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]





function-construction evaluation hold







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
    Dec 4 at 15:47












  • I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
    – ablmf
    Dec 4 at 16:36










  • 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
    Dec 4 at 17:02

















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]





function-construction evaluation hold







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
    Dec 4 at 15:47












  • I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
    – ablmf
    Dec 4 at 16:36










  • 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
    Dec 4 at 17:02















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]





function-construction evaluation hold







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




function-construction evaluation hold






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 4 at 13:41

























asked Dec 4 at 11:02









ablmf

24617




24617








  • 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
    Dec 4 at 15:47












  • I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
    – ablmf
    Dec 4 at 16:36










  • 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
    Dec 4 at 17:02
















  • 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
    Dec 4 at 15:47












  • I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
    – ablmf
    Dec 4 at 16:36










  • 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
    Dec 4 at 17:02










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
Dec 4 at 15:47






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
Dec 4 at 15:47














I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
– ablmf
Dec 4 at 16:36




I was expecting to have f[a+b]==Expand[(a+b)^2]=a^2+2 a b+b^2
– ablmf
Dec 4 at 16:36












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
Dec 4 at 17:02






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
Dec 4 at 17:02












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
    Dec 4 at 12:36






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba♦
    Dec 4 at 12:37












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    Dec 4 at 14:49


















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♦
    Dec 5 at 9:24


















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
    Dec 4 at 12:09






  • 1




    Sorry, this is not what I asked.
    – ablmf
    Dec 4 at 13:07











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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
    Dec 4 at 12:36






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba♦
    Dec 4 at 12:37












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    Dec 4 at 14:49















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
    Dec 4 at 12:36






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba♦
    Dec 4 at 12:37












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    Dec 4 at 14:49













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 Dec 4 at 11:39









Kuba♦

103k12201515




103k12201515












  • why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
    – Ali Hashmi
    Dec 4 at 12:36






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba♦
    Dec 4 at 12:37












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    Dec 4 at 14:49


















  • why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
    – Ali Hashmi
    Dec 4 at 12:36






  • 1




    @AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
    – Kuba♦
    Dec 4 at 12:37












  • sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
    – Ali Hashmi
    Dec 4 at 14:49
















why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
Dec 4 at 12:36




why not With[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
Dec 4 at 12:36




1




1




@AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
– Kuba♦
Dec 4 at 12:37






@AliHashmi Because the point is not to evaluate Expand. Compare ?f in my and your case.
– Kuba♦
Dec 4 at 12:37














sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
– Ali Hashmi
Dec 4 at 14:49




sorry i misinterpreted the question. I thought the purpose was for Evaluate to do its job before the definitions are saved.
– Ali Hashmi
Dec 4 at 14:49










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♦
    Dec 5 at 9:24















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♦
    Dec 5 at 9:24













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 Dec 4 at 11:48









xzczd

25.7k469246




25.7k469246












  • FYI, I'd accept that. This is what I usually do anyway :p
    – Kuba♦
    Dec 5 at 9:24


















  • FYI, I'd accept that. This is what I usually do anyway :p
    – Kuba♦
    Dec 5 at 9:24
















FYI, I'd accept that. This is what I usually do anyway :p
– Kuba♦
Dec 5 at 9:24




FYI, I'd accept that. This is what I usually do anyway :p
– Kuba♦
Dec 5 at 9:24










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
    Dec 4 at 12:09






  • 1




    Sorry, this is not what I asked.
    – ablmf
    Dec 4 at 13:07















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
    Dec 4 at 12:09






  • 1




    Sorry, this is not what I asked.
    – ablmf
    Dec 4 at 13:07













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 Dec 4 at 12:04









Hadi Sobhani

32417




32417








  • 1




    I'm sorry, but this isn't what OP asks for, is it?
    – xzczd
    Dec 4 at 12:09






  • 1




    Sorry, this is not what I asked.
    – ablmf
    Dec 4 at 13:07














  • 1




    I'm sorry, but this isn't what OP asks for, is it?
    – xzczd
    Dec 4 at 12:09






  • 1




    Sorry, this is not what I asked.
    – ablmf
    Dec 4 at 13:07








1




1




I'm sorry, but this isn't what OP asks for, is it?
– xzczd
Dec 4 at 12:09




I'm sorry, but this isn't what OP asks for, is it?
– xzczd
Dec 4 at 12:09




1




1




Sorry, this is not what I asked.
– ablmf
Dec 4 at 13:07




Sorry, this is not what I asked.
– ablmf
Dec 4 at 13:07


















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