n-th derivative of an arbitrary polynomial / power seriesCompose two special power series expansionsSymbolic linear algebra gradients/matrix calculusPartial derivative: Understanding outputterm derivationSimplify sum of unnatural conditional functions generated by derivativeTaking derivative of general orderRelations between two reciprocal partial derivatives?

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n-th derivative of an arbitrary polynomial / power series


Compose two special power series expansionsSymbolic linear algebra gradients/matrix calculusPartial derivative: Understanding outputterm derivationSimplify sum of unnatural conditional functions generated by derivativeTaking derivative of general orderRelations between two reciprocal partial derivatives?






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2














$begingroup$


How can I tell mathematica to give me the symbolic derivative of the following sum
$$ fracpartial^npartial x^nsum_j=0^Ngamma_jx^j=sum_j=n^Nfracj!(j-n)! gamma_jx^j-n $$
I only get the left hand side with my simple but wrong code:



G=Sum[Subscript[[Gamma], i]*(x)^i, i, 0, N]
Assuming[n <= N, D[G, x, n]


Quick followup question: How do I get $gamma_nn!$ for x=0?










share|improve this question







New contributor



the.polo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$






















    2














    $begingroup$


    How can I tell mathematica to give me the symbolic derivative of the following sum
    $$ fracpartial^npartial x^nsum_j=0^Ngamma_jx^j=sum_j=n^Nfracj!(j-n)! gamma_jx^j-n $$
    I only get the left hand side with my simple but wrong code:



    G=Sum[Subscript[[Gamma], i]*(x)^i, i, 0, N]
    Assuming[n <= N, D[G, x, n]


    Quick followup question: How do I get $gamma_nn!$ for x=0?










    share|improve this question







    New contributor



    the.polo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$


















      2












      2








      2





      $begingroup$


      How can I tell mathematica to give me the symbolic derivative of the following sum
      $$ fracpartial^npartial x^nsum_j=0^Ngamma_jx^j=sum_j=n^Nfracj!(j-n)! gamma_jx^j-n $$
      I only get the left hand side with my simple but wrong code:



      G=Sum[Subscript[[Gamma], i]*(x)^i, i, 0, N]
      Assuming[n <= N, D[G, x, n]


      Quick followup question: How do I get $gamma_nn!$ for x=0?










      share|improve this question







      New contributor



      the.polo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      How can I tell mathematica to give me the symbolic derivative of the following sum
      $$ fracpartial^npartial x^nsum_j=0^Ngamma_jx^j=sum_j=n^Nfracj!(j-n)! gamma_jx^j-n $$
      I only get the left hand side with my simple but wrong code:



      G=Sum[Subscript[[Gamma], i]*(x)^i, i, 0, N]
      Assuming[n <= N, D[G, x, n]


      Quick followup question: How do I get $gamma_nn!$ for x=0?







      calculus-and-analysis symbolic polynomials






      share|improve this question







      New contributor



      the.polo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share|improve this question







      New contributor



      the.polo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      share|improve this question




      share|improve this question



      share|improve this question






      New contributor



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      asked 8 hours ago









      the.polothe.polo

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      1112 bronze badges




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          1 Answer
          1






          active

          oldest

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          3
















          $begingroup$

          Clear["Global`*"]

          rule = D[Sum[expr_, k_Symbol, kmin_, kmax_], x_, n_] :>
          Sum[D[expr, x, n], k, kmin, kmax];

          Format[γ[j_]] := Subscript[γ, j]

          expr = Sum[γ[j] x^j, j, 0, n];

          (expr2 = D[expr, x, n] /. rule) // TraditionalForm


          enter image description here



          where FactorialPower is used and expr2 equivalent to



          TraditionalForm[
          expr3 = ReplacePart[expr2, 1 -> FunctionExpand[expr2[[1]]]] /.
          Gamma[z_] :> (z - 1)!]


          enter image description here






          share|improve this answer










          $endgroup$
















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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3
















            $begingroup$

            Clear["Global`*"]

            rule = D[Sum[expr_, k_Symbol, kmin_, kmax_], x_, n_] :>
            Sum[D[expr, x, n], k, kmin, kmax];

            Format[γ[j_]] := Subscript[γ, j]

            expr = Sum[γ[j] x^j, j, 0, n];

            (expr2 = D[expr, x, n] /. rule) // TraditionalForm


            enter image description here



            where FactorialPower is used and expr2 equivalent to



            TraditionalForm[
            expr3 = ReplacePart[expr2, 1 -> FunctionExpand[expr2[[1]]]] /.
            Gamma[z_] :> (z - 1)!]


            enter image description here






            share|improve this answer










            $endgroup$



















              3
















              $begingroup$

              Clear["Global`*"]

              rule = D[Sum[expr_, k_Symbol, kmin_, kmax_], x_, n_] :>
              Sum[D[expr, x, n], k, kmin, kmax];

              Format[γ[j_]] := Subscript[γ, j]

              expr = Sum[γ[j] x^j, j, 0, n];

              (expr2 = D[expr, x, n] /. rule) // TraditionalForm


              enter image description here



              where FactorialPower is used and expr2 equivalent to



              TraditionalForm[
              expr3 = ReplacePart[expr2, 1 -> FunctionExpand[expr2[[1]]]] /.
              Gamma[z_] :> (z - 1)!]


              enter image description here






              share|improve this answer










              $endgroup$

















                3














                3










                3







                $begingroup$

                Clear["Global`*"]

                rule = D[Sum[expr_, k_Symbol, kmin_, kmax_], x_, n_] :>
                Sum[D[expr, x, n], k, kmin, kmax];

                Format[γ[j_]] := Subscript[γ, j]

                expr = Sum[γ[j] x^j, j, 0, n];

                (expr2 = D[expr, x, n] /. rule) // TraditionalForm


                enter image description here



                where FactorialPower is used and expr2 equivalent to



                TraditionalForm[
                expr3 = ReplacePart[expr2, 1 -> FunctionExpand[expr2[[1]]]] /.
                Gamma[z_] :> (z - 1)!]


                enter image description here






                share|improve this answer










                $endgroup$



                Clear["Global`*"]

                rule = D[Sum[expr_, k_Symbol, kmin_, kmax_], x_, n_] :>
                Sum[D[expr, x, n], k, kmin, kmax];

                Format[γ[j_]] := Subscript[γ, j]

                expr = Sum[γ[j] x^j, j, 0, n];

                (expr2 = D[expr, x, n] /. rule) // TraditionalForm


                enter image description here



                where FactorialPower is used and expr2 equivalent to



                TraditionalForm[
                expr3 = ReplacePart[expr2, 1 -> FunctionExpand[expr2[[1]]]] /.
                Gamma[z_] :> (z - 1)!]


                enter image description here







                share|improve this answer













                share|improve this answer




                share|improve this answer



                share|improve this answer










                answered 7 hours ago









                Bob HanlonBob Hanlon

                66.2k3 gold badges37 silver badges102 bronze badges




                66.2k3 gold badges37 silver badges102 bronze badges
























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