Change arguments of a product of functions

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3












$begingroup$


I am trying to do the following with rule-based pattern matching



(f1[t] f2[t] f3[t]) /. x__ -> n[
Product[x[[i, 0]][om[i]], i, 3]] // Quiet


which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages



"Part specification x[[1,0]] is longer than depth of object"


I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?










share|improve this question









$endgroup$
















    3












    $begingroup$


    I am trying to do the following with rule-based pattern matching



    (f1[t] f2[t] f3[t]) /. x__ -> n[
    Product[x[[i, 0]][om[i]], i, 3]] // Quiet


    which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages



    "Part specification x[[1,0]] is longer than depth of object"


    I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?










    share|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      I am trying to do the following with rule-based pattern matching



      (f1[t] f2[t] f3[t]) /. x__ -> n[
      Product[x[[i, 0]][om[i]], i, 3]] // Quiet


      which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages



      "Part specification x[[1,0]] is longer than depth of object"


      I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?










      share|improve this question









      $endgroup$




      I am trying to do the following with rule-based pattern matching



      (f1[t] f2[t] f3[t]) /. x__ -> n[
      Product[x[[i, 0]][om[i]], i, 3]] // Quiet


      which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages



      "Part specification x[[1,0]] is longer than depth of object"


      I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?







      pattern-matching






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 17 at 9:45









      Display NameDisplay Name

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      57438




















          2 Answers
          2






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          3












          $begingroup$

          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[i = 1, f1[t] f2[t] f3[t] /. t :> om[i++] // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, Inactivate[f1[t] f1[t] f1[t]] /. t :> om[i++] // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Update 2: "... if one replaces e.g. f1[t] to become (x1[t]-x2[t])"



          The first method works if the level specification -2 is used in MapIndexed:



          Inactivate[f1[t] f2[t] f3[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Inactivate[f1[t] f1[t] f1[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])] /. 
          Inactive[Times][x_, y__] :>
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]




          The second method can also be modified (so that ReplaceAll is mapped onto the terms of the product) to work for all three cases:



          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f2[t] f3[t]]]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f1[t] f1[t]]]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])]]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            Jan 17 at 11:56











          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            Jan 17 at 12:05










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            Jan 17 at 12:18











          • $begingroup$
            I think the code must be edited once more if one replaces e.g. f1[t] to become (x1[t]-x2[t]).
            $endgroup$
            – Display Name
            Jan 17 at 14:14










          • $begingroup$
            @DisplayName, please see update 2.
            $endgroup$
            – kglr
            Jan 18 at 10:52


















          1












          $begingroup$

          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]

          changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]







          share|improve this answer









          $endgroup$












          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            Jan 17 at 12:32










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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          3












          $begingroup$

          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[i = 1, f1[t] f2[t] f3[t] /. t :> om[i++] // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, Inactivate[f1[t] f1[t] f1[t]] /. t :> om[i++] // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Update 2: "... if one replaces e.g. f1[t] to become (x1[t]-x2[t])"



          The first method works if the level specification -2 is used in MapIndexed:



          Inactivate[f1[t] f2[t] f3[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Inactivate[f1[t] f1[t] f1[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])] /. 
          Inactive[Times][x_, y__] :>
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]




          The second method can also be modified (so that ReplaceAll is mapped onto the terms of the product) to work for all three cases:



          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f2[t] f3[t]]]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f1[t] f1[t]]]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])]]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            Jan 17 at 11:56











          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            Jan 17 at 12:05










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            Jan 17 at 12:18











          • $begingroup$
            I think the code must be edited once more if one replaces e.g. f1[t] to become (x1[t]-x2[t]).
            $endgroup$
            – Display Name
            Jan 17 at 14:14










          • $begingroup$
            @DisplayName, please see update 2.
            $endgroup$
            – kglr
            Jan 18 at 10:52















          3












          $begingroup$

          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[i = 1, f1[t] f2[t] f3[t] /. t :> om[i++] // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, Inactivate[f1[t] f1[t] f1[t]] /. t :> om[i++] // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Update 2: "... if one replaces e.g. f1[t] to become (x1[t]-x2[t])"



          The first method works if the level specification -2 is used in MapIndexed:



          Inactivate[f1[t] f2[t] f3[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Inactivate[f1[t] f1[t] f1[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])] /. 
          Inactive[Times][x_, y__] :>
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]




          The second method can also be modified (so that ReplaceAll is mapped onto the terms of the product) to work for all three cases:



          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f2[t] f3[t]]]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f1[t] f1[t]]]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])]]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            Jan 17 at 11:56











          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            Jan 17 at 12:05










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            Jan 17 at 12:18











          • $begingroup$
            I think the code must be edited once more if one replaces e.g. f1[t] to become (x1[t]-x2[t]).
            $endgroup$
            – Display Name
            Jan 17 at 14:14










          • $begingroup$
            @DisplayName, please see update 2.
            $endgroup$
            – kglr
            Jan 18 at 10:52













          3












          3








          3





          $begingroup$

          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[i = 1, f1[t] f2[t] f3[t] /. t :> om[i++] // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, Inactivate[f1[t] f1[t] f1[t]] /. t :> om[i++] // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Update 2: "... if one replaces e.g. f1[t] to become (x1[t]-x2[t])"



          The first method works if the level specification -2 is used in MapIndexed:



          Inactivate[f1[t] f2[t] f3[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Inactivate[f1[t] f1[t] f1[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])] /. 
          Inactive[Times][x_, y__] :>
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]




          The second method can also be modified (so that ReplaceAll is mapped onto the terms of the product) to work for all three cases:



          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f2[t] f3[t]]]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f1[t] f1[t]]]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])]]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]







          share|improve this answer











          $endgroup$



          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[i = 1, f1[t] f2[t] f3[t] /. t :> om[i++] // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, Inactivate[f1[t] f1[t] f1[t]] /. t :> om[i++] // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Update 2: "... if one replaces e.g. f1[t] to become (x1[t]-x2[t])"



          The first method works if the level specification -2 is used in MapIndexed:



          Inactivate[f1[t] f2[t] f3[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Inactivate[f1[t] f1[t] f1[t]] /. Inactive[Times][x_, y__] :> 
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])] /. 
          Inactive[Times][x_, y__] :>
          Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, x, y, -2]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]




          The second method can also be modified (so that ReplaceAll is mapped onto the terms of the product) to work for all three cases:



          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f2[t] f3[t]]]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[f1[t] f1[t] f1[t]]]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[i = 1, n[With[j = i++, # /. t :> om[j]] & /@ 
          Inactivate[(x1[t] - x2[t]) (x1[t] - x2[t]) (x1[t] - x2[t])]]]



          n[(x1[om[1]] - x2[om[1]]) (x1[om[2]] - x2[om[2]]) (x1[om[3]] - x2[om[3]])]








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Jan 18 at 11:07

























          answered Jan 17 at 11:43









          kglrkglr

          181k10200413




          181k10200413











          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            Jan 17 at 11:56











          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            Jan 17 at 12:05










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            Jan 17 at 12:18











          • $begingroup$
            I think the code must be edited once more if one replaces e.g. f1[t] to become (x1[t]-x2[t]).
            $endgroup$
            – Display Name
            Jan 17 at 14:14










          • $begingroup$
            @DisplayName, please see update 2.
            $endgroup$
            – kglr
            Jan 18 at 10:52
















          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            Jan 17 at 11:56











          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            Jan 17 at 12:05










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            Jan 17 at 12:18











          • $begingroup$
            I think the code must be edited once more if one replaces e.g. f1[t] to become (x1[t]-x2[t]).
            $endgroup$
            – Display Name
            Jan 17 at 14:14










          • $begingroup$
            @DisplayName, please see update 2.
            $endgroup$
            – kglr
            Jan 18 at 10:52















          $begingroup$
          Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
          $endgroup$
          – Display Name
          Jan 17 at 11:56





          $begingroup$
          Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
          $endgroup$
          – Display Name
          Jan 17 at 11:56













          $begingroup$
          @DisplayName, please see the update.
          $endgroup$
          – kglr
          Jan 17 at 12:05




          $begingroup$
          @DisplayName, please see the update.
          $endgroup$
          – kglr
          Jan 17 at 12:05












          $begingroup$
          Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
          $endgroup$
          – Display Name
          Jan 17 at 12:18





          $begingroup$
          Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
          $endgroup$
          – Display Name
          Jan 17 at 12:18













          $begingroup$
          I think the code must be edited once more if one replaces e.g. f1[t] to become (x1[t]-x2[t]).
          $endgroup$
          – Display Name
          Jan 17 at 14:14




          $begingroup$
          I think the code must be edited once more if one replaces e.g. f1[t] to become (x1[t]-x2[t]).
          $endgroup$
          – Display Name
          Jan 17 at 14:14












          $begingroup$
          @DisplayName, please see update 2.
          $endgroup$
          – kglr
          Jan 18 at 10:52




          $begingroup$
          @DisplayName, please see update 2.
          $endgroup$
          – kglr
          Jan 18 at 10:52











          1












          $begingroup$

          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]

          changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]







          share|improve this answer









          $endgroup$












          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            Jan 17 at 12:32















          1












          $begingroup$

          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]

          changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]







          share|improve this answer









          $endgroup$












          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            Jan 17 at 12:32













          1












          1








          1





          $begingroup$

          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]

          changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]







          share|improve this answer









          $endgroup$



          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]

          changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 17 at 12:17









          m_goldbergm_goldberg

          85.6k872196




          85.6k872196











          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            Jan 17 at 12:32
















          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            Jan 17 at 12:32















          $begingroup$
          Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
          $endgroup$
          – Display Name
          Jan 17 at 12:32




          $begingroup$
          Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
          $endgroup$
          – Display Name
          Jan 17 at 12:32

















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