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

f[x_]:=Evaluate[expr]

as suggested in this question.

However, when we do

f[x_]=Expand[expr]

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]

edited Dec 4 at 13:41

asked Dec 4 at 11:02

ablmf

23917

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

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

f[x_]:=Evaluate[expr]

as suggested in this question.

However, when we do

f[x_]=Expand[expr]

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]

edited Dec 4 at 13:41

asked Dec 4 at 11:02

ablmf

23917

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

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

f[x_]:=Evaluate[expr]

as suggested in this question.

However, when we do

f[x_]=Expand[expr]

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]

edited Dec 4 at 13:41

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

f[x_]:=Evaluate[expr]

as suggested in this question.

However, when we do

f[x_]=Expand[expr]

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

edited Dec 4 at 13:41

asked Dec 4 at 11:02

ablmf

23917

edited Dec 4 at 13:41

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

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

add a comment |

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

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

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 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

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

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 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

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 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

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

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

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

add a comment |

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

@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

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

(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

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

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

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

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

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

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

add a comment |

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