The (hyper)ball has been dropped

Geometry Level 3

Let V n V_n be the n n -volume of the unit hyperball (the solid hypersphere) in R n \mathbb{R}^{n} . It turns out that V 2019 V 2018 \frac{V_{2019}}{V_{2018}} is a rational number. If we write this number as p q \frac{p}{q} , in lowest terms, find the sum of the smallest prime factor of p p and the largest prime factor of q q .


The answer is 2019.

Otto Bretscher by Otto Bretscher , 65, USA

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2 solutions

Otto Bretscher
Jan 1, 2019

Mark Hennings has submitted a "fancy" solution, based on a formula for V n V_n .

For the sake of variety, let me propose a solution "from first principles."

A cross section through a unit n n -hyperball at x n = u x_n=u , where ∣ u ∣ ≤ 1 |u|\leq 1 , is an ( n − 1 ) (n-1) -hyperball with radius r = 1 − u 2 r=\sqrt{1-u^2} and ( n − 1 ) (n-1) -volume r n − 1 V n − 1 r^{n-1}V_{n-1} . Thus V n = ∫ − 1 1 ( 1 − u 2 ) n − 1 2 V n − 1 d u = 2 V n − 1 ∫ 0 π 2 cos ⁡ n t d t V_n=\int_{-1}^{1} (1-u^2)^{\frac{n-1}{2}}V_{n-1}du=2V_{n-1}\int_0^\frac{\pi}{2}\cos^nt\ dt . From introductory calculus we recall that ∫ 0 π 2 cos ⁡ n t d t = ( n − 1 ) ! ! n ! ! \int_0^\frac{\pi}{2}\cos^nt\ dt=\frac{(n-1)!!}{n!!} for odd n n , where n ! ! n!! denotes the double factorial. Thus V 2019 V 2018 = 2 ( 2018 ) ! ! 2019 ! ! \frac{V_{2019}}{V_{2018}}=\frac{2(2018)!!}{2019!!} . Since the numerator is even and the largest prime factor of the (odd) denominator is 2017, the answer is, fittingly for the occasion, 2019 \boxed{2019} .

1 Helpful 0 Interesting 2 Brilliant 0 Confused

Wow Sir!!!! I am amazed by your talent of transforming solutions which may seem pretty far fetched and use heavy machinery, into those which are simpler and easy to understand for people who are only introduced to calculus!!!! And also, a very Happy New Year, Sir.......!!

P.S. I found this a good read for today.......!!! Just thought to share it......:)

Aaghaz Mahajan - 2 years, 5 months ago
Mark Hennings
Jan 1, 2019

The volume of the n n -dimensional unit ball is V n = π 1 2 n Γ ( 1 2 n + 1 ) V_n \; = \; \frac{\pi^{\frac12n}}{\Gamma(\frac12n+1)} and hence V n + 1 V n = π n + 1 2 Γ ( 1 2 n + 1 ) π 1 2 n Γ ( 1 2 ( n + 1 ) + 1 ) = π Γ ( 1 2 n + 1 ) Γ ( 1 2 n + 3 2 ) \frac{V_{n+1}}{V_{n}} \; = \; \frac{\pi^{\frac{n+1}{2}}\Gamma(\frac12n+1)}{\pi^{\frac12n}\Gamma(\frac12(n+1)+1)} \; = \; \frac{\sqrt{\pi}\Gamma(\frac12n+1)}{\Gamma(\frac12n+\frac32)} In particular V 2 n + 1 V 2 n = π Γ ( n + 1 ) Γ ( n + 3 2 ) = 2 2 n + 1 ( n ! ) 2 ( 2 n + 1 ) ! \frac{V_{2n+1}}{V_{2n}} \; = \; \frac{\sqrt{\pi}\Gamma(n+1)}{\Gamma(n+\frac32)} \; = \; \frac{2^{2n+1}(n!)^2}{(2n+1)!} In the case n = 1009 n=1009 , we see that V 2019 V 2018 = p q \frac{V_{2019}}{V_{2018}} = \frac{p}{q} where q q is a factor of 2019 ! 2019! . Since 2 2019 ( 1009 ! ) 2 2^{2019}(1009!)^2 is not divisible by 2017 2017 , we deduce that the largest prime factor of q q is 2017 2017 . On the other hand, the index of 2 2 in 1009 ! 1009! is 1002 1002 , while the index of 2 2 in 2019 ! 2019! is 2011 2011 . Thus it is clear that p p is even while q q is odd, and so the smallest prime factor of p p is 2 2 , making the answer 2 + 2017 = 2019 2 + 2017 = \boxed{2019} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes, indeed, once you have a formula for V n V_n , everything falls into place. Happy 2019, Mark!

Otto Bretscher - 2 years, 5 months ago

n 0 = 1 & n_0=1\&

Table [ n i = Evaluate [ FullSimplify [ Evaluate [ ∫ − $#$1 $#$1 n i − 1 ( $#$1 2 − x 2 ) d x ] , $#$1 ∈ R ∧ $#$1 ≥ 0 ] ] & , { i , Range [ 1 , 10 ] } ] \text{Table}\left[n_i=\text{Evaluate}\left[\text{FullSimplify}\left[\text{Evaluate}\left[\int_{-\text{\$\#\$1}}^{\text{\$\#\$1}} n_{i-1}\left(\sqrt{\text{\$\#\$1}^2-x^2}\right) \, dx\right],\text{\$\#\$1}\in \mathbb{R}\land \text{\$\#\$1}\geq 0\right]\right]\&,\{i,\text{Range}[1,10]\}\right]

{ 2 $#$1 & , π $#$1 2 & , 4 π $#$1 3 3 & , π 2 $#$1 4 2 & , 8 π 2 $#$1 5 15 & , π 3 $#$1 6 6 & , 16 π 3 $#$1 7 105 & , π 4 $#$1 8 24 & , 32 π 4 $#$1 9 945 & , π 5 $#$1 10 120 & } \left\{2 \text{\$\#\$1}\&,\pi \text{\$\#\$1}^2\&,\frac{4 \pi \text{\$\#\$1}^3}{3}\&,\frac{\pi ^2 \text{\$\#\$1}^4}{2}\&,\frac{8 \pi ^2 \text{\$\#\$1}^5}{15}\&,\frac{\pi ^3 \text{\$\#\$1}^6}{6}\&,\frac{16 \pi ^3 \text{\$\#\$1}^7}{105}\&,\frac{\pi ^4 \text{\$\#\$1}^8}{24}\&,\frac{32 \pi ^4 \text{\$\#\$1}^9}{945}\&,\frac{\pi ^5 \text{\$\#\$1}^{10}}{120}\&\right\}

Proof by induction gives:

v = π $#$1 2 $#$2 $#$1 Γ ( $#$1 2 + 1 ) & ; v=\frac{\pi ^{\frac{\text{\$\#\$1}}{2}} \text{\$\#\$2}^{\text{\$\#\$1}}}{\Gamma \left(\frac{\text{\$\#\$1}}{2}+1\right)}\&;

Then, as you said, the rest is simple:

v ( 2019 , 1 ) v ( 2018 , 1 ) \frac{v(2019,1)}{v(2018,1)} is

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A Former Brilliant Member - 2 years, 5 months ago

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