Can the Assuming function be used with ContourPlot or DensityPlot?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












3












$begingroup$


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_] :=
(x*y)^(p - 1)/(α + β*x + γ*y + δ*x*y)^(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[(x*y)^(p - 1)/(α + β*x + γ*y + δ*x*y)^(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.










share|improve this question











$endgroup$







  • 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 slightly pensive
    Mar 4 at 2:47










  • $begingroup$
    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
    Mar 4 at 3:09






  • 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
    Mar 4 at 3:59











  • $begingroup$
    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
    Mar 4 at 4:08










  • $begingroup$
    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
    Mar 4 at 4:11















3












$begingroup$


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_] :=
(x*y)^(p - 1)/(α + β*x + γ*y + δ*x*y)^(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[(x*y)^(p - 1)/(α + β*x + γ*y + δ*x*y)^(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.










share|improve this question











$endgroup$







  • 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 slightly pensive
    Mar 4 at 2:47










  • $begingroup$
    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
    Mar 4 at 3:09






  • 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
    Mar 4 at 3:59











  • $begingroup$
    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
    Mar 4 at 4:08










  • $begingroup$
    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
    Mar 4 at 4:11













3












3








3





$begingroup$


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_] :=
(x*y)^(p - 1)/(α + β*x + γ*y + δ*x*y)^(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[(x*y)^(p - 1)/(α + β*x + γ*y + δ*x*y)^(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.










share|improve this question











$endgroup$




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_] :=
(x*y)^(p - 1)/(α + β*x + γ*y + δ*x*y)^(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[(x*y)^(p - 1)/(α + β*x + γ*y + δ*x*y)^(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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 4 at 18:47







Banks Osborne

















asked Mar 4 at 2:40









Banks OsborneBanks Osborne

183




183







  • 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 slightly pensive
    Mar 4 at 2:47










  • $begingroup$
    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
    Mar 4 at 3:09






  • 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
    Mar 4 at 3:59











  • $begingroup$
    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
    Mar 4 at 4:08










  • $begingroup$
    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
    Mar 4 at 4:11












  • 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 slightly pensive
    Mar 4 at 2:47










  • $begingroup$
    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
    Mar 4 at 3:09






  • 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
    Mar 4 at 3:59











  • $begingroup$
    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
    Mar 4 at 4:08










  • $begingroup$
    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
    Mar 4 at 4:11







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 slightly pensive
Mar 4 at 2:47




$begingroup$
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 slightly pensive
Mar 4 at 2:47












$begingroup$
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
Mar 4 at 3:09




$begingroup$
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
Mar 4 at 3:09




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
Mar 4 at 3:59





$begingroup$
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
Mar 4 at 3:59













$begingroup$
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
Mar 4 at 4:08




$begingroup$
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
Mar 4 at 4:08












$begingroup$
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
Mar 4 at 4:11




$begingroup$
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
Mar 4 at 4:11










2 Answers
2






active

oldest

votes


















4












$begingroup$

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"]]


demo



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.






share|improve this answer











$endgroup$












  • $begingroup$
    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
    Mar 4 at 5:33










  • $begingroup$
    @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 slightly pensive
    Mar 4 at 5:46










  • $begingroup$
    @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
    Mar 4 at 5:50











  • $begingroup$
    @mjw. I use this style more for reasons of clarity than convenience.
    $endgroup$
    – m_goldberg
    Mar 4 at 5:54










  • $begingroup$
    @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
    Mar 4 at 5:57


















0












$begingroup$

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]


enter image description here






share|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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"]]


    demo



    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.






    share|improve this answer











    $endgroup$












    • $begingroup$
      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
      Mar 4 at 5:33










    • $begingroup$
      @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 slightly pensive
      Mar 4 at 5:46










    • $begingroup$
      @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
      Mar 4 at 5:50











    • $begingroup$
      @mjw. I use this style more for reasons of clarity than convenience.
      $endgroup$
      – m_goldberg
      Mar 4 at 5:54










    • $begingroup$
      @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
      Mar 4 at 5:57















    4












    $begingroup$

    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"]]


    demo



    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.






    share|improve this answer











    $endgroup$












    • $begingroup$
      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
      Mar 4 at 5:33










    • $begingroup$
      @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 slightly pensive
      Mar 4 at 5:46










    • $begingroup$
      @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
      Mar 4 at 5:50











    • $begingroup$
      @mjw. I use this style more for reasons of clarity than convenience.
      $endgroup$
      – m_goldberg
      Mar 4 at 5:54










    • $begingroup$
      @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
      Mar 4 at 5:57













    4












    4








    4





    $begingroup$

    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"]]


    demo



    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.






    share|improve this answer











    $endgroup$



    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"]]


    demo



    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.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 4 at 5:58

























    answered Mar 4 at 4:42









    m_goldbergm_goldberg

    88k872199




    88k872199











    • $begingroup$
      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
      Mar 4 at 5:33










    • $begingroup$
      @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 slightly pensive
      Mar 4 at 5:46










    • $begingroup$
      @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
      Mar 4 at 5:50











    • $begingroup$
      @mjw. I use this style more for reasons of clarity than convenience.
      $endgroup$
      – m_goldberg
      Mar 4 at 5:54










    • $begingroup$
      @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
      Mar 4 at 5:57
















    • $begingroup$
      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
      Mar 4 at 5:33










    • $begingroup$
      @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 slightly pensive
      Mar 4 at 5:46










    • $begingroup$
      @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
      Mar 4 at 5:50











    • $begingroup$
      @mjw. I use this style more for reasons of clarity than convenience.
      $endgroup$
      – m_goldberg
      Mar 4 at 5:54










    • $begingroup$
      @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
      Mar 4 at 5:57















    $begingroup$
    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
    Mar 4 at 5:33




    $begingroup$
    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
    Mar 4 at 5:33












    $begingroup$
    @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 slightly pensive
    Mar 4 at 5:46




    $begingroup$
    @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 slightly pensive
    Mar 4 at 5:46












    $begingroup$
    @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
    Mar 4 at 5:50





    $begingroup$
    @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
    Mar 4 at 5:50













    $begingroup$
    @mjw. I use this style more for reasons of clarity than convenience.
    $endgroup$
    – m_goldberg
    Mar 4 at 5:54




    $begingroup$
    @mjw. I use this style more for reasons of clarity than convenience.
    $endgroup$
    – m_goldberg
    Mar 4 at 5:54












    $begingroup$
    @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
    Mar 4 at 5:57




    $begingroup$
    @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
    Mar 4 at 5:57











    0












    $begingroup$

    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]


    enter image description here






    share|improve this answer











    $endgroup$

















      0












      $begingroup$

      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]


      enter image description here






      share|improve this answer











      $endgroup$















        0












        0








        0





        $begingroup$

        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]


        enter image description here






        share|improve this answer











        $endgroup$



        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]


        enter image description here







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 4 at 6:14









        m_goldberg

        88k872199




        88k872199










        answered Mar 4 at 4:20









        mjwmjw

        1,21810




        1,21810



























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