Finding the volume of an n-dimensional simplex (recursion)I want to find the volume of an n-dimensional simplex, i.e. determine σ n = ∫ A n d μ , A n = { ( x 1 , … , x n ) ∈ R n ∣ ∀ i : x i ⩾ 0 , x 1 + … + x n ⩽ 1 } . I came up with the following informal computation: We consider A ′ = [ 0 , 1 ] , A x 1 = { ( x 2 , … , x n ) ∣ x 2 + … + x n ⩽ 1 − x 1 } and conclude σ n = ∫ 0 1 ( ∫ A x 1 d μ ( x 2 , … , x n ) ) d x 1 = ( ∗ ) ∫ 0 1 σ n − 1 ( 1 − x 1 ) n − 1 d x 1 = σ n − 1 [ − 1 n ( 1 − x 1 ) n ] 0 1 = σ n − 1 n . Solving this recursion we find σ 1 = ∫ 0 1 d x 1 = 1 ⟹ σ n = σ 1 n ! = 1 n ! . How I can formalise the (∗) step?

Question

Finding the volume of an n-dimensional simplex (recursion)I want to find the volume of an n-dimensional simplex, i.e. determine                               σ                      n                          =                  ∫                                    A                              n                                                                                             d                μ        ,                          A                      n                          =        {        (                  x                      1                          ,        …        ,                  x                      n                          )        ∈                              R                                n                          ∣        ∀        i        :                  x                      i                          ⩾        0        ,                          x                      1                          +        …        +                  x                      n                          ⩽        1        }        .            I came up with the following informal computation: We consider                              A          ′                =        [        0        ,        1        ]        ,                          A                                    x                              1                                                    =        {        (                  x                      2                          ,        …        ,                  x                      n                          )        ∣                  x                      2                          +        …        +                  x                      n                          ⩽        1        −                  x                      1                          }            and conclude                              σ                      n                                              =                  ∫                      0                                1                                    (                      ∫                                          A                                                      x                                          1                                                                                                                                                     d                    μ          (                      x                          2                                ,          …          ,                      x                          n                                )          )                                     d                          x                      1                                                                                                =                                      (              ∗              )                                                ∫                      0                                1                                    σ                      n            −            1                          (        1        −                  x                      1                                    )                      n            −            1                                               d                          x                      1                                                            =                  σ                      n            −            1                                                [            −                          1              n                        (            1            −                          x                              1                                                    )              n                        ]                                0                                1                          =                              σ                          n              −              1                                n                .            Solving this recursion we find                              σ                      1                          =                  ∫                      0                                1                                               d                          x                      1                          =        1                ⟹                          σ                      n                          =                              σ                          1                                            n            !                          =                  1                      n            !                          .            How I can formalise the (∗) step?

Janiah Hoffman · Accepted Answer

Step 1The interval can be iteratively splitted up, i.e. first you have                                    A          ′                =        {        (                  x                      1                          ,        …        ,                  x                      n            −            1                          )        ∣                  x                      1                          +        ⋯        +                  x                      n            −            1                          ⩽        1        }                                            A                      (                          x                              1                                      ,            …            ,                          x                              n                −                1                                      )                          =        {                  x                      n                          ∣                  x                      n                          ⩽        1        −                  x                      1                          −        ⋯        −                  x                      n            −            1                          }            yielding                              ∫                                    A                              n                                                                                             d                μ                            =                  ∫                                    A              ′                                                                    (                      ∫                          A              (                              x                                  1                                            ,              …              ,                              x                                  n                  −                  1                                            )                                                                                 d                                x                          n                                )                                     d                μ        (                  x                      n            −            1                          ,        …        ,                  x                      1                          )                                          =                  ∫                                    A              ′                                                                    ∫                      0                                1            −                          x                              1                                      −            ⋯            −                          x                              n                −                1                                                                         d                          x                      n                                               d                μ        (                  x                      n            −            1                          ,        …        ,                  x                      1                          )        .            Continuing this procedure with A′ one ends up with                              σ                      n                          =                  ∫                      0                                1                                    ∫                      0                                1            −                          x                              1                                                    ⋯                  ∫                      0                                1            −                          x                              1                                      −            ⋯            −                          x                              n                −                1                                                                         d                          x                      n                          ⋯                             d                          x                      2                                               d                          x                      1                          .            Step 2To compute this, we write                              f                      n                          (        r        )        =                  ∫                      0                                r                                    ∫                      0                                r            −                          x                              1                                                    ⋯                  ∫                      0                                r            −                          x                              1                                      −            ⋯            −                          x                              n                −                1                                                                         d                          x                      n                          ⋯                             d                          x                      2                                               d                          x                      1                          .            This allows us to establish a recursion, namely                              f                      n            +            1                          (        r        )                            =                  ∫                      0                                r                          ⋯                  ∫                      0                                r            −                          x                              1                                      −            ⋯            −                          x                              n                                                                         d                          x                      n            +            1                          ⋯                             d                          x                      1                          =                  ∫                      0                                r                                    f                      n                          (        r        −                  x                      1                          )                             d                          x                      1                                                            =        −                  ∫                      r                                0                                    f                      n                          (                  x                      1                          )                             d                          x                      1                          =                  ∫                      0                                r                                    f                      n                          (                  x                      1                          )                             d                          x                      1                                                                    ⟹                          f                      n            +            1                    ′                (        r        )        =                  f                      n                          (        r        )        .            With the base case       f          1        (  r  )  =  r this yields                              f                      n                          (        r        )        =                              r                          n                                            n            !                                  ⟹                          f                      n                          (        1        )        =                  1                      n            !                          .

Find the volume of an n-dimensional simplex, i.e. determine sigma_n=int_{A_{n}} d mu, A_n={(x_1,...,x_n) in mathbb{R}^n| forall i:x_i geq 0, x_1+...+x_n leq 1}

Answered question

Answer & Explanation

New Questions in High school geometry