An equal expression to $sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$












0












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Is there a way to rewrite an equal expression to the following



$$sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$$



but without the sum?



$z$ is integer and $xgeq 0$.






real-analysis calculus sequences-and-series summation







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Is there a way to rewrite an equal expression to the following



    $$sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$$



    but without the sum?



    $z$ is integer and $xgeq 0$.






    real-analysis calculus sequences-and-series summation







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Is there a way to rewrite an equal expression to the following



      $$sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$$



      but without the sum?



      $z$ is integer and $xgeq 0$.






      real-analysis calculus sequences-and-series summation







      share|cite|improve this question









      $endgroup$




      Is there a way to rewrite an equal expression to the following



      $$sum^{z}_{n=0}nfrac{x^{z-n}}{(z-n)!}$$



      but without the sum?



      $z$ is integer and $xgeq 0$.






      real-analysis calculus sequences-and-series summation




      real-analysis calculus sequences-and-series summation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 3 at 22:31









      Y.LY.L

      597




      597






















          2 Answers
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          $begingroup$

          The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
            $endgroup$
            – Y.L
            Jan 4 at 0:28



















          0












          $begingroup$

          Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)



          begin{align}
          sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
          &= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
          = nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
          &= nP_n(x) - xfrac{d}{dx}P_n(x)
          end{align}



          where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as



          $$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
          hence,
          $$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
          We may then compute



          begin{align}
          frac{d}{dx}P_n(x)
          &= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
          &= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
          &= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
          end{align}



          Solving and simplifying, we get



          begin{align}
          sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
          &= nP_n(x) - xfrac{d}{dx}P_n(x)\
          &= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
          end{align}






          share|cite|improve this answer









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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
              $endgroup$
              – Y.L
              Jan 4 at 0:28
















            0












            $begingroup$

            The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
              $endgroup$
              – Y.L
              Jan 4 at 0:28














            0












            0








            0





            $begingroup$

            The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.






            share|cite|improve this answer









            $endgroup$



            The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)sum_{k=1}^{z-1} frac {x^{k}} {k!} +frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 3 at 23:38









            Kavi Rama MurthyKavi Rama Murthy

            54.2k32055




            54.2k32055












            • $begingroup$
              Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
              $endgroup$
              – Y.L
              Jan 4 at 0:28


















            • $begingroup$
              Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
              $endgroup$
              – Y.L
              Jan 4 at 0:28
















            $begingroup$
            Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
            $endgroup$
            – Y.L
            Jan 4 at 0:28




            $begingroup$
            Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression.
            $endgroup$
            – Y.L
            Jan 4 at 0:28











            0












            $begingroup$

            Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)



            begin{align}
            sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
            &= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
            = nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
            &= nP_n(x) - xfrac{d}{dx}P_n(x)
            end{align}



            where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as



            $$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
            hence,
            $$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
            We may then compute



            begin{align}
            frac{d}{dx}P_n(x)
            &= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
            &= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
            &= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
            end{align}



            Solving and simplifying, we get



            begin{align}
            sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
            &= nP_n(x) - xfrac{d}{dx}P_n(x)\
            &= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
            end{align}






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)



              begin{align}
              sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
              &= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
              = nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
              &= nP_n(x) - xfrac{d}{dx}P_n(x)
              end{align}



              where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as



              $$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
              hence,
              $$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
              We may then compute



              begin{align}
              frac{d}{dx}P_n(x)
              &= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
              &= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
              &= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
              end{align}



              Solving and simplifying, we get



              begin{align}
              sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
              &= nP_n(x) - xfrac{d}{dx}P_n(x)\
              &= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
              end{align}






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)



                begin{align}
                sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
                &= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
                = nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
                &= nP_n(x) - xfrac{d}{dx}P_n(x)
                end{align}



                where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as



                $$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
                hence,
                $$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
                We may then compute



                begin{align}
                frac{d}{dx}P_n(x)
                &= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
                &= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
                &= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
                end{align}



                Solving and simplifying, we get



                begin{align}
                sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
                &= nP_n(x) - xfrac{d}{dx}P_n(x)\
                &= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
                end{align}






                share|cite|improve this answer









                $endgroup$



                Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)



                begin{align}
                sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
                &= sum_{k=0}^{n} (n-k) frac{x^k}{k!}
                = nsum_{k=0}^n frac{x^k}{k!} - sum_{k=0}^n xfrac{d}{dx} frac{x^k}{k!}\
                &= nP_n(x) - xfrac{d}{dx}P_n(x)
                end{align}



                where $P_n(x) = sum_{k=0}^n frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $Gamma(n,x) = int_x^{infty} x^{n-1}e^{-x};dx$ may be expressed as



                $$Gamma(n,x) = (n-1)! e^{-x} sum_{k=0}^{n-1}frac{x^k}{k!}$$
                hence,
                $$P_n(x) = frac{Gamma(n+1,x)}{n!} e^x$$
                We may then compute



                begin{align}
                frac{d}{dx}P_n(x)
                &= P_n(x) frac{d}{dx} ln P_n(x) = P_n(x) left[,frac{frac{partial}{partial{x}}Gamma(n+1,x)}{Gamma(n+1,x)} + 1,right]\
                &= frac{Gamma(n+1,x)}{n!} e^x left[,frac{-x^ne^{-x}}{Gamma(n+1,x)} + 1,right]\
                &= frac{Gamma(n+1,x)}{n!} e^x - frac{x^n}{n!}
                end{align}



                Solving and simplifying, we get



                begin{align}
                sum_{k=0}^n k frac{x^{n-k}}{(n-k)!}
                &= nP_n(x) - xfrac{d}{dx}P_n(x)\
                &= frac{1}{n!} left[,(n-x)Gamma(n+1,x)e^x + x^{n+1},right]
                end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 1:18









                adfriedmanadfriedman

                3,161169




                3,161169






























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