Basic substitution

up vote
1
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I am relatively new to Mathematica, so I am sure this is basic, but I lack the basic terminology to be able to google it.

How can I make this command work?

Plot[x / a, x, 0, a] /. a->10

I would expect this to plot the same line for all values of a. But I get

Plot::plln: Limiting value a in Charting`Private`pvar$10408,0,a is not a machine-sized real number.

I understand that Plot expects numbers not variables, but what is the correct syntax to resolve this since a is ultimately a number?

edited 2 hours ago

Michael E2

142k11192458

asked 4 hours ago

Tohiko

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1

I wonder whether "to google sth" really should replace "to look sth up"? Are we getting paid for that kind of word of mouth advertising - especially for a monopolist? ;-)
â€“Â gwr
3 hours ago

add a commentÂ |Â

up vote
1
down vote

favorite

I am relatively new to Mathematica, so I am sure this is basic, but I lack the basic terminology to be able to google it.

How can I make this command work?

Plot[x / a, x, 0, a] /. a->10

I would expect this to plot the same line for all values of a. But I get

Plot::plln: Limiting value a in Charting`Private`pvar$10408,0,a is not a machine-sized real number.

I understand that Plot expects numbers not variables, but what is the correct syntax to resolve this since a is ultimately a number?

edited 2 hours ago

Michael E2

142k11192458

asked 4 hours ago

Tohiko

1083

New contributor

1

I wonder whether "to google sth" really should replace "to look sth up"? Are we getting paid for that kind of word of mouth advertising - especially for a monopolist? ;-)
â€“Â gwr
3 hours ago

add a commentÂ |Â

up vote
1
down vote

favorite

I am relatively new to Mathematica, so I am sure this is basic, but I lack the basic terminology to be able to google it.

How can I make this command work?

Plot[x / a, x, 0, a] /. a->10

I would expect this to plot the same line for all values of a. But I get

Plot::plln: Limiting value a in Charting`Private`pvar$10408,0,a is not a machine-sized real number.

I understand that Plot expects numbers not variables, but what is the correct syntax to resolve this since a is ultimately a number?

edited 2 hours ago

Michael E2

142k11192458

asked 4 hours ago

Tohiko

1083

New contributor

I am relatively new to Mathematica, so I am sure this is basic, but I lack the basic terminology to be able to google it.

How can I make this command work?

Plot[x / a, x, 0, a] /. a->10

I would expect this to plot the same line for all values of a. But I get

Plot::plln: Limiting value a in Charting`Private`pvar$10408,0,a is not a machine-sized real number.

I understand that Plot expects numbers not variables, but what is the correct syntax to resolve this since a is ultimately a number?

evaluation replacement

edited 2 hours ago

Michael E2

142k11192458

asked 4 hours ago

Tohiko

1083

New contributor

edited 2 hours ago

Michael E2

142k11192458

asked 4 hours ago

Tohiko

1083

New contributor

edited 2 hours ago

Michael E2

142k11192458

edited 2 hours ago

Michael E2

142k11192458

edited 2 hours ago

Michael E2

142k11192458

asked 4 hours ago

Tohiko

1083

New contributor

asked 4 hours ago

Tohiko

1083

asked 4 hours ago

Tohiko

1083

New contributor

Tohiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

1

I wonder whether "to google sth" really should replace "to look sth up"? Are we getting paid for that kind of word of mouth advertising - especially for a monopolist? ;-)
â€“Â gwr
3 hours ago

add a commentÂ |Â

1

I wonder whether "to google sth" really should replace "to look sth up"? Are we getting paid for that kind of word of mouth advertising - especially for a monopolist? ;-)
â€“Â gwr
3 hours ago

I wonder whether "to google sth" really should replace "to look sth up"? Are we getting paid for that kind of word of mouth advertising - especially for a monopolist? ;-)
â€“Â gwr
3 hours ago

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
1
down vote

accepted

So you need to get your substitution in before Plot evaluates its arguments. Next to defining the Plot inside a function with an argument a as shown by LouisB you can do the following:

Either

plotargs = x/a, x,0,a ;
Plot @@ ( plotargs /. a-> 10 )

inactive = Inactivate @ Plot[ x/a, x,0,a ];
inactive /. a -> 10 // Activate

To make this more general:

parameters = Association[
 a -> 10
]; (* could be a whole bunch of parameters... *)

SetAttributes[ withParameters, HoldFirst];
withParameters[ func_ , pars_Association ] := Inactivate[ func ] /. pars // Activate

withParameters[ Plot[ x/a, x, 0, a] , parameters ]

Update:

If you want to pass your function as an argument to withParameters it is safest to use Inactivate also. So:

myParametricFunc = Inactivate @ Plot[ x/a, x,0,a];
withParameters[ myParametricFunc, parameters]

As a word of caution: As you can see in the documentation, the inactive form is different from the FullForm. So replacement will work for simple parameters and even options, but not for more complex substitution patters with regard to functional form.

edited 1 hour ago

answered 3 hours ago

gwr

6,89622457

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up vote
2
down vote

You could do this:

Clear[p]
p[a_] := Plot[x/a, x, 0, a]
p[10]

answered 4 hours ago

LouisB

4,2341616

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up vote
2
down vote

With[a = 10,
 Plot[
 x/a,
 x, 0, a
 ]
 ]

answered 2 hours ago

rhermans

21.8k439104

add a commentÂ |Â

up vote
2
down vote

You can delay the evaluation of Plot with Unevaluated:

Unevaluated@Plot[x/a, x, 0, a] /. a -> 10

If all occurrences of a are not literally present in the Plot code, Block can be used in place of ReplaceAll:

function = x/a;
x = 2; (* to show that this is ignored in the Plot command *)
Block[a = 10,
 Plot[function, x, 0, a]
 ]

The issue to keep in mind is that Plot is HoldAll, so that the value for function is not substituted (it is said to be "held") before Plot is called. This means, for instance, that is x also has a value, it won't be used because Plot localizes (via Block).

A note on the second example:
The definitions of function and x is pretty bad programming practice. The order of the definition is crucial, and it's bad to have such a dependency in Mathematica's notebook environment in which any cell can be evaluated at any time. Later if I decide I need to fix the definition of function, say function = 2 x/a, then the value of x will be substituted and function will be equal to 4/a and no longer depend on x. It's not as bad if it's a quick, one-off calculation, but it will still leave your current session with a latent definition of x. (Personally, my eyesight has trouble distinguishing Mathematica's undefined blue from the defined black on single-letter variables.) If you're writing code that will be reevaluated several times, you're better off making all dependencies functional:

function[x_, a_] := x/a

It doesn't help with the OP's substitution problem, in which a appears in more than just the function, so this or one of the other solutions will be needed for that.

edited 2 hours ago

answered 2 hours ago

Michael E2

142k11192458

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
1
down vote

accepted

So you need to get your substitution in before Plot evaluates its arguments. Next to defining the Plot inside a function with an argument a as shown by LouisB you can do the following:

Either

plotargs = x/a, x,0,a ;
Plot @@ ( plotargs /. a-> 10 )

inactive = Inactivate @ Plot[ x/a, x,0,a ];
inactive /. a -> 10 // Activate

To make this more general:

parameters = Association[
 a -> 10
]; (* could be a whole bunch of parameters... *)

SetAttributes[ withParameters, HoldFirst];
withParameters[ func_ , pars_Association ] := Inactivate[ func ] /. pars // Activate

withParameters[ Plot[ x/a, x, 0, a] , parameters ]

Update:

If you want to pass your function as an argument to withParameters it is safest to use Inactivate also. So:

myParametricFunc = Inactivate @ Plot[ x/a, x,0,a];
withParameters[ myParametricFunc, parameters]

edited 1 hour ago

answered 3 hours ago

gwr

6,89622457

add a commentÂ |Â

up vote
1
down vote

accepted

So you need to get your substitution in before Plot evaluates its arguments. Next to defining the Plot inside a function with an argument a as shown by LouisB you can do the following:

Either

plotargs = x/a, x,0,a ;
Plot @@ ( plotargs /. a-> 10 )

inactive = Inactivate @ Plot[ x/a, x,0,a ];
inactive /. a -> 10 // Activate

To make this more general:

parameters = Association[
 a -> 10
]; (* could be a whole bunch of parameters... *)

SetAttributes[ withParameters, HoldFirst];
withParameters[ func_ , pars_Association ] := Inactivate[ func ] /. pars // Activate

withParameters[ Plot[ x/a, x, 0, a] , parameters ]

Update:

If you want to pass your function as an argument to withParameters it is safest to use Inactivate also. So:

myParametricFunc = Inactivate @ Plot[ x/a, x,0,a];
withParameters[ myParametricFunc, parameters]

edited 1 hour ago

answered 3 hours ago

gwr

6,89622457

add a commentÂ |Â

up vote
1
down vote

accepted

So you need to get your substitution in before Plot evaluates its arguments. Next to defining the Plot inside a function with an argument a as shown by LouisB you can do the following:

Either

plotargs = x/a, x,0,a ;
Plot @@ ( plotargs /. a-> 10 )

inactive = Inactivate @ Plot[ x/a, x,0,a ];
inactive /. a -> 10 // Activate

To make this more general:

parameters = Association[
 a -> 10
]; (* could be a whole bunch of parameters... *)

SetAttributes[ withParameters, HoldFirst];
withParameters[ func_ , pars_Association ] := Inactivate[ func ] /. pars // Activate

withParameters[ Plot[ x/a, x, 0, a] , parameters ]

Update:

If you want to pass your function as an argument to withParameters it is safest to use Inactivate also. So:

myParametricFunc = Inactivate @ Plot[ x/a, x,0,a];
withParameters[ myParametricFunc, parameters]

edited 1 hour ago

answered 3 hours ago

gwr

6,89622457

So you need to get your substitution in before Plot evaluates its arguments. Next to defining the Plot inside a function with an argument a as shown by LouisB you can do the following:

Either

plotargs = x/a, x,0,a ;
Plot @@ ( plotargs /. a-> 10 )

inactive = Inactivate @ Plot[ x/a, x,0,a ];
inactive /. a -> 10 // Activate

To make this more general:

parameters = Association[
 a -> 10
]; (* could be a whole bunch of parameters... *)

SetAttributes[ withParameters, HoldFirst];
withParameters[ func_ , pars_Association ] := Inactivate[ func ] /. pars // Activate

withParameters[ Plot[ x/a, x, 0, a] , parameters ]

Update:

If you want to pass your function as an argument to withParameters it is safest to use Inactivate also. So:

myParametricFunc = Inactivate @ Plot[ x/a, x,0,a];
withParameters[ myParametricFunc, parameters]

edited 1 hour ago

answered 3 hours ago

gwr

6,89622457

edited 1 hour ago

answered 3 hours ago

gwr

6,89622457

answered 3 hours ago

gwr

6,89622457

answered 3 hours ago

gwr

6,89622457

add a commentÂ |Â

up vote
2
down vote

You could do this:

Clear[p]
p[a_] := Plot[x/a, x, 0, a]
p[10]

answered 4 hours ago

LouisB

4,2341616

add a commentÂ |Â

up vote
2
down vote

You could do this:

Clear[p]
p[a_] := Plot[x/a, x, 0, a]
p[10]

answered 4 hours ago

LouisB

4,2341616

add a commentÂ |Â

up vote
2
down vote

You could do this:

Clear[p]
p[a_] := Plot[x/a, x, 0, a]
p[10]

answered 4 hours ago

LouisB

4,2341616

You could do this:

Clear[p]
p[a_] := Plot[x/a, x, 0, a]
p[10]

answered 4 hours ago

LouisB

4,2341616

answered 4 hours ago

LouisB

4,2341616

answered 4 hours ago

LouisB

4,2341616

answered 4 hours ago

LouisB

4,2341616

add a commentÂ |Â

up vote
2
down vote

With[a = 10,
 Plot[
 x/a,
 x, 0, a
 ]
 ]

answered 2 hours ago

rhermans

21.8k439104

add a commentÂ |Â

up vote
2
down vote

With[a = 10,
 Plot[
 x/a,
 x, 0, a
 ]
 ]

answered 2 hours ago

rhermans

21.8k439104

add a commentÂ |Â

up vote
2
down vote

With[a = 10,
 Plot[
 x/a,
 x, 0, a
 ]
 ]

answered 2 hours ago

rhermans

21.8k439104

With[a = 10,
 Plot[
 x/a,
 x, 0, a
 ]
 ]

answered 2 hours ago

rhermans

21.8k439104

answered 2 hours ago

rhermans

21.8k439104

answered 2 hours ago

rhermans

21.8k439104

answered 2 hours ago

rhermans

21.8k439104

add a commentÂ |Â

up vote
2
down vote

You can delay the evaluation of Plot with Unevaluated:

Unevaluated@Plot[x/a, x, 0, a] /. a -> 10

If all occurrences of a are not literally present in the Plot code, Block can be used in place of ReplaceAll:

function = x/a;
x = 2; (* to show that this is ignored in the Plot command *)
Block[a = 10,
 Plot[function, x, 0, a]
 ]

function[x_, a_] := x/a

It doesn't help with the OP's substitution problem, in which a appears in more than just the function, so this or one of the other solutions will be needed for that.

edited 2 hours ago

answered 2 hours ago

Michael E2

142k11192458

add a commentÂ |Â

up vote
2
down vote

You can delay the evaluation of Plot with Unevaluated:

Unevaluated@Plot[x/a, x, 0, a] /. a -> 10

If all occurrences of a are not literally present in the Plot code, Block can be used in place of ReplaceAll:

function = x/a;
x = 2; (* to show that this is ignored in the Plot command *)
Block[a = 10,
 Plot[function, x, 0, a]
 ]

function[x_, a_] := x/a

It doesn't help with the OP's substitution problem, in which a appears in more than just the function, so this or one of the other solutions will be needed for that.

edited 2 hours ago

answered 2 hours ago

Michael E2

142k11192458

add a commentÂ |Â

up vote
2
down vote

You can delay the evaluation of Plot with Unevaluated:

Unevaluated@Plot[x/a, x, 0, a] /. a -> 10

If all occurrences of a are not literally present in the Plot code, Block can be used in place of ReplaceAll:

function = x/a;
x = 2; (* to show that this is ignored in the Plot command *)
Block[a = 10,
 Plot[function, x, 0, a]
 ]

function[x_, a_] := x/a

It doesn't help with the OP's substitution problem, in which a appears in more than just the function, so this or one of the other solutions will be needed for that.

edited 2 hours ago

answered 2 hours ago

Michael E2

142k11192458

You can delay the evaluation of Plot with Unevaluated:

Unevaluated@Plot[x/a, x, 0, a] /. a -> 10

If all occurrences of a are not literally present in the Plot code, Block can be used in place of ReplaceAll:

function = x/a;
x = 2; (* to show that this is ignored in the Plot command *)
Block[a = 10,
 Plot[function, x, 0, a]
 ]

function[x_, a_] := x/a

It doesn't help with the OP's substitution problem, in which a appears in more than just the function, so this or one of the other solutions will be needed for that.

edited 2 hours ago

answered 2 hours ago

Michael E2

142k11192458

edited 2 hours ago

answered 2 hours ago

Michael E2

142k11192458

answered 2 hours ago

Michael E2

142k11192458

answered 2 hours ago

Michael E2

142k11192458

add a commentÂ |Â

Tohiko is a new contributor. Be nice, and check out our Code of Conduct.

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