What is the the proper way to define a function from an expression?











up vote
5
down vote

favorite
2












Say after some long computation we get an expression



expr=x^2


We do not know what the value of expr beforehand.
Now we want to turn this in a function. We can use either



f[x_]=expr


or



f[x_]:=Evaluate[expr]


as suggested in this question.



However, when we do



f[x_]=Expand[expr]


or



f[x_]:=Evaluate[Expand[expr]]


The Expand will not have any effect.



Is there any way to make this work? Of course, we can define another function



g[x_]:=Expand[f[x]]


But is there any way to do it a bit more concisely?



Update:



If you do what suggested by Kuba in the comment, you get



In[89]:= With[{expr = expr}, f11[x_] := Expand[expr]]
In[90]:= f11[a + b]
Out[90]= x^2
In[91]:= ?? f11
Notebook$$34$907690`f11
f11[x$_]:=Expand[x^2]









share|improve this question




















  • 1




    I think f[x_] = Expand[expr] actually works fine. It expands expr and then turns x into a function slot. Just try it with expr = (1 + x)^10 and then evaluate f[y] after defining f. Or were you expecting something different?
    – Sjoerd Smit
    yesterday












  • I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
    – ablmf
    yesterday










  • In that case I would say that the g[x] := Expand[f[x]] method is really the way to go here, because it makes the evaluation process easiest to follow. Any other method is just going to be confusing one way or another.
    – Sjoerd Smit
    yesterday

















up vote
5
down vote

favorite
2












Say after some long computation we get an expression



expr=x^2


We do not know what the value of expr beforehand.
Now we want to turn this in a function. We can use either



f[x_]=expr


or



f[x_]:=Evaluate[expr]


as suggested in this question.



However, when we do



f[x_]=Expand[expr]


or



f[x_]:=Evaluate[Expand[expr]]


The Expand will not have any effect.



Is there any way to make this work? Of course, we can define another function



g[x_]:=Expand[f[x]]


But is there any way to do it a bit more concisely?



Update:



If you do what suggested by Kuba in the comment, you get



In[89]:= With[{expr = expr}, f11[x_] := Expand[expr]]
In[90]:= f11[a + b]
Out[90]= x^2
In[91]:= ?? f11
Notebook$$34$907690`f11
f11[x$_]:=Expand[x^2]









share|improve this question




















  • 1




    I think f[x_] = Expand[expr] actually works fine. It expands expr and then turns x into a function slot. Just try it with expr = (1 + x)^10 and then evaluate f[y] after defining f. Or were you expecting something different?
    – Sjoerd Smit
    yesterday












  • I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
    – ablmf
    yesterday










  • In that case I would say that the g[x] := Expand[f[x]] method is really the way to go here, because it makes the evaluation process easiest to follow. Any other method is just going to be confusing one way or another.
    – Sjoerd Smit
    yesterday















up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
2






2





Say after some long computation we get an expression



expr=x^2


We do not know what the value of expr beforehand.
Now we want to turn this in a function. We can use either



f[x_]=expr


or



f[x_]:=Evaluate[expr]


as suggested in this question.



However, when we do



f[x_]=Expand[expr]


or



f[x_]:=Evaluate[Expand[expr]]


The Expand will not have any effect.



Is there any way to make this work? Of course, we can define another function



g[x_]:=Expand[f[x]]


But is there any way to do it a bit more concisely?



Update:



If you do what suggested by Kuba in the comment, you get



In[89]:= With[{expr = expr}, f11[x_] := Expand[expr]]
In[90]:= f11[a + b]
Out[90]= x^2
In[91]:= ?? f11
Notebook$$34$907690`f11
f11[x$_]:=Expand[x^2]









share|improve this question















Say after some long computation we get an expression



expr=x^2


We do not know what the value of expr beforehand.
Now we want to turn this in a function. We can use either



f[x_]=expr


or



f[x_]:=Evaluate[expr]


as suggested in this question.



However, when we do



f[x_]=Expand[expr]


or



f[x_]:=Evaluate[Expand[expr]]


The Expand will not have any effect.



Is there any way to make this work? Of course, we can define another function



g[x_]:=Expand[f[x]]


But is there any way to do it a bit more concisely?



Update:



If you do what suggested by Kuba in the comment, you get



In[89]:= With[{expr = expr}, f11[x_] := Expand[expr]]
In[90]:= f11[a + b]
Out[90]= x^2
In[91]:= ?? f11
Notebook$$34$907690`f11
f11[x$_]:=Expand[x^2]






function-construction evaluation hold






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday

























asked yesterday









ablmf

21717




21717








  • 1




    I think f[x_] = Expand[expr] actually works fine. It expands expr and then turns x into a function slot. Just try it with expr = (1 + x)^10 and then evaluate f[y] after defining f. Or were you expecting something different?
    – Sjoerd Smit
    yesterday












  • I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
    – ablmf
    yesterday










  • In that case I would say that the g[x] := Expand[f[x]] method is really the way to go here, because it makes the evaluation process easiest to follow. Any other method is just going to be confusing one way or another.
    – Sjoerd Smit
    yesterday
















  • 1




    I think f[x_] = Expand[expr] actually works fine. It expands expr and then turns x into a function slot. Just try it with expr = (1 + x)^10 and then evaluate f[y] after defining f. Or were you expecting something different?
    – Sjoerd Smit
    yesterday












  • I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
    – ablmf
    yesterday










  • In that case I would say that the g[x] := Expand[f[x]] method is really the way to go here, because it makes the evaluation process easiest to follow. Any other method is just going to be confusing one way or another.
    – Sjoerd Smit
    yesterday










1




1




I think f[x_] = Expand[expr] actually works fine. It expands expr and then turns x into a function slot. Just try it with expr = (1 + x)^10 and then evaluate f[y] after defining f. Or were you expecting something different?
– Sjoerd Smit
yesterday






I think f[x_] = Expand[expr] actually works fine. It expands expr and then turns x into a function slot. Just try it with expr = (1 + x)^10 and then evaluate f[y] after defining f. Or were you expecting something different?
– Sjoerd Smit
yesterday














I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
– ablmf
yesterday




I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
– ablmf
yesterday












In that case I would say that the g[x] := Expand[f[x]] method is really the way to go here, because it makes the evaluation process easiest to follow. Any other method is just going to be confusing one way or another.
– Sjoerd Smit
yesterday






In that case I would say that the g[x] := Expand[f[x]] method is really the way to go here, because it makes the evaluation process easiest to follow. Any other method is just going to be confusing one way or another.
– Sjoerd Smit
yesterday












3 Answers
3






active

oldest

votes

















up vote
6
down vote



accepted










Sorry, I was too hasty in comments:



With[{expr = expr}, SetDelayed @@ Hold[f[x_], Expand[expr]]]


SetDelayed @@ Hold is needed because of: Enforcing correct variable bindings and avoiding renamings for conflicting variables in nested scoping constructs






share|improve this answer





















  • why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
    – Ali Hashmi
    yesterday






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba
    yesterday












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    yesterday


















up vote
4
down vote













(f[x_] := Expand@#) &@expr





share|improve this answer





















  • FYI, I'd accept that. This is what I usually do anyway :p
    – Kuba
    19 hours ago


















up vote
0
down vote













Dear @ablmf you can use the following syntax in Mathematica.



f[x_]:=x^2;


When you input any desired value for x, Mathematica gives you its value in f[x]. Say, you want f[2], Mathematica gives you 4



f[2]



4




You can also use it to obtain a list. Consider the following code



ClearAll["Global`*"];
f[x_] := x^2;

Table[
f[x]
, {x, 0, 10}]


it gives you




{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100}







share|improve this answer

















  • 1




    I'm sorry, but this isn't what OP asks for, is it?
    – xzczd
    yesterday






  • 1




    Sorry, this is not what I asked.
    – ablmf
    yesterday











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote



accepted










Sorry, I was too hasty in comments:



With[{expr = expr}, SetDelayed @@ Hold[f[x_], Expand[expr]]]


SetDelayed @@ Hold is needed because of: Enforcing correct variable bindings and avoiding renamings for conflicting variables in nested scoping constructs






share|improve this answer





















  • why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
    – Ali Hashmi
    yesterday






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba
    yesterday












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    yesterday















up vote
6
down vote



accepted










Sorry, I was too hasty in comments:



With[{expr = expr}, SetDelayed @@ Hold[f[x_], Expand[expr]]]


SetDelayed @@ Hold is needed because of: Enforcing correct variable bindings and avoiding renamings for conflicting variables in nested scoping constructs






share|improve this answer





















  • why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
    – Ali Hashmi
    yesterday






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba
    yesterday












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    yesterday













up vote
6
down vote



accepted







up vote
6
down vote



accepted






Sorry, I was too hasty in comments:



With[{expr = expr}, SetDelayed @@ Hold[f[x_], Expand[expr]]]


SetDelayed @@ Hold is needed because of: Enforcing correct variable bindings and avoiding renamings for conflicting variables in nested scoping constructs






share|improve this answer












Sorry, I was too hasty in comments:



With[{expr = expr}, SetDelayed @@ Hold[f[x_], Expand[expr]]]


SetDelayed @@ Hold is needed because of: Enforcing correct variable bindings and avoiding renamings for conflicting variables in nested scoping constructs







share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









Kuba

103k12201512




103k12201512












  • why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
    – Ali Hashmi
    yesterday






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba
    yesterday












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    yesterday


















  • why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
    – Ali Hashmi
    yesterday






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba
    yesterday












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    yesterday
















why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
yesterday




why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
yesterday




1




1




@AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
– Kuba
yesterday






@AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
– Kuba
yesterday














sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
– Ali Hashmi
yesterday




sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
– Ali Hashmi
yesterday










up vote
4
down vote













(f[x_] := Expand@#) &@expr





share|improve this answer





















  • FYI, I'd accept that. This is what I usually do anyway :p
    – Kuba
    19 hours ago















up vote
4
down vote













(f[x_] := Expand@#) &@expr





share|improve this answer





















  • FYI, I'd accept that. This is what I usually do anyway :p
    – Kuba
    19 hours ago













up vote
4
down vote










up vote
4
down vote









(f[x_] := Expand@#) &@expr





share|improve this answer












(f[x_] := Expand@#) &@expr






share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









xzczd

25.7k468244




25.7k468244












  • FYI, I'd accept that. This is what I usually do anyway :p
    – Kuba
    19 hours ago


















  • FYI, I'd accept that. This is what I usually do anyway :p
    – Kuba
    19 hours ago
















FYI, I'd accept that. This is what I usually do anyway :p
– Kuba
19 hours ago




FYI, I'd accept that. This is what I usually do anyway :p
– Kuba
19 hours ago










up vote
0
down vote













Dear @ablmf you can use the following syntax in Mathematica.



f[x_]:=x^2;


When you input any desired value for x, Mathematica gives you its value in f[x]. Say, you want f[2], Mathematica gives you 4



f[2]



4




You can also use it to obtain a list. Consider the following code



ClearAll["Global`*"];
f[x_] := x^2;

Table[
f[x]
, {x, 0, 10}]


it gives you




{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100}







share|improve this answer

















  • 1




    I'm sorry, but this isn't what OP asks for, is it?
    – xzczd
    yesterday






  • 1




    Sorry, this is not what I asked.
    – ablmf
    yesterday















up vote
0
down vote













Dear @ablmf you can use the following syntax in Mathematica.



f[x_]:=x^2;


When you input any desired value for x, Mathematica gives you its value in f[x]. Say, you want f[2], Mathematica gives you 4



f[2]



4




You can also use it to obtain a list. Consider the following code



ClearAll["Global`*"];
f[x_] := x^2;

Table[
f[x]
, {x, 0, 10}]


it gives you




{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100}







share|improve this answer

















  • 1




    I'm sorry, but this isn't what OP asks for, is it?
    – xzczd
    yesterday






  • 1




    Sorry, this is not what I asked.
    – ablmf
    yesterday













up vote
0
down vote










up vote
0
down vote









Dear @ablmf you can use the following syntax in Mathematica.



f[x_]:=x^2;


When you input any desired value for x, Mathematica gives you its value in f[x]. Say, you want f[2], Mathematica gives you 4



f[2]



4




You can also use it to obtain a list. Consider the following code



ClearAll["Global`*"];
f[x_] := x^2;

Table[
f[x]
, {x, 0, 10}]


it gives you




{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100}







share|improve this answer












Dear @ablmf you can use the following syntax in Mathematica.



f[x_]:=x^2;


When you input any desired value for x, Mathematica gives you its value in f[x]. Say, you want f[2], Mathematica gives you 4



f[2]



4




You can also use it to obtain a list. Consider the following code



ClearAll["Global`*"];
f[x_] := x^2;

Table[
f[x]
, {x, 0, 10}]


it gives you




{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100}








share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









Hadi Sobhani

32417




32417








  • 1




    I'm sorry, but this isn't what OP asks for, is it?
    – xzczd
    yesterday






  • 1




    Sorry, this is not what I asked.
    – ablmf
    yesterday














  • 1




    I'm sorry, but this isn't what OP asks for, is it?
    – xzczd
    yesterday






  • 1




    Sorry, this is not what I asked.
    – ablmf
    yesterday








1




1




I'm sorry, but this isn't what OP asks for, is it?
– xzczd
yesterday




I'm sorry, but this isn't what OP asks for, is it?
– xzczd
yesterday




1




1




Sorry, this is not what I asked.
– ablmf
yesterday




Sorry, this is not what I asked.
– ablmf
yesterday


















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