What is the the proper way to define a function from an expression?
up vote
5
down vote
favorite
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$$34$907690`f11
f11[x$_]:=Expand[x^2]
function-construction evaluation hold
asked Dec 4 at 11:02
ablmf
23917
add a comment |
up vote
5
down vote
favorite
Say after some long computation we get an expression
expr=x^2
We do not know what the value of expr beforehand.
Now we want to turn this in a function. We can use either
f[x_]=expr
or
f[x_]:=Evaluate[expr]
as suggested in this question.
However, when we do
f[x_]=Expand[expr]
or
f[x_]:=Evaluate[Expand[expr]]
The Expand will not have any effect.
Is there any way to make this work? Of course, we can define another function
g[x_]:=Expand[f[x]]
But is there any way to do it a bit more concisely?
Update:
If you do what suggested by Kuba in the comment, you get
In[89]:= With[{expr = expr}, f11[x_] := Expand[expr]]
In[90]:= f11[a + b]
Out[90]= x^2
In[91]:= ?? f11
Notebook$$34$907690`f11
f11[x$_]:=Expand[x^2]
function-construction evaluation hold
asked Dec 4 at 11:02
ablmf
23917
1
I thinkf[x_] = Expand[expr]actually works fine. It expandsexprand then turnsxinto a function slot. Just try it withexpr = (1 + x)^10and then evaluatef[y]after definingf. Or were you expecting something different?
– Sjoerd Smit
Dec 4 at 15:47
I was expecting to havef[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 theg[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
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
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$$34$907690`f11
f11[x$_]:=Expand[x^2]
function-construction evaluation hold
asked Dec 4 at 11:02
ablmf
23917
Say after some long computation we get an expression
expr=x^2
We do not know what the value of expr beforehand.
Now we want to turn this in a function. We can use either
f[x_]=expr
or
f[x_]:=Evaluate[expr]
as suggested in this question.
However, when we do
f[x_]=Expand[expr]
or
f[x_]:=Evaluate[Expand[expr]]
The Expand will not have any effect.
Is there any way to make this work? Of course, we can define another function
g[x_]:=Expand[f[x]]
But is there any way to do it a bit more concisely?
Update:
If you do what suggested by Kuba in the comment, you get
In[89]:= With[{expr = expr}, f11[x_] := Expand[expr]]
In[90]:= f11[a + b]
Out[90]= x^2
In[91]:= ?? f11
Notebook$$34$907690`f11
f11[x$_]:=Expand[x^2]
function-construction evaluation hold
function-construction evaluation hold
asked Dec 4 at 11:02
ablmf
23917
asked Dec 4 at 11:02
ablmf
23917
edited Dec 4 at 13:41
asked Dec 4 at 11:02
ablmf
23917
asked Dec 4 at 11:02
ablmf
23917
asked Dec 4 at 11:02
ablmf
23917
23917
1
I thinkf[x_] = Expand[expr]actually works fine. It expandsexprand then turnsxinto a function slot. Just try it withexpr = (1 + x)^10and then evaluatef[y]after definingf. Or were you expecting something different?
– Sjoerd Smit
Dec 4 at 15:47
I was expecting to havef[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 theg[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
add a comment |
1
I thinkf[x_] = Expand[expr]actually works fine. It expandsexprand then turnsxinto a function slot. Just try it withexpr = (1 + x)^10and then evaluatef[y]after definingf. Or were you expecting something different?
– Sjoerd Smit
Dec 4 at 15:47
I was expecting to havef[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 theg[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
add a comment |
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
answered Dec 4 at 11:39
Kuba♦
103k12201515
why notWith[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
Dec 4 at 12:36
1
@AliHashmi Because the point is not to evaluateExpand. Compare?fin 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
add a comment |
up vote
4
down vote
(f[x_] := Expand@#) &@expr
answered Dec 4 at 11:48
xzczd
25.7k469245
FYI, I'd accept that. This is what I usually do anyway :p
– Kuba♦
Dec 5 at 9:24
add a comment |
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}
answered Dec 4 at 12:04
Hadi Sobhani
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
add a comment |
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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
answered Dec 4 at 11:39
Kuba♦
103k12201515
why notWith[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
Dec 4 at 12:36
1
@AliHashmi Because the point is not to evaluateExpand. Compare?fin 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
add a comment |
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
answered Dec 4 at 11:39
Kuba♦
103k12201515
why notWith[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
Dec 4 at 12:36
1
@AliHashmi Because the point is not to evaluateExpand. Compare?fin 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
add a comment |
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
answered Dec 4 at 11:39
Kuba♦
103k12201515
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
answered Dec 4 at 11:39
Kuba♦
103k12201515
answered Dec 4 at 11:39
Kuba♦
103k12201515
answered Dec 4 at 11:39
Kuba♦
103k12201515
answered Dec 4 at 11:39
Kuba♦
103k12201515
103k12201515
why notWith[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
Dec 4 at 12:36
1
@AliHashmi Because the point is not to evaluateExpand. Compare?fin 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
add a comment |
why notWith[{expr = Expand[expr]}, SetDelayed @@ Hold[f[x_], expr]]
– Ali Hashmi
Dec 4 at 12:36
1
@AliHashmi Because the point is not to evaluateExpand. Compare?fin 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
add a comment |
up vote
4
down vote
(f[x_] := Expand@#) &@expr
answered Dec 4 at 11:48
xzczd
25.7k469245
FYI, I'd accept that. This is what I usually do anyway :p
– Kuba♦
Dec 5 at 9:24
add a comment |
up vote
4
down vote
(f[x_] := Expand@#) &@expr
answered Dec 4 at 11:48
xzczd
25.7k469245
FYI, I'd accept that. This is what I usually do anyway :p
– Kuba♦
Dec 5 at 9:24
add a comment |
up vote
4
down vote
up vote
4
down vote
(f[x_] := Expand@#) &@expr
answered Dec 4 at 11:48
xzczd
25.7k469245
(f[x_] := Expand@#) &@expr
answered Dec 4 at 11:48
xzczd
25.7k469245
answered Dec 4 at 11:48
xzczd
25.7k469245
answered Dec 4 at 11:48
xzczd
25.7k469245
answered Dec 4 at 11:48
xzczd
25.7k469245
25.7k469245
FYI, I'd accept that. This is what I usually do anyway :p
– Kuba♦
Dec 5 at 9:24
add a comment |
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
add a comment |
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}
answered Dec 4 at 12:04
Hadi Sobhani
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
add a comment |
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}
answered Dec 4 at 12:04
Hadi Sobhani
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
add a comment |
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}
answered Dec 4 at 12:04
Hadi Sobhani
32417
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}
answered Dec 4 at 12:04
Hadi Sobhani
32417
answered Dec 4 at 12:04
Hadi Sobhani
32417
answered Dec 4 at 12:04
Hadi Sobhani
32417
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
add a comment |
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
add a comment |
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1
I think
f[x_] = Expand[expr]actually works fine. It expandsexprand then turnsxinto a function slot. Just try it withexpr = (1 + x)^10and then evaluatef[y]after definingf. 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