Change arguments of a product of functions
Clash Royale CLAN TAG#URR8PPP
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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
asked Jan 17 at 9:45
Display NameDisplay Name
57438
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add a comment |
$begingroup$
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?
pattern-matching
asked Jan 17 at 9:45
Display NameDisplay Name
57438
$endgroup$
add a comment |
$begingroup$
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?
pattern-matching
asked Jan 17 at 9:45
Display NameDisplay Name
57438
$endgroup$
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
pattern-matching
asked Jan 17 at 9:45
Display NameDisplay Name
57438
asked Jan 17 at 9:45
Display NameDisplay Name
57438
asked Jan 17 at 9:45
Display NameDisplay Name
57438
asked Jan 17 at 9:45
Display NameDisplay Name
57438
asked Jan 17 at 9:45
Display NameDisplay Name
57438
57438
add a comment |
add a comment |
2 Answers
2
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oldest
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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]])]
answered Jan 17 at 11:43
kglrkglr
181k10200413
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Thank you for your answer. It works forf1 != f2 != f3but stops working if all the functions are equal.
$endgroup$
– Display Name
Jan 17 at 11:56
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@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 theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– 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
|
show 1 more comment
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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]]]
answered Jan 17 at 12:17
m_goldbergm_goldberg
85.6k872196
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Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
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– Display Name
Jan 17 at 12:32
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
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$begingroup$
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]])]
answered Jan 17 at 11:43
kglrkglr
181k10200413
$endgroup$
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but 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 theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– 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
|
show 1 more comment
$begingroup$
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]])]
answered Jan 17 at 11:43
kglrkglr
181k10200413
$endgroup$
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but 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 theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– 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
|
show 1 more comment
$begingroup$
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]])]
answered Jan 17 at 11:43
kglrkglr
181k10200413
$endgroup$
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]])]
answered Jan 17 at 11:43
kglrkglr
181k10200413
edited Jan 18 at 11:07
answered Jan 17 at 11:43
kglrkglr
181k10200413
answered Jan 17 at 11:43
kglrkglr
181k10200413
answered Jan 17 at 11:43
kglrkglr
181k10200413
181k10200413
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but 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 theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– 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
|
show 1 more comment
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but 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 theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– 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
|
show 1 more comment
$begingroup$
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]]]
answered Jan 17 at 12:17
m_goldbergm_goldberg
85.6k872196
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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]]]
answered Jan 17 at 12:17
m_goldbergm_goldberg
85.6k872196
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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]]]
answered Jan 17 at 12:17
m_goldbergm_goldberg
85.6k872196
$endgroup$
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]]]
answered Jan 17 at 12:17
m_goldbergm_goldberg
85.6k872196
answered Jan 17 at 12:17
m_goldbergm_goldberg
85.6k872196
answered Jan 17 at 12:17
m_goldbergm_goldberg
85.6k872196
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
add a comment |
$begingroup$
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
add a comment |
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