Can the Assuming function be used with ContourPlot or DensityPlot?
$begingroup$
I'm new to Mathematica, and for most purposes the program has served me well and been straightforward. However, I'm hitting a snag while trying to create a contour plot for the distribution function
$qquad f(x,y) = (x,y)^{p-1}/(alpha + beta,x + gamma,y + delta,x,y)^{p + q}$
Notice $x,y$ are variables, and $alpha,beta,gamma,delta, p,$ and $q$ are constants. I need to set a list of assumptions for constants in the function, but my attempts have been fruitless. Every command yields a graph without an image.
I first tried assigning my function with its assumptions by:
Assuming[
{x > 0, y > 0, p > 0, α > 0, β > 0, γ > 0, δ > 0},
f[x_, y_] :=
(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
After the assignment, I tried plotting with ContourPlot
and DensityPlot
.
I'll provide just the ContourPlot
expression below because not much changes across them:
ContourPlot[f[x, y], {x, 0, 200}, {y, 0, 200}]
In regards to the ContourPlot
code, I've changed the domain to both larger and smaller numbers to no avail. Neither ContourPlot
nor DensityPlot
provides an image. I then try the code without assigning the function beforehand, while including ContourPlot within the Assuming command:
Assuming[
{α > 0, β > 0, γ > 0, δ > 0, p > 0},
ContourPlot[(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q), {x, 0, 3}, {y, 0, 3}]]
I know this equation should produce some sort of image since it's simply a type of truncated distribution function. I believe I've narrowed down the issue to one of the following: Mathematica does not allow assumptions to be used with ContourPlot/DensityPlot
, the distribution function is too complicated for Mathematica, or my user error is hindering me. My next step is to try creating different plots on the same graph for various pre-determined values of the parameters.
Any help is much appreciated. As previously mentioned, I'm not very experienced with Mathematica, so I'm more than willing to learn something new or help further explain my goals.
plotting assumptions
New contributor
$endgroup$
add a comment |
$begingroup$
I'm new to Mathematica, and for most purposes the program has served me well and been straightforward. However, I'm hitting a snag while trying to create a contour plot for the distribution function
$qquad f(x,y) = (x,y)^{p-1}/(alpha + beta,x + gamma,y + delta,x,y)^{p + q}$
Notice $x,y$ are variables, and $alpha,beta,gamma,delta, p,$ and $q$ are constants. I need to set a list of assumptions for constants in the function, but my attempts have been fruitless. Every command yields a graph without an image.
I first tried assigning my function with its assumptions by:
Assuming[
{x > 0, y > 0, p > 0, α > 0, β > 0, γ > 0, δ > 0},
f[x_, y_] :=
(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
After the assignment, I tried plotting with ContourPlot
and DensityPlot
.
I'll provide just the ContourPlot
expression below because not much changes across them:
ContourPlot[f[x, y], {x, 0, 200}, {y, 0, 200}]
In regards to the ContourPlot
code, I've changed the domain to both larger and smaller numbers to no avail. Neither ContourPlot
nor DensityPlot
provides an image. I then try the code without assigning the function beforehand, while including ContourPlot within the Assuming command:
Assuming[
{α > 0, β > 0, γ > 0, δ > 0, p > 0},
ContourPlot[(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q), {x, 0, 3}, {y, 0, 3}]]
I know this equation should produce some sort of image since it's simply a type of truncated distribution function. I believe I've narrowed down the issue to one of the following: Mathematica does not allow assumptions to be used with ContourPlot/DensityPlot
, the distribution function is too complicated for Mathematica, or my user error is hindering me. My next step is to try creating different plots on the same graph for various pre-determined values of the parameters.
Any help is much appreciated. As previously mentioned, I'm not very experienced with Mathematica, so I'm more than willing to learn something new or help further explain my goals.
plotting assumptions
New contributor
$endgroup$
1
$begingroup$
MemberQ[Keys[Options[ContourPlot]], Assumptions]
returnsFalse
, so you can't use assumptions onContourPlot
. Your more pressing problem is that you have neglected to provide concrete values for your parameters, so there really is nothing for the plotter to do. (Also,xy
andx y
are very different things, which contributes to why you can't plot.)
$endgroup$
– J. M. is computer-less♦
5 hours ago
$begingroup$
That makes sense. I'll stop trying to useAssumptions
withContourPlot
now. I don't know how I didn't catch myself sooner, but I now realize I had typedxy
instead ofx*y
. This actually fixes another, unrelated issue I was having with the code. That said, thank you so much for your help!
$endgroup$
– Banks Osborne
4 hours ago
1
$begingroup$
Shouldn't(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
be(x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)]
The additional spaces make a big difference.
$endgroup$
– m_goldberg
3 hours ago
$begingroup$
Yes, was thinking the same thing, otherwise Mathematica thinks each term is one variable for examplexy
and not the product of these.
$endgroup$
– mjw
3 hours ago
$begingroup$
You also need to set the constants to some values to plot your function. To take a simpler example, to plotExp[-x^2 / (2 sigma^2)] / (sigma Sqrt[2 pi]
, you would need to specifysigma
.
$endgroup$
– mjw
3 hours ago
add a comment |
$begingroup$
I'm new to Mathematica, and for most purposes the program has served me well and been straightforward. However, I'm hitting a snag while trying to create a contour plot for the distribution function
$qquad f(x,y) = (x,y)^{p-1}/(alpha + beta,x + gamma,y + delta,x,y)^{p + q}$
Notice $x,y$ are variables, and $alpha,beta,gamma,delta, p,$ and $q$ are constants. I need to set a list of assumptions for constants in the function, but my attempts have been fruitless. Every command yields a graph without an image.
I first tried assigning my function with its assumptions by:
Assuming[
{x > 0, y > 0, p > 0, α > 0, β > 0, γ > 0, δ > 0},
f[x_, y_] :=
(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
After the assignment, I tried plotting with ContourPlot
and DensityPlot
.
I'll provide just the ContourPlot
expression below because not much changes across them:
ContourPlot[f[x, y], {x, 0, 200}, {y, 0, 200}]
In regards to the ContourPlot
code, I've changed the domain to both larger and smaller numbers to no avail. Neither ContourPlot
nor DensityPlot
provides an image. I then try the code without assigning the function beforehand, while including ContourPlot within the Assuming command:
Assuming[
{α > 0, β > 0, γ > 0, δ > 0, p > 0},
ContourPlot[(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q), {x, 0, 3}, {y, 0, 3}]]
I know this equation should produce some sort of image since it's simply a type of truncated distribution function. I believe I've narrowed down the issue to one of the following: Mathematica does not allow assumptions to be used with ContourPlot/DensityPlot
, the distribution function is too complicated for Mathematica, or my user error is hindering me. My next step is to try creating different plots on the same graph for various pre-determined values of the parameters.
Any help is much appreciated. As previously mentioned, I'm not very experienced with Mathematica, so I'm more than willing to learn something new or help further explain my goals.
plotting assumptions
New contributor
$endgroup$
I'm new to Mathematica, and for most purposes the program has served me well and been straightforward. However, I'm hitting a snag while trying to create a contour plot for the distribution function
$qquad f(x,y) = (x,y)^{p-1}/(alpha + beta,x + gamma,y + delta,x,y)^{p + q}$
Notice $x,y$ are variables, and $alpha,beta,gamma,delta, p,$ and $q$ are constants. I need to set a list of assumptions for constants in the function, but my attempts have been fruitless. Every command yields a graph without an image.
I first tried assigning my function with its assumptions by:
Assuming[
{x > 0, y > 0, p > 0, α > 0, β > 0, γ > 0, δ > 0},
f[x_, y_] :=
(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
After the assignment, I tried plotting with ContourPlot
and DensityPlot
.
I'll provide just the ContourPlot
expression below because not much changes across them:
ContourPlot[f[x, y], {x, 0, 200}, {y, 0, 200}]
In regards to the ContourPlot
code, I've changed the domain to both larger and smaller numbers to no avail. Neither ContourPlot
nor DensityPlot
provides an image. I then try the code without assigning the function beforehand, while including ContourPlot within the Assuming command:
Assuming[
{α > 0, β > 0, γ > 0, δ > 0, p > 0},
ContourPlot[(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q), {x, 0, 3}, {y, 0, 3}]]
I know this equation should produce some sort of image since it's simply a type of truncated distribution function. I believe I've narrowed down the issue to one of the following: Mathematica does not allow assumptions to be used with ContourPlot/DensityPlot
, the distribution function is too complicated for Mathematica, or my user error is hindering me. My next step is to try creating different plots on the same graph for various pre-determined values of the parameters.
Any help is much appreciated. As previously mentioned, I'm not very experienced with Mathematica, so I'm more than willing to learn something new or help further explain my goals.
plotting assumptions
plotting assumptions
New contributor
New contributor
edited 3 hours ago
m_goldberg
87k872197
87k872197
New contributor
asked 5 hours ago
Banks OsborneBanks Osborne
61
61
New contributor
New contributor
1
$begingroup$
MemberQ[Keys[Options[ContourPlot]], Assumptions]
returnsFalse
, so you can't use assumptions onContourPlot
. Your more pressing problem is that you have neglected to provide concrete values for your parameters, so there really is nothing for the plotter to do. (Also,xy
andx y
are very different things, which contributes to why you can't plot.)
$endgroup$
– J. M. is computer-less♦
5 hours ago
$begingroup$
That makes sense. I'll stop trying to useAssumptions
withContourPlot
now. I don't know how I didn't catch myself sooner, but I now realize I had typedxy
instead ofx*y
. This actually fixes another, unrelated issue I was having with the code. That said, thank you so much for your help!
$endgroup$
– Banks Osborne
4 hours ago
1
$begingroup$
Shouldn't(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
be(x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)]
The additional spaces make a big difference.
$endgroup$
– m_goldberg
3 hours ago
$begingroup$
Yes, was thinking the same thing, otherwise Mathematica thinks each term is one variable for examplexy
and not the product of these.
$endgroup$
– mjw
3 hours ago
$begingroup$
You also need to set the constants to some values to plot your function. To take a simpler example, to plotExp[-x^2 / (2 sigma^2)] / (sigma Sqrt[2 pi]
, you would need to specifysigma
.
$endgroup$
– mjw
3 hours ago
add a comment |
1
$begingroup$
MemberQ[Keys[Options[ContourPlot]], Assumptions]
returnsFalse
, so you can't use assumptions onContourPlot
. Your more pressing problem is that you have neglected to provide concrete values for your parameters, so there really is nothing for the plotter to do. (Also,xy
andx y
are very different things, which contributes to why you can't plot.)
$endgroup$
– J. M. is computer-less♦
5 hours ago
$begingroup$
That makes sense. I'll stop trying to useAssumptions
withContourPlot
now. I don't know how I didn't catch myself sooner, but I now realize I had typedxy
instead ofx*y
. This actually fixes another, unrelated issue I was having with the code. That said, thank you so much for your help!
$endgroup$
– Banks Osborne
4 hours ago
1
$begingroup$
Shouldn't(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
be(x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)]
The additional spaces make a big difference.
$endgroup$
– m_goldberg
3 hours ago
$begingroup$
Yes, was thinking the same thing, otherwise Mathematica thinks each term is one variable for examplexy
and not the product of these.
$endgroup$
– mjw
3 hours ago
$begingroup$
You also need to set the constants to some values to plot your function. To take a simpler example, to plotExp[-x^2 / (2 sigma^2)] / (sigma Sqrt[2 pi]
, you would need to specifysigma
.
$endgroup$
– mjw
3 hours ago
1
1
$begingroup$
MemberQ[Keys[Options[ContourPlot]], Assumptions]
returns False
, so you can't use assumptions on ContourPlot
. Your more pressing problem is that you have neglected to provide concrete values for your parameters, so there really is nothing for the plotter to do. (Also, xy
and x y
are very different things, which contributes to why you can't plot.)$endgroup$
– J. M. is computer-less♦
5 hours ago
$begingroup$
MemberQ[Keys[Options[ContourPlot]], Assumptions]
returns False
, so you can't use assumptions on ContourPlot
. Your more pressing problem is that you have neglected to provide concrete values for your parameters, so there really is nothing for the plotter to do. (Also, xy
and x y
are very different things, which contributes to why you can't plot.)$endgroup$
– J. M. is computer-less♦
5 hours ago
$begingroup$
That makes sense. I'll stop trying to use
Assumptions
with ContourPlot
now. I don't know how I didn't catch myself sooner, but I now realize I had typed xy
instead of x*y
. This actually fixes another, unrelated issue I was having with the code. That said, thank you so much for your help!$endgroup$
– Banks Osborne
4 hours ago
$begingroup$
That makes sense. I'll stop trying to use
Assumptions
with ContourPlot
now. I don't know how I didn't catch myself sooner, but I now realize I had typed xy
instead of x*y
. This actually fixes another, unrelated issue I was having with the code. That said, thank you so much for your help!$endgroup$
– Banks Osborne
4 hours ago
1
1
$begingroup$
Shouldn't
(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
be (x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)]
The additional spaces make a big difference.$endgroup$
– m_goldberg
3 hours ago
$begingroup$
Shouldn't
(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
be (x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)]
The additional spaces make a big difference.$endgroup$
– m_goldberg
3 hours ago
$begingroup$
Yes, was thinking the same thing, otherwise Mathematica thinks each term is one variable for example
xy
and not the product of these.$endgroup$
– mjw
3 hours ago
$begingroup$
Yes, was thinking the same thing, otherwise Mathematica thinks each term is one variable for example
xy
and not the product of these.$endgroup$
– mjw
3 hours ago
$begingroup$
You also need to set the constants to some values to plot your function. To take a simpler example, to plot
Exp[-x^2 / (2 sigma^2)] / (sigma Sqrt[2 pi]
, you would need to specify sigma
.$endgroup$
– mjw
3 hours ago
$begingroup$
You also need to set the constants to some values to plot your function. To take a simpler example, to plot
Exp[-x^2 / (2 sigma^2)] / (sigma Sqrt[2 pi]
, you would need to specify sigma
.$endgroup$
– mjw
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As has been said in the comments to your question, because all plotting functions are based on strictly numerical calculations, you must give definite values to all six parameters. If you are in the position where you have no good idea how the function behaves as the parameters vary, you can explore the situation with Manipulate
. Here is an example.
f[α_, β_, γ_, δ_, p_, q_][x_, y_] := (x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)
With[{ϵ = .0001},
Manipulate[
ContourPlot[f[α, β, γ, δ, p, q][x, y], {x, 0, 2}, {y, 0, 2}],
{α, ϵ, 1, Appearance -> "Labeled"},
{β, ϵ, 1, Appearance -> "Labeled"},
{γ, ϵ, 1, Appearance -> "Labeled"},
{δ, ϵ, 1, Appearance -> "Labeled"},
{p, 1, 4, 1, Appearance -> "Labeled"},
{q, 1, 4, 1, Appearance -> "Labeled"}]]
Note: I have no clue about what comprise good ranges for either the parameters or the variables. I made some simple assumptions about them. You should revise these assumptions to suit your needs.
$endgroup$
$begingroup$
It looks like you separated out the "constants" and the "variables" in your function definition. Is there any significance to that other than convenience of notation? Where is this documented (just did a quick search for defining functions and did not see it)? Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
@mjw, this tutorial might be of interest. It is not necessary to separate parameters and variables in this way, but it is definitely convenient.
$endgroup$
– J. M. is computer-less♦
2 hours ago
$begingroup$
@mjw. It is documented as J.M. points out. A higher level reference which a list of links about topics concerning functions is this one, which included the link given by J.M.
$endgroup$
– m_goldberg
2 hours ago
$begingroup$
@mjw. I use this style more for reasons of clarity than convenience.
$endgroup$
– m_goldberg
1 hour ago
$begingroup$
@J.M. Thank you! So what isf[a_,b_][x_,y_]
, a function? Or a function of a function? I guess that I understand that now the head of the expression is f[a,b].
$endgroup$
– mjw
1 hour ago
|
show 3 more comments
$begingroup$
One way is to simply set the constants as variables in your function definition, and then set them to the values you want when you call the function:
f[x_, y_, α_, β_, γ_, δ_, p_, q_] := (x y)^(p -1)/(α + β x + γ y + δ x y)^(p + q);
ContourPlot[f[x, y, 2, 3, 4, 5, .5, .5], {x, 0, 200}, {y, 0, 200}]
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
2
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votes
$begingroup$
As has been said in the comments to your question, because all plotting functions are based on strictly numerical calculations, you must give definite values to all six parameters. If you are in the position where you have no good idea how the function behaves as the parameters vary, you can explore the situation with Manipulate
. Here is an example.
f[α_, β_, γ_, δ_, p_, q_][x_, y_] := (x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)
With[{ϵ = .0001},
Manipulate[
ContourPlot[f[α, β, γ, δ, p, q][x, y], {x, 0, 2}, {y, 0, 2}],
{α, ϵ, 1, Appearance -> "Labeled"},
{β, ϵ, 1, Appearance -> "Labeled"},
{γ, ϵ, 1, Appearance -> "Labeled"},
{δ, ϵ, 1, Appearance -> "Labeled"},
{p, 1, 4, 1, Appearance -> "Labeled"},
{q, 1, 4, 1, Appearance -> "Labeled"}]]
Note: I have no clue about what comprise good ranges for either the parameters or the variables. I made some simple assumptions about them. You should revise these assumptions to suit your needs.
$endgroup$
$begingroup$
It looks like you separated out the "constants" and the "variables" in your function definition. Is there any significance to that other than convenience of notation? Where is this documented (just did a quick search for defining functions and did not see it)? Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
@mjw, this tutorial might be of interest. It is not necessary to separate parameters and variables in this way, but it is definitely convenient.
$endgroup$
– J. M. is computer-less♦
2 hours ago
$begingroup$
@mjw. It is documented as J.M. points out. A higher level reference which a list of links about topics concerning functions is this one, which included the link given by J.M.
$endgroup$
– m_goldberg
2 hours ago
$begingroup$
@mjw. I use this style more for reasons of clarity than convenience.
$endgroup$
– m_goldberg
1 hour ago
$begingroup$
@J.M. Thank you! So what isf[a_,b_][x_,y_]
, a function? Or a function of a function? I guess that I understand that now the head of the expression is f[a,b].
$endgroup$
– mjw
1 hour ago
|
show 3 more comments
$begingroup$
As has been said in the comments to your question, because all plotting functions are based on strictly numerical calculations, you must give definite values to all six parameters. If you are in the position where you have no good idea how the function behaves as the parameters vary, you can explore the situation with Manipulate
. Here is an example.
f[α_, β_, γ_, δ_, p_, q_][x_, y_] := (x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)
With[{ϵ = .0001},
Manipulate[
ContourPlot[f[α, β, γ, δ, p, q][x, y], {x, 0, 2}, {y, 0, 2}],
{α, ϵ, 1, Appearance -> "Labeled"},
{β, ϵ, 1, Appearance -> "Labeled"},
{γ, ϵ, 1, Appearance -> "Labeled"},
{δ, ϵ, 1, Appearance -> "Labeled"},
{p, 1, 4, 1, Appearance -> "Labeled"},
{q, 1, 4, 1, Appearance -> "Labeled"}]]
Note: I have no clue about what comprise good ranges for either the parameters or the variables. I made some simple assumptions about them. You should revise these assumptions to suit your needs.
$endgroup$
$begingroup$
It looks like you separated out the "constants" and the "variables" in your function definition. Is there any significance to that other than convenience of notation? Where is this documented (just did a quick search for defining functions and did not see it)? Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
@mjw, this tutorial might be of interest. It is not necessary to separate parameters and variables in this way, but it is definitely convenient.
$endgroup$
– J. M. is computer-less♦
2 hours ago
$begingroup$
@mjw. It is documented as J.M. points out. A higher level reference which a list of links about topics concerning functions is this one, which included the link given by J.M.
$endgroup$
– m_goldberg
2 hours ago
$begingroup$
@mjw. I use this style more for reasons of clarity than convenience.
$endgroup$
– m_goldberg
1 hour ago
$begingroup$
@J.M. Thank you! So what isf[a_,b_][x_,y_]
, a function? Or a function of a function? I guess that I understand that now the head of the expression is f[a,b].
$endgroup$
– mjw
1 hour ago
|
show 3 more comments
$begingroup$
As has been said in the comments to your question, because all plotting functions are based on strictly numerical calculations, you must give definite values to all six parameters. If you are in the position where you have no good idea how the function behaves as the parameters vary, you can explore the situation with Manipulate
. Here is an example.
f[α_, β_, γ_, δ_, p_, q_][x_, y_] := (x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)
With[{ϵ = .0001},
Manipulate[
ContourPlot[f[α, β, γ, δ, p, q][x, y], {x, 0, 2}, {y, 0, 2}],
{α, ϵ, 1, Appearance -> "Labeled"},
{β, ϵ, 1, Appearance -> "Labeled"},
{γ, ϵ, 1, Appearance -> "Labeled"},
{δ, ϵ, 1, Appearance -> "Labeled"},
{p, 1, 4, 1, Appearance -> "Labeled"},
{q, 1, 4, 1, Appearance -> "Labeled"}]]
Note: I have no clue about what comprise good ranges for either the parameters or the variables. I made some simple assumptions about them. You should revise these assumptions to suit your needs.
$endgroup$
As has been said in the comments to your question, because all plotting functions are based on strictly numerical calculations, you must give definite values to all six parameters. If you are in the position where you have no good idea how the function behaves as the parameters vary, you can explore the situation with Manipulate
. Here is an example.
f[α_, β_, γ_, δ_, p_, q_][x_, y_] := (x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)
With[{ϵ = .0001},
Manipulate[
ContourPlot[f[α, β, γ, δ, p, q][x, y], {x, 0, 2}, {y, 0, 2}],
{α, ϵ, 1, Appearance -> "Labeled"},
{β, ϵ, 1, Appearance -> "Labeled"},
{γ, ϵ, 1, Appearance -> "Labeled"},
{δ, ϵ, 1, Appearance -> "Labeled"},
{p, 1, 4, 1, Appearance -> "Labeled"},
{q, 1, 4, 1, Appearance -> "Labeled"}]]
Note: I have no clue about what comprise good ranges for either the parameters or the variables. I made some simple assumptions about them. You should revise these assumptions to suit your needs.
edited 1 hour ago
answered 3 hours ago
m_goldbergm_goldberg
87k872197
87k872197
$begingroup$
It looks like you separated out the "constants" and the "variables" in your function definition. Is there any significance to that other than convenience of notation? Where is this documented (just did a quick search for defining functions and did not see it)? Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
@mjw, this tutorial might be of interest. It is not necessary to separate parameters and variables in this way, but it is definitely convenient.
$endgroup$
– J. M. is computer-less♦
2 hours ago
$begingroup$
@mjw. It is documented as J.M. points out. A higher level reference which a list of links about topics concerning functions is this one, which included the link given by J.M.
$endgroup$
– m_goldberg
2 hours ago
$begingroup$
@mjw. I use this style more for reasons of clarity than convenience.
$endgroup$
– m_goldberg
1 hour ago
$begingroup$
@J.M. Thank you! So what isf[a_,b_][x_,y_]
, a function? Or a function of a function? I guess that I understand that now the head of the expression is f[a,b].
$endgroup$
– mjw
1 hour ago
|
show 3 more comments
$begingroup$
It looks like you separated out the "constants" and the "variables" in your function definition. Is there any significance to that other than convenience of notation? Where is this documented (just did a quick search for defining functions and did not see it)? Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
@mjw, this tutorial might be of interest. It is not necessary to separate parameters and variables in this way, but it is definitely convenient.
$endgroup$
– J. M. is computer-less♦
2 hours ago
$begingroup$
@mjw. It is documented as J.M. points out. A higher level reference which a list of links about topics concerning functions is this one, which included the link given by J.M.
$endgroup$
– m_goldberg
2 hours ago
$begingroup$
@mjw. I use this style more for reasons of clarity than convenience.
$endgroup$
– m_goldberg
1 hour ago
$begingroup$
@J.M. Thank you! So what isf[a_,b_][x_,y_]
, a function? Or a function of a function? I guess that I understand that now the head of the expression is f[a,b].
$endgroup$
– mjw
1 hour ago
$begingroup$
It looks like you separated out the "constants" and the "variables" in your function definition. Is there any significance to that other than convenience of notation? Where is this documented (just did a quick search for defining functions and did not see it)? Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
It looks like you separated out the "constants" and the "variables" in your function definition. Is there any significance to that other than convenience of notation? Where is this documented (just did a quick search for defining functions and did not see it)? Thanks!
$endgroup$
– mjw
2 hours ago
$begingroup$
@mjw, this tutorial might be of interest. It is not necessary to separate parameters and variables in this way, but it is definitely convenient.
$endgroup$
– J. M. is computer-less♦
2 hours ago
$begingroup$
@mjw, this tutorial might be of interest. It is not necessary to separate parameters and variables in this way, but it is definitely convenient.
$endgroup$
– J. M. is computer-less♦
2 hours ago
$begingroup$
@mjw. It is documented as J.M. points out. A higher level reference which a list of links about topics concerning functions is this one, which included the link given by J.M.
$endgroup$
– m_goldberg
2 hours ago
$begingroup$
@mjw. It is documented as J.M. points out. A higher level reference which a list of links about topics concerning functions is this one, which included the link given by J.M.
$endgroup$
– m_goldberg
2 hours ago
$begingroup$
@mjw. I use this style more for reasons of clarity than convenience.
$endgroup$
– m_goldberg
1 hour ago
$begingroup$
@mjw. I use this style more for reasons of clarity than convenience.
$endgroup$
– m_goldberg
1 hour ago
$begingroup$
@J.M. Thank you! So what is
f[a_,b_][x_,y_]
, a function? Or a function of a function? I guess that I understand that now the head of the expression is f[a,b].$endgroup$
– mjw
1 hour ago
$begingroup$
@J.M. Thank you! So what is
f[a_,b_][x_,y_]
, a function? Or a function of a function? I guess that I understand that now the head of the expression is f[a,b].$endgroup$
– mjw
1 hour ago
|
show 3 more comments
$begingroup$
One way is to simply set the constants as variables in your function definition, and then set them to the values you want when you call the function:
f[x_, y_, α_, β_, γ_, δ_, p_, q_] := (x y)^(p -1)/(α + β x + γ y + δ x y)^(p + q);
ContourPlot[f[x, y, 2, 3, 4, 5, .5, .5], {x, 0, 200}, {y, 0, 200}]
$endgroup$
add a comment |
$begingroup$
One way is to simply set the constants as variables in your function definition, and then set them to the values you want when you call the function:
f[x_, y_, α_, β_, γ_, δ_, p_, q_] := (x y)^(p -1)/(α + β x + γ y + δ x y)^(p + q);
ContourPlot[f[x, y, 2, 3, 4, 5, .5, .5], {x, 0, 200}, {y, 0, 200}]
$endgroup$
add a comment |
$begingroup$
One way is to simply set the constants as variables in your function definition, and then set them to the values you want when you call the function:
f[x_, y_, α_, β_, γ_, δ_, p_, q_] := (x y)^(p -1)/(α + β x + γ y + δ x y)^(p + q);
ContourPlot[f[x, y, 2, 3, 4, 5, .5, .5], {x, 0, 200}, {y, 0, 200}]
$endgroup$
One way is to simply set the constants as variables in your function definition, and then set them to the values you want when you call the function:
f[x_, y_, α_, β_, γ_, δ_, p_, q_] := (x y)^(p -1)/(α + β x + γ y + δ x y)^(p + q);
ContourPlot[f[x, y, 2, 3, 4, 5, .5, .5], {x, 0, 200}, {y, 0, 200}]
edited 1 hour ago
m_goldberg
87k872197
87k872197
answered 3 hours ago
mjwmjw
3867
3867
add a comment |
add a comment |
Banks Osborne is a new contributor. Be nice, and check out our Code of Conduct.
Banks Osborne is a new contributor. Be nice, and check out our Code of Conduct.
Banks Osborne is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
MemberQ[Keys[Options[ContourPlot]], Assumptions]
returnsFalse
, so you can't use assumptions onContourPlot
. Your more pressing problem is that you have neglected to provide concrete values for your parameters, so there really is nothing for the plotter to do. (Also,xy
andx y
are very different things, which contributes to why you can't plot.)$endgroup$
– J. M. is computer-less♦
5 hours ago
$begingroup$
That makes sense. I'll stop trying to use
Assumptions
withContourPlot
now. I don't know how I didn't catch myself sooner, but I now realize I had typedxy
instead ofx*y
. This actually fixes another, unrelated issue I was having with the code. That said, thank you so much for your help!$endgroup$
– Banks Osborne
4 hours ago
1
$begingroup$
Shouldn't
(xy)^(p - 1)/(α + βx + γy + δxy)^(p + q)]
be(x y)^(p - 1)/(α + β x + γ y + δ x y)^(p + q)]
The additional spaces make a big difference.$endgroup$
– m_goldberg
3 hours ago
$begingroup$
Yes, was thinking the same thing, otherwise Mathematica thinks each term is one variable for example
xy
and not the product of these.$endgroup$
– mjw
3 hours ago
$begingroup$
You also need to set the constants to some values to plot your function. To take a simpler example, to plot
Exp[-x^2 / (2 sigma^2)] / (sigma Sqrt[2 pi]
, you would need to specifysigma
.$endgroup$
– mjw
3 hours ago