So we are dealing with a function (1) f ( t ) = ∑ n = 1 ∞ ( t − T ) n n ! ∑ m = 1 n ( m − 1 ) ! ( m + c ) ! ,

Question

So we are dealing with a function                    (1)                    f        (        t        )        =                  ∑                      n            =            1                                ∞                                                (            t            −            T                          )              n                                            n            !                                    ∑                      m            =            1                                n                                                (            m            −            1            )            !                                (            m            +            c            )            !                          ,

Kristina Petty · Accepted Answer

f  (  t  )  =      ∑          n      =      1              ∞                  (      t      −      T              )        n                    n      !            ∑          m      =      1              n                  (      m      −      1      )      !              (      m      +      c      )      !      For the time being, let   x  =  t  −  T and use the fact that      ∑          m      =      1              n                  (      m      −      1      )      !              (      m      +      c      )      !        =      1          c      2            (          1              Γ        (        c        )              −    c                  Γ        (        n        +        1        )                    Γ        (        c        +        n        +        1        )              )    g  (  x  )  =                    e        x            −      1                      c        2            Γ      (      c      )        −                    e        x                    x                  −          c                                    c        2            Γ      (      c      )                  [        Γ  (  c  +  1  )  −  Γ  (  c  +  1  ,  x  )            ]      Expanded as series around   x  =  0  g  (  x  )  =                    c        2            +      5      c      +      6              Γ      (      c      +      4      )        x            [        1  +                    c        2            +      6      c      +      9              2              (                  c          2                +        5        c        +        6        )              x  +                    c        2            +      6      c      +      11              6              (                  c          2                +        5        c        +        6        )                  x    2    +  O      (          x      3        )              ]      which does not seem to make any problem

So we are dealing with a function <mtable displaystyle="true"> <mlabeledtr> <mtd id="m

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