A sharp Stirling's approximation form states that n ! ∼ ( n e ) n 2 π n .Use that form to show that: ( 2 m m ) = Θ ( 2 2 m m ) .

Question

A sharp Stirling&#039;s approximation form states that  n  !  ∼            (              n        e            )        n        2    π    n    .Use that form to show that:                    (                            2        m            m                      )              =  Θ      (                  2                  2          m                            m              )    .

Braylon Perez · Accepted Answer

Using Stirling’s approximation, we have                                                        (                                                          2              n                        n                                              )                                                          =                              (            2            n            )            !                                n                          !              2                                                                        ≈                                            2              π              (              2              n              )                        (            2            n                          /                        e                          )                              2                n                                                                        (                              2                π                n                            (              n                              /                            e                              )                n                            )                        2                                                            =                              2                          π              n                                      2                              2                n                                      (            n                          /                        e                          )                              2                n                                                          2            π            n            (            n                          /                        e                          )                              2                n                                                                                      =                              2                          2              n                                            π            n                                  ;            it only remains to show that this approximation is good enough to justify the claim that                     (                            2        n            n                      )             is   Θ      (                  2                  2          n                            π        n              )   For this you have to understand that   f  (  n  )  ∼  g  (  n  ) means that       lim          n      →      ∞                  f      (      n      )              g      (      n      )        =  1. Thus, we’re given not just that n! is approximately       2    π    n              (              n        e            )        n  , but that      lim          n      →      ∞                  n      !                      2        π        n                              (                      n            e                    )                n              =  1    .Thus,                    (        2        n        )        !                    n                  !          2                                              2          π          (          2          n          )                (        2        n                  /                e                  )                      2            n                                                (                      2            π            n                    (          n                      /                    e                      )            n                    )                2              =            (      2      n      )      !                      2        π        (        2        n        )            (      2      n              /            e              )                  2          n                      ⋅                    (                  2          π          n                (        n                  /                e                  )          n                )            2              n              !        2              →      1    2    =  1as   n  →  ∞, i.e.,                    (                            2        n            n                      )              ∼            2              2        n                    π      n          .You shouldn’t have much trouble showing that if   f  (  n  )  ∼  g  (  n  ), then   f  (  n  ) is   Θ            (        g  (  n  )            )      , which gives you your result; just use the definitions of limit and   Θ.

Averi Mitchell · Accepted Answer

(                                      2          m                m                              )                      =            (      2      m      )      !              (      m      !              )        2              ∼                              (                                      2          m                e                                          )                                    2          m                            4        π        m                    (                                    (                                    m          e                                                    )                                m                          2          π          m                            )        2            Since      0.5              2              2        m                    m        ≤            2              2        m                    π      m        ≤      2              2              2        m                    π      m

A sharp Stirling's approximation form states that n ! ∼

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