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General::ivar 0 is not a valid variable when using Series in Plot
General::ivar is not a valid variable when plotting - what actually causes this and how to avoid it?Series and that old ivar errorGeneral::ivar is not a valid variable when plotting - what actually causes this and how to avoid it?Problems with For statementAdding plot legends in version 8Series and that old ivar errorHow to define a function that is related to derivative of Jacobi theta functionQuartic function from dataImplementing an integral test for series convergenceGeneral: 0 is not a valid variableWhy is ww[[1]] not a valid variable in FindMinValue and PlotEvaluation of an Error function
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I have been having some trouble graphing a Taylor series approximation (Series).
f[x] := Series[E^(-x/4) Sin[3 * x], x, 1, 4]
Plot[f[x], x, 0, 6]
General::ivar: 0.00012257142857142857` is not a valid variable.
How can I fix this error?
plotting error variable
edited Aug 2 at 9:38
rhermans
23.8k4 gold badges42 silver badges110 bronze badges
asked Aug 1 at 5:29
Oliver MurfettOliver Murfett
161 bronze badge
$endgroup$
|
show 1 more comment
$begingroup$
I have been having some trouble graphing a Taylor series approximation (Series).
f[x] := Series[E^(-x/4) Sin[3 * x], x, 1, 4]
Plot[f[x], x, 0, 6]
General::ivar: 0.00012257142857142857` is not a valid variable.
How can I fix this error?
plotting error variable
edited Aug 2 at 9:38
rhermans
23.8k4 gold badges42 silver badges110 bronze badges
asked Aug 1 at 5:29
Oliver MurfettOliver Murfett
161 bronze badge
$endgroup$
5
$begingroup$
Usef[x_] := ...notf[x] := .... Please see Defining functions.
$endgroup$
– Szabolcs
Aug 1 at 5:39
1
$begingroup$
@Szabolcs and others, I reopened the question because that comment is not an answer. As shown in the last code block in my answer it does not even need to be a part of the answer. I believe voters were influenced by your comment but I won't reopen it again if it gets closed again.
$endgroup$
– Kuba♦
Aug 1 at 9:14
$begingroup$
Please include the message name to make it easier for others who have the same problem to search for and find this question
$endgroup$
– Michael E2
Aug 1 at 11:47
$begingroup$
@Kuba Duplicate?: mathematica.stackexchange.com/q/48980/4999
$endgroup$
– Michael E2
Aug 1 at 11:49
$begingroup$
Related: mathematica.stackexchange.com/q/1301/4999
$endgroup$
– Michael E2
Aug 1 at 11:53
|
show 1 more comment
$begingroup$
I have been having some trouble graphing a Taylor series approximation (Series).
f[x] := Series[E^(-x/4) Sin[3 * x], x, 1, 4]
Plot[f[x], x, 0, 6]
General::ivar: 0.00012257142857142857` is not a valid variable.
How can I fix this error?
plotting error variable
edited Aug 2 at 9:38
rhermans
23.8k4 gold badges42 silver badges110 bronze badges
asked Aug 1 at 5:29
Oliver MurfettOliver Murfett
161 bronze badge
$endgroup$
I have been having some trouble graphing a Taylor series approximation (Series).
f[x] := Series[E^(-x/4) Sin[3 * x], x, 1, 4]
Plot[f[x], x, 0, 6]
General::ivar: 0.00012257142857142857` is not a valid variable.
How can I fix this error?
plotting error variable
plotting error variable
edited Aug 2 at 9:38
rhermans
23.8k4 gold badges42 silver badges110 bronze badges
asked Aug 1 at 5:29
Oliver MurfettOliver Murfett
161 bronze badge
edited Aug 2 at 9:38
rhermans
23.8k4 gold badges42 silver badges110 bronze badges
asked Aug 1 at 5:29
Oliver MurfettOliver Murfett
161 bronze badge
edited Aug 2 at 9:38
rhermans
23.8k4 gold badges42 silver badges110 bronze badges
edited Aug 2 at 9:38
rhermans
23.8k4 gold badges42 silver badges110 bronze badges
edited Aug 2 at 9:38
rhermans
23.8k4 gold badges42 silver badges110 bronze badges
23.8k4 gold badges42 silver badges110 bronze badges
asked Aug 1 at 5:29
Oliver MurfettOliver Murfett
161 bronze badge
asked Aug 1 at 5:29
Oliver MurfettOliver Murfett
161 bronze badge
asked Aug 1 at 5:29
Oliver MurfettOliver Murfett
161 bronze badge
161 bronze badge
5
$begingroup$
Usef[x_] := ...notf[x] := .... Please see Defining functions.
$endgroup$
– Szabolcs
Aug 1 at 5:39
1
$begingroup$
@Szabolcs and others, I reopened the question because that comment is not an answer. As shown in the last code block in my answer it does not even need to be a part of the answer. I believe voters were influenced by your comment but I won't reopen it again if it gets closed again.
$endgroup$
– Kuba♦
Aug 1 at 9:14
$begingroup$
Please include the message name to make it easier for others who have the same problem to search for and find this question
$endgroup$
– Michael E2
Aug 1 at 11:47
$begingroup$
@Kuba Duplicate?: mathematica.stackexchange.com/q/48980/4999
$endgroup$
– Michael E2
Aug 1 at 11:49
$begingroup$
Related: mathematica.stackexchange.com/q/1301/4999
$endgroup$
– Michael E2
Aug 1 at 11:53
|
show 1 more comment
5
$begingroup$
Usef[x_] := ...notf[x] := .... Please see Defining functions.
$endgroup$
– Szabolcs
Aug 1 at 5:39
1
$begingroup$
@Szabolcs and others, I reopened the question because that comment is not an answer. As shown in the last code block in my answer it does not even need to be a part of the answer. I believe voters were influenced by your comment but I won't reopen it again if it gets closed again.
$endgroup$
– Kuba♦
Aug 1 at 9:14
$begingroup$
Please include the message name to make it easier for others who have the same problem to search for and find this question
$endgroup$
– Michael E2
Aug 1 at 11:47
$begingroup$
@Kuba Duplicate?: mathematica.stackexchange.com/q/48980/4999
$endgroup$
– Michael E2
Aug 1 at 11:49
$begingroup$
Related: mathematica.stackexchange.com/q/1301/4999
$endgroup$
– Michael E2
Aug 1 at 11:53
5
5
$begingroup$
Use
f[x_] := ... not f[x] := .... Please see Defining functions.$endgroup$
– Szabolcs
Aug 1 at 5:39
$begingroup$
Use
f[x_] := ... not f[x] := .... Please see Defining functions.$endgroup$
– Szabolcs
Aug 1 at 5:39
1
1
$begingroup$
@Szabolcs and others, I reopened the question because that comment is not an answer. As shown in the last code block in my answer it does not even need to be a part of the answer. I believe voters were influenced by your comment but I won't reopen it again if it gets closed again.
$endgroup$
– Kuba♦
Aug 1 at 9:14
$begingroup$
@Szabolcs and others, I reopened the question because that comment is not an answer. As shown in the last code block in my answer it does not even need to be a part of the answer. I believe voters were influenced by your comment but I won't reopen it again if it gets closed again.
$endgroup$
– Kuba♦
Aug 1 at 9:14
$begingroup$
Please include the message name to make it easier for others who have the same problem to search for and find this question
$endgroup$
– Michael E2
Aug 1 at 11:47
$begingroup$
Please include the message name to make it easier for others who have the same problem to search for and find this question
$endgroup$
– Michael E2
Aug 1 at 11:47
$begingroup$
@Kuba Duplicate?: mathematica.stackexchange.com/q/48980/4999
$endgroup$
– Michael E2
Aug 1 at 11:49
$begingroup$
@Kuba Duplicate?: mathematica.stackexchange.com/q/48980/4999
$endgroup$
– Michael E2
Aug 1 at 11:49
$begingroup$
Related: mathematica.stackexchange.com/q/1301/4999
$endgroup$
– Michael E2
Aug 1 at 11:53
$begingroup$
Related: mathematica.stackexchange.com/q/1301/4999
$endgroup$
– Michael E2
Aug 1 at 11:53
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
There are couple of problems with your code.
- It is an image, it would be way nicer not to have to rewrite it.
Plotting
SeriesIf you go to ref / Series / Application you will see that
Normalis used to plot aSeriesas otherwiseO[x]^nwill makePlotconfused.Function definition
Functions are defined more or less like that
f[x_]:=...but ifxis an argument to your function thenSeriesspecx,1,4will become invalid asxwill be replace with the passed value.You need to create series before you pass the value. One way to do this is to create series once for all by doing it during definition:
=vs:=:
So a 'proper way' to get what you need is:
ClearAll[f];
f[x_] = Normal @ Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[f[x], x, 0, 6]
Coincidentally you original code was close to working, if you know what is going on:
ClearAll[f];
f[x] := Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[Evaluate@Normal@f[x], x, 0, 6]
But this is not a way to go anyway.
answered Aug 1 at 7:40
Kuba♦Kuba
111k12 gold badges220 silver badges563 bronze badges
$endgroup$
1
$begingroup$
ClearAll[f,x]? If you are ensuring there are no lingering definitions, then probably address all the symbols involved?
$endgroup$
– rhermans
Aug 2 at 9:23
add a comment |
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
There are couple of problems with your code.
- It is an image, it would be way nicer not to have to rewrite it.
Plotting
SeriesIf you go to ref / Series / Application you will see that
Normalis used to plot aSeriesas otherwiseO[x]^nwill makePlotconfused.Function definition
Functions are defined more or less like that
f[x_]:=...but ifxis an argument to your function thenSeriesspecx,1,4will become invalid asxwill be replace with the passed value.You need to create series before you pass the value. One way to do this is to create series once for all by doing it during definition:
=vs:=:
So a 'proper way' to get what you need is:
ClearAll[f];
f[x_] = Normal @ Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[f[x], x, 0, 6]
Coincidentally you original code was close to working, if you know what is going on:
ClearAll[f];
f[x] := Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[Evaluate@Normal@f[x], x, 0, 6]
But this is not a way to go anyway.
answered Aug 1 at 7:40
Kuba♦Kuba
111k12 gold badges220 silver badges563 bronze badges
$endgroup$
1
$begingroup$
ClearAll[f,x]? If you are ensuring there are no lingering definitions, then probably address all the symbols involved?
$endgroup$
– rhermans
Aug 2 at 9:23
add a comment |
$begingroup$
There are couple of problems with your code.
- It is an image, it would be way nicer not to have to rewrite it.
Plotting
SeriesIf you go to ref / Series / Application you will see that
Normalis used to plot aSeriesas otherwiseO[x]^nwill makePlotconfused.Function definition
Functions are defined more or less like that
f[x_]:=...but ifxis an argument to your function thenSeriesspecx,1,4will become invalid asxwill be replace with the passed value.You need to create series before you pass the value. One way to do this is to create series once for all by doing it during definition:
=vs:=:
So a 'proper way' to get what you need is:
ClearAll[f];
f[x_] = Normal @ Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[f[x], x, 0, 6]
Coincidentally you original code was close to working, if you know what is going on:
ClearAll[f];
f[x] := Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[Evaluate@Normal@f[x], x, 0, 6]
But this is not a way to go anyway.
answered Aug 1 at 7:40
Kuba♦Kuba
111k12 gold badges220 silver badges563 bronze badges
$endgroup$
1
$begingroup$
ClearAll[f,x]? If you are ensuring there are no lingering definitions, then probably address all the symbols involved?
$endgroup$
– rhermans
Aug 2 at 9:23
add a comment |
$begingroup$
There are couple of problems with your code.
- It is an image, it would be way nicer not to have to rewrite it.
Plotting
SeriesIf you go to ref / Series / Application you will see that
Normalis used to plot aSeriesas otherwiseO[x]^nwill makePlotconfused.Function definition
Functions are defined more or less like that
f[x_]:=...but ifxis an argument to your function thenSeriesspecx,1,4will become invalid asxwill be replace with the passed value.You need to create series before you pass the value. One way to do this is to create series once for all by doing it during definition:
=vs:=:
So a 'proper way' to get what you need is:
ClearAll[f];
f[x_] = Normal @ Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[f[x], x, 0, 6]
Coincidentally you original code was close to working, if you know what is going on:
ClearAll[f];
f[x] := Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[Evaluate@Normal@f[x], x, 0, 6]
But this is not a way to go anyway.
answered Aug 1 at 7:40
Kuba♦Kuba
111k12 gold badges220 silver badges563 bronze badges
$endgroup$
There are couple of problems with your code.
- It is an image, it would be way nicer not to have to rewrite it.
Plotting
SeriesIf you go to ref / Series / Application you will see that
Normalis used to plot aSeriesas otherwiseO[x]^nwill makePlotconfused.Function definition
Functions are defined more or less like that
f[x_]:=...but ifxis an argument to your function thenSeriesspecx,1,4will become invalid asxwill be replace with the passed value.You need to create series before you pass the value. One way to do this is to create series once for all by doing it during definition:
=vs:=:
So a 'proper way' to get what you need is:
ClearAll[f];
f[x_] = Normal @ Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[f[x], x, 0, 6]
Coincidentally you original code was close to working, if you know what is going on:
ClearAll[f];
f[x] := Series[E^(-x/4) Sin[3 x], x, 1, 4]
Plot[Evaluate@Normal@f[x], x, 0, 6]
But this is not a way to go anyway.
answered Aug 1 at 7:40
Kuba♦Kuba
111k12 gold badges220 silver badges563 bronze badges
answered Aug 1 at 7:40
Kuba♦Kuba
111k12 gold badges220 silver badges563 bronze badges
answered Aug 1 at 7:40
Kuba♦Kuba
111k12 gold badges220 silver badges563 bronze badges
answered Aug 1 at 7:40
Kuba♦Kuba
111k12 gold badges220 silver badges563 bronze badges
111k12 gold badges220 silver badges563 bronze badges
1
$begingroup$
ClearAll[f,x]? If you are ensuring there are no lingering definitions, then probably address all the symbols involved?
$endgroup$
– rhermans
Aug 2 at 9:23
add a comment |
1
$begingroup$
ClearAll[f,x]? If you are ensuring there are no lingering definitions, then probably address all the symbols involved?
$endgroup$
– rhermans
Aug 2 at 9:23
1
1
$begingroup$
ClearAll[f,x]? If you are ensuring there are no lingering definitions, then probably address all the symbols involved?$endgroup$
– rhermans
Aug 2 at 9:23
$begingroup$
ClearAll[f,x]? If you are ensuring there are no lingering definitions, then probably address all the symbols involved?$endgroup$
– rhermans
Aug 2 at 9:23
add a comment |
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5
$begingroup$
Use
f[x_] := ...notf[x] := .... Please see Defining functions.$endgroup$
– Szabolcs
Aug 1 at 5:39
1
$begingroup$
@Szabolcs and others, I reopened the question because that comment is not an answer. As shown in the last code block in my answer it does not even need to be a part of the answer. I believe voters were influenced by your comment but I won't reopen it again if it gets closed again.
$endgroup$
– Kuba♦
Aug 1 at 9:14
$begingroup$
Please include the message name to make it easier for others who have the same problem to search for and find this question
$endgroup$
– Michael E2
Aug 1 at 11:47
$begingroup$
@Kuba Duplicate?: mathematica.stackexchange.com/q/48980/4999
$endgroup$
– Michael E2
Aug 1 at 11:49
$begingroup$
Related: mathematica.stackexchange.com/q/1301/4999
$endgroup$
– Michael E2
Aug 1 at 11:53