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nth_dim_ball

A proof for the volume of an Nth dimensional ball

https://github.com/nsp13737/nth_dim_ball

Science Score: 31.0%

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Repository

A proof for the volume of an Nth dimensional ball

Basic Info
  • Host: GitHub
  • Owner: NSP13737
  • Language: TeX
  • Default Branch: main
  • Size: 11.7 KB
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  • Stars: 0
  • Watchers: 1
  • Forks: 0
  • Open Issues: 0
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Created over 1 year ago · Last pushed over 1 year ago
Metadata Files
Readme Citation

README.md

This is an expository explanation finding the hyper volume of an nth dimensional ball. Please view the pdf for an easily viewable version of the paper.

The paper was written as a final research project for Math 147 (Honors Calculus 3), and is intended to be read by those with a similar experience in math.

Owner

  • Name: Nathan Phillips
  • Login: NSP13737
  • Kind: user
  • Company: University of Kansas

Citation (citations.txt) 

Citations:

proving the Beta function from the Gamma function: https://www.youtube.com/watch?v=E9sc1FnJF9k&ab_channel=PhysicsandMathLectures

https://math.libretexts.org/Bookshelves/Analysis/Complex_Variables_with_Applications_(Orloff)/14%3A_Analytic_Continuation_and_the_Gamma_Function/14.02%3A_Definition_and_properties_of_the_Gamma_function



AI Explanation of how gamma function accounts for different even and odd dimentions


'''
Ah, this proof actually does handle the even/odd dimension cases implicitly through the Gamma function! The final result:
Vₙ(R) = πⁿ/²Rⁿ/Γ(n/2+1)
contains Γ(n/2+1), where n/2 will be either an integer (for even n) or a half-integer (for odd n). The Gamma function behaves differently in these two cases:

For even n, n/2 is an integer k, so Γ(k+1) = k!
For odd n, n/2 is a half-integer k+1/2, so Γ(k+3/2) involves factors of √π

The apparent uniformity of the formula masks this even/odd distinction because it's all handled automatically by the properties of the Gamma function. The distinction becomes more visible if you try to evaluate specific cases:
For n=2: V₂(R) = πR²/Γ(2) = πR² (as Γ(2)=1)
For n=3: V₃(R) = π^(3/2)R³/Γ(5/2) = 4πR³/3
For n=4: V₄(R) = π²R⁴/Γ(3) = π²R⁴/2
For n=5: V₅(R) = π^(5/2)R⁵/Γ(7/2) = 8π²R⁵/15
So while other proofs might make the even/odd distinction explicit in their approach, this proof elegantly unifies both cases through the Gamma function. Would you like me to show how the Gamma function evaluations work out differently for even versus odd dimensions?
'''

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