ ( ⋅ ) \Gamma(\cdot) denotes the gamma function . Therefore,

S = V 0 ( 1 ) + V 2 ( 1 ) + V 4 ( 1 ) + ⋯ = 1 + ∑ k = 1 ∞ S 2 k ( 1 2 ) 2 k = 1 + ∑ k = 1 ∞ π k k Γ ( k ) = 1 + ∑ k = 1 ∞ π k k ! = ∑ k = 0 ∞ π k k ! = e π \begin{aligned} S & = V_0(1) + V_2(1) + V_4(1) + \cdots \\ & = 1 + \sum_{k=1}^\infty \frac {S_{2k}(1^2)}{2k} \\ & = 1 + \sum_{k=1}^\infty \frac {\pi^k}{k\Gamma(k)} \\ & = 1 + \sum_{\color{#3D99F6}k=1}^\infty \frac {\pi^k}{k!} \\\ & = \sum_{\color{#D61F06}k=0}^\infty \frac {\pi^k}{k!} \\ & = e^\pi \end{aligned}

⟹ ⌊ S ⌋ = ⌊ e π ⌋ = ⌊ 23.140... ⌋ = 23 \implies \lfloor S \rfloor = \lfloor e^\pi \rfloor = \lfloor 23.140... \rfloor = \boxed{23}

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