I have this problem where I need to rewrite the logarithmic form of Stirling's series ln ⁡ ( z ! ) = 1 2 ln ⁡ ( 2 π ) + ( z + 1 2 ) ln ⁡ ( z ) − z + 1 12 z − 1 360 z 3 + 1 1260 z 5 − . . .

Question

I have this problem where I need to rewrite the logarithmic form of Stirling&#039;s series  ln  ⁡  (  z  !  )  =      1    2    ln  ⁡  (  2  π  )  +  (  z  +      1    2    )  ln  ⁡  (  z  )  −  z  +      1          12      z        −      1          360              z        3              +      1          1260              z        5              −  .  .  .

Esteban Johnson · Accepted Answer

We have  log  ⁡  (  z  !  )  ∼      1    2    log  ⁡  (  2  π  )  +      (          z      +              1        2              )    log  ⁡  z  −  z  +      1          12      z        −      1          360              z        3              +  …  .Taking the exponential of both sides gives                    z        !                            ∼        exp        ⁡                  (                                    1              2                        log            ⁡            (            2            π            )            +                          (                              z                +                                  1                  2                                            )                        log            ⁡            z            −            z            +                          1                              12                z                                      −                          1                              360                                  z                  3                                                      +            …                    )                                                  =        exp        ⁡                  (                      log            ⁡                          (                                                2                  π                                                  z                                      z                    +                    1                                          /                                        2                                                                    e                                      −                    z                                                              )                                )                exp        ⁡                  (                                    1                              12                z                                      −                          1                              360                                  z                  3                                                      +            …                    )                                                  =                  2          π                          z                      z            +            1                          /                        2                                    e                      −            z                          exp        ⁡                  (                                    1                              12                z                                      −                          1                              360                                  z                  3                                                      +            …                    )                .            To finish the derivation, we use the Maclaurin series of the exponential:                    exp        ⁡                  (                                    1                              12                z                                      −                          1                              360                                  z                  3                                                      +            …                    )                                    =        1        +                  (                                    1                              12                z                                      −                          1                              360                                  z                  3                                                      +            …                    )                +                                                            (                                                      1                                          12                      z                                                        −                                      1                                          360                                              z                        3                                                                              +                  …                                )                            2                                2                +        …                                          =        1        +                  1                      12            z                          +                  1                      288                          z              2                                      −                              139                                51840                          z              3                                      −        …        .

I have this problem where I need to rewrite the logarithmic form of Stirling's series ln &#x20

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