Given that x 0 is the unique positive solution of ( 2 − x ) n + 1 = x ( x + 1 ) ⋯ ( x + n ), try to find the asymptotic value of M = ∏ k = 0 n ( k + 2 k + x 0 ) k + 2 with absolute error o ( 1 ) as n → ∞, where H n denotes n-th harmonic number ∑ k = 1 n 1 / k

Question

Given that       x    0   is the unique positive solution of   (  2  −  x      )          n      +      1        =  x  (  x  +  1  )  ⋯  (  x  +  n  ), try to find the asymptotic value of  M  =      ∏          k      =      0        n              (                        k          +          2                          k          +                      x            0                              )              k      +      2      with absolute error   o  (  1  ) as   n  →  ∞, where       H    n   denotes n-th harmonic number       ∑          k      =      1        n    1      /    k

periasemdy · Accepted Answer

All 0-notations and o-notations work for   n  →  ∞First we have       2          n      +      1        ≥  n  !      x    0  , hence       x    0    ≤      2          n      +      1            /    n  !  =  O  (      2    n        /    n  !  ). Take logarithm, we derive that                    (        n        +        1        )        ln        ⁡        (        2        −                  x          0                )                            =        ln        ⁡                  x          0                +                  ∑                      k            =            1                    n                ln        ⁡        (                  x          0                +        k        )                                          =        ln        ⁡        n        !        +        ln        ⁡                  x          0                +                  ∑                      k            =            1                    n                ln        ⁡                  (          1          +                                    x              0                        k                    )                    therefore      x    0    =                    2                  n          +          1                    (      1      +      O      (      n              x        0            )      )              n      !      We have enough stuff to estimate   ln  ⁡  M now.                    ln        ⁡        M        =        (        n        +        1        )        ln        ⁡        (        n        +        1        )        +        (        n        +        2        )        ln        ⁡        (        n        +        2        )        −        2        (        n        +        1        )        ln        ⁡        2                                    +                              2                          n              +              1                                            n            !                          −                              4            n                                n                          !              2                                      (        n        +        2                  H          n                +        1        )        +        O        (                  n          3                          x          0          3                )            Finally, we have                    exp        ⁡                  (                                    2                              n                +                1                                                    n              !                                )                =        1        +                              2                          n              +              1                                            n            !                          +                              2            ⋅                          4              n                                            n                          !              2                                      +        O        (                  x          0          3                )                            exp        ⁡                  (          −                                    4              n                                      n                              !                2                                              (          n          +          2                      H            n                    +          1          )          )                =        1        −                              4            n                                n                          !              2                                      (        n        +        2                  H          n                +        1        )        +        O        (                  n          2                          x          0          4                )            and                    M                            =                              (            n            +            1                          )                              n                +                1                                      (            n            +            2                          )                              n                +                2                                                          4                          n              +              1                                                                                          (          1          +                                    2                              n                +                1                                                    n              !                                −                                    4              n                                      n                              !                2                                              (          n          +          2                      H            n                    −          1          )          )                                                                  (        1        +        O        (                  n          3                          x          0          3                )        )            Notice that the absolute error  O      (                            n          3                (        n        +        1                  )                      n            +            1                          (        n        +        2                  )                      n            +            2                                      4                  n          +          1                            x      0      3        )  approaches 0 when   n  →  ∞

Given that x_0 is the unique positive solution of (2−x)^(n+1)=x(x+1)⋯(x+n), try to find the asymptotic value of M=prod^n_(k=0)((k+2)/(k+x_0))^(k+2)

Answered question

Answer & Explanation

New Questions in Precalculus