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How can a “non-function” be graphed?
Graphing inverse of parabola, should the domain be restricted or not?Moriarty's calculator: some bizarre and deceptive graphical anomaliesCan trigonometric equations be graphed?Best program for creating educational math animations?Visualising High Powers of Sin and Cos without GraphingBehaviour of multivariate functionsFunctional Square Root Grapher3D Graphing TI-Nspire CXHow to sketch an $xy$ graph from a $yu$ and $xu$ graph.Better method for graphing functions with greater complexity
$begingroup$
I am trying to create a program that requires the graphing of a "non-function", i.e, a function where "x" isn't strictly dependent on "y".
Let's answer the obvious question: Is it even possible?
Yes, it is possible, as demonstrated by online graphing calculators such as Desmos and GeoGebra, as well as several other open-source softwares. However, I haven't had success in figuring out how these graphing calculators graph non-functions. I have done some research, and have found nothing helpful.
Now, I'm looking for means to graph any non-functions, not just simple ones such as a circle or a sideways parabola, which simply require the graphs of +f(x) and -f(x). Maybe more complicated ones such as sin(x) + sin(y)= 1. The following graph is of the said non-function, as graphed by Desmos.

Any help on how to graph non-functions is appreciated. Thanks in advance.
graphing-functions
$endgroup$
add a comment |
$begingroup$
I am trying to create a program that requires the graphing of a "non-function", i.e, a function where "x" isn't strictly dependent on "y".
Let's answer the obvious question: Is it even possible?
Yes, it is possible, as demonstrated by online graphing calculators such as Desmos and GeoGebra, as well as several other open-source softwares. However, I haven't had success in figuring out how these graphing calculators graph non-functions. I have done some research, and have found nothing helpful.
Now, I'm looking for means to graph any non-functions, not just simple ones such as a circle or a sideways parabola, which simply require the graphs of +f(x) and -f(x). Maybe more complicated ones such as sin(x) + sin(y)= 1. The following graph is of the said non-function, as graphed by Desmos.

Any help on how to graph non-functions is appreciated. Thanks in advance.
graphing-functions
$endgroup$
add a comment |
$begingroup$
I am trying to create a program that requires the graphing of a "non-function", i.e, a function where "x" isn't strictly dependent on "y".
Let's answer the obvious question: Is it even possible?
Yes, it is possible, as demonstrated by online graphing calculators such as Desmos and GeoGebra, as well as several other open-source softwares. However, I haven't had success in figuring out how these graphing calculators graph non-functions. I have done some research, and have found nothing helpful.
Now, I'm looking for means to graph any non-functions, not just simple ones such as a circle or a sideways parabola, which simply require the graphs of +f(x) and -f(x). Maybe more complicated ones such as sin(x) + sin(y)= 1. The following graph is of the said non-function, as graphed by Desmos.

Any help on how to graph non-functions is appreciated. Thanks in advance.
graphing-functions
$endgroup$
I am trying to create a program that requires the graphing of a "non-function", i.e, a function where "x" isn't strictly dependent on "y".
Let's answer the obvious question: Is it even possible?
Yes, it is possible, as demonstrated by online graphing calculators such as Desmos and GeoGebra, as well as several other open-source softwares. However, I haven't had success in figuring out how these graphing calculators graph non-functions. I have done some research, and have found nothing helpful.
Now, I'm looking for means to graph any non-functions, not just simple ones such as a circle or a sideways parabola, which simply require the graphs of +f(x) and -f(x). Maybe more complicated ones such as sin(x) + sin(y)= 1. The following graph is of the said non-function, as graphed by Desmos.

Any help on how to graph non-functions is appreciated. Thanks in advance.
graphing-functions
graphing-functions
asked Sep 18 '17 at 3:16
Rapid ReadersRapid Readers
12516
12516
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A lot of the time examples like these "look" like functions locally. Therefore they can broken up into little pieces by using a valid set of initial conditions or a function which is valid and describes part of the overall relation you are trying to graph.
In your example we can convert the equation to look like:
$$
y = sin^-1(1-sin(x))
$$
Which gives a starting point looking like the bottom half of a single warped sphere. Then we see that since $sin(y)=sin(pi-y)$ we can cap the warped sphere. Then we can argue that the equation is clearly unchanged if we were to replace $(x,y)$ with $(x+2pi n ,y+2pi m)$ (for integer values of m,n) and so we can repeat our pattern throughout the entire grid due to the periodicity of the $sin$ function.
Another good example is $y^2 = x$ where we have two sets of solutions corresponding to $y=pm sqrtx $.
$endgroup$
$begingroup$
I really like your explanation. However, there is a problem. Some non-functions cannot be simplified exclusively to have y be dependent on x. For instance,x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy, which isn't even symmetric like the other examples, but can still be graphed by Desmos.
$endgroup$
– Rapid Readers
Sep 18 '17 at 4:11
$begingroup$
If you are alright with approximate solutions then you can differentiate w.r.t. x to isolate for $fracdydx$ and then expand around as many sets of valid initial conditions as you can find. I'm sure there are some fancier tricks for finding a complete graph for an equation but I was just trying to point out that such an equation acts like a function somewhere around every set of valid initial conditions.
$endgroup$
– Ryan Watt
Sep 19 '17 at 6:36
add a comment |
$begingroup$
An aproach for this would be to take some values.
x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy
For example x=5. We substitute this value. And later we try different values for y until we have a equality (left part is some as right side). Some mechanisms help to do this, looking on how quick we get a approximation. I would first convert this to a equation equal to 0 :
x arcsin( x/y) + sqrt( y^2 - x^2) - 4xy = 0
This is trivial, and now for each x-value, we can get a y value (or many y-values) using a aproximation methode like Newton, which use derivates.
Daniel
P.S. see https://en.wikipedia.org/wiki/Newton%27s_method
$endgroup$
add a comment |
$begingroup$
According from what you asked, let's see images below and take conclusions.
https://i.ytimg.com/vi/y7N0Pr3N1RY/maxresdefault.jpg
(points tracing)
https://i.stack.imgur.com/SAubS.png
(from the same equation in 3D graph)
To trace a XY graph which is non-function, it is actually calculate the equation with every X and Y in the graph in range.
And it is determined by intersection between XY plane.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A lot of the time examples like these "look" like functions locally. Therefore they can broken up into little pieces by using a valid set of initial conditions or a function which is valid and describes part of the overall relation you are trying to graph.
In your example we can convert the equation to look like:
$$
y = sin^-1(1-sin(x))
$$
Which gives a starting point looking like the bottom half of a single warped sphere. Then we see that since $sin(y)=sin(pi-y)$ we can cap the warped sphere. Then we can argue that the equation is clearly unchanged if we were to replace $(x,y)$ with $(x+2pi n ,y+2pi m)$ (for integer values of m,n) and so we can repeat our pattern throughout the entire grid due to the periodicity of the $sin$ function.
Another good example is $y^2 = x$ where we have two sets of solutions corresponding to $y=pm sqrtx $.
$endgroup$
$begingroup$
I really like your explanation. However, there is a problem. Some non-functions cannot be simplified exclusively to have y be dependent on x. For instance,x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy, which isn't even symmetric like the other examples, but can still be graphed by Desmos.
$endgroup$
– Rapid Readers
Sep 18 '17 at 4:11
$begingroup$
If you are alright with approximate solutions then you can differentiate w.r.t. x to isolate for $fracdydx$ and then expand around as many sets of valid initial conditions as you can find. I'm sure there are some fancier tricks for finding a complete graph for an equation but I was just trying to point out that such an equation acts like a function somewhere around every set of valid initial conditions.
$endgroup$
– Ryan Watt
Sep 19 '17 at 6:36
add a comment |
$begingroup$
A lot of the time examples like these "look" like functions locally. Therefore they can broken up into little pieces by using a valid set of initial conditions or a function which is valid and describes part of the overall relation you are trying to graph.
In your example we can convert the equation to look like:
$$
y = sin^-1(1-sin(x))
$$
Which gives a starting point looking like the bottom half of a single warped sphere. Then we see that since $sin(y)=sin(pi-y)$ we can cap the warped sphere. Then we can argue that the equation is clearly unchanged if we were to replace $(x,y)$ with $(x+2pi n ,y+2pi m)$ (for integer values of m,n) and so we can repeat our pattern throughout the entire grid due to the periodicity of the $sin$ function.
Another good example is $y^2 = x$ where we have two sets of solutions corresponding to $y=pm sqrtx $.
$endgroup$
$begingroup$
I really like your explanation. However, there is a problem. Some non-functions cannot be simplified exclusively to have y be dependent on x. For instance,x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy, which isn't even symmetric like the other examples, but can still be graphed by Desmos.
$endgroup$
– Rapid Readers
Sep 18 '17 at 4:11
$begingroup$
If you are alright with approximate solutions then you can differentiate w.r.t. x to isolate for $fracdydx$ and then expand around as many sets of valid initial conditions as you can find. I'm sure there are some fancier tricks for finding a complete graph for an equation but I was just trying to point out that such an equation acts like a function somewhere around every set of valid initial conditions.
$endgroup$
– Ryan Watt
Sep 19 '17 at 6:36
add a comment |
$begingroup$
A lot of the time examples like these "look" like functions locally. Therefore they can broken up into little pieces by using a valid set of initial conditions or a function which is valid and describes part of the overall relation you are trying to graph.
In your example we can convert the equation to look like:
$$
y = sin^-1(1-sin(x))
$$
Which gives a starting point looking like the bottom half of a single warped sphere. Then we see that since $sin(y)=sin(pi-y)$ we can cap the warped sphere. Then we can argue that the equation is clearly unchanged if we were to replace $(x,y)$ with $(x+2pi n ,y+2pi m)$ (for integer values of m,n) and so we can repeat our pattern throughout the entire grid due to the periodicity of the $sin$ function.
Another good example is $y^2 = x$ where we have two sets of solutions corresponding to $y=pm sqrtx $.
$endgroup$
A lot of the time examples like these "look" like functions locally. Therefore they can broken up into little pieces by using a valid set of initial conditions or a function which is valid and describes part of the overall relation you are trying to graph.
In your example we can convert the equation to look like:
$$
y = sin^-1(1-sin(x))
$$
Which gives a starting point looking like the bottom half of a single warped sphere. Then we see that since $sin(y)=sin(pi-y)$ we can cap the warped sphere. Then we can argue that the equation is clearly unchanged if we were to replace $(x,y)$ with $(x+2pi n ,y+2pi m)$ (for integer values of m,n) and so we can repeat our pattern throughout the entire grid due to the periodicity of the $sin$ function.
Another good example is $y^2 = x$ where we have two sets of solutions corresponding to $y=pm sqrtx $.
answered Sep 18 '17 at 3:36
Ryan WattRyan Watt
262
262
$begingroup$
I really like your explanation. However, there is a problem. Some non-functions cannot be simplified exclusively to have y be dependent on x. For instance,x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy, which isn't even symmetric like the other examples, but can still be graphed by Desmos.
$endgroup$
– Rapid Readers
Sep 18 '17 at 4:11
$begingroup$
If you are alright with approximate solutions then you can differentiate w.r.t. x to isolate for $fracdydx$ and then expand around as many sets of valid initial conditions as you can find. I'm sure there are some fancier tricks for finding a complete graph for an equation but I was just trying to point out that such an equation acts like a function somewhere around every set of valid initial conditions.
$endgroup$
– Ryan Watt
Sep 19 '17 at 6:36
add a comment |
$begingroup$
I really like your explanation. However, there is a problem. Some non-functions cannot be simplified exclusively to have y be dependent on x. For instance,x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy, which isn't even symmetric like the other examples, but can still be graphed by Desmos.
$endgroup$
– Rapid Readers
Sep 18 '17 at 4:11
$begingroup$
If you are alright with approximate solutions then you can differentiate w.r.t. x to isolate for $fracdydx$ and then expand around as many sets of valid initial conditions as you can find. I'm sure there are some fancier tricks for finding a complete graph for an equation but I was just trying to point out that such an equation acts like a function somewhere around every set of valid initial conditions.
$endgroup$
– Ryan Watt
Sep 19 '17 at 6:36
$begingroup$
I really like your explanation. However, there is a problem. Some non-functions cannot be simplified exclusively to have y be dependent on x. For instance,
x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy, which isn't even symmetric like the other examples, but can still be graphed by Desmos.$endgroup$
– Rapid Readers
Sep 18 '17 at 4:11
$begingroup$
I really like your explanation. However, there is a problem. Some non-functions cannot be simplified exclusively to have y be dependent on x. For instance,
x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy, which isn't even symmetric like the other examples, but can still be graphed by Desmos.$endgroup$
– Rapid Readers
Sep 18 '17 at 4:11
$begingroup$
If you are alright with approximate solutions then you can differentiate w.r.t. x to isolate for $fracdydx$ and then expand around as many sets of valid initial conditions as you can find. I'm sure there are some fancier tricks for finding a complete graph for an equation but I was just trying to point out that such an equation acts like a function somewhere around every set of valid initial conditions.
$endgroup$
– Ryan Watt
Sep 19 '17 at 6:36
$begingroup$
If you are alright with approximate solutions then you can differentiate w.r.t. x to isolate for $fracdydx$ and then expand around as many sets of valid initial conditions as you can find. I'm sure there are some fancier tricks for finding a complete graph for an equation but I was just trying to point out that such an equation acts like a function somewhere around every set of valid initial conditions.
$endgroup$
– Ryan Watt
Sep 19 '17 at 6:36
add a comment |
$begingroup$
An aproach for this would be to take some values.
x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy
For example x=5. We substitute this value. And later we try different values for y until we have a equality (left part is some as right side). Some mechanisms help to do this, looking on how quick we get a approximation. I would first convert this to a equation equal to 0 :
x arcsin( x/y) + sqrt( y^2 - x^2) - 4xy = 0
This is trivial, and now for each x-value, we can get a y value (or many y-values) using a aproximation methode like Newton, which use derivates.
Daniel
P.S. see https://en.wikipedia.org/wiki/Newton%27s_method
$endgroup$
add a comment |
$begingroup$
An aproach for this would be to take some values.
x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy
For example x=5. We substitute this value. And later we try different values for y until we have a equality (left part is some as right side). Some mechanisms help to do this, looking on how quick we get a approximation. I would first convert this to a equation equal to 0 :
x arcsin( x/y) + sqrt( y^2 - x^2) - 4xy = 0
This is trivial, and now for each x-value, we can get a y value (or many y-values) using a aproximation methode like Newton, which use derivates.
Daniel
P.S. see https://en.wikipedia.org/wiki/Newton%27s_method
$endgroup$
add a comment |
$begingroup$
An aproach for this would be to take some values.
x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy
For example x=5. We substitute this value. And later we try different values for y until we have a equality (left part is some as right side). Some mechanisms help to do this, looking on how quick we get a approximation. I would first convert this to a equation equal to 0 :
x arcsin( x/y) + sqrt( y^2 - x^2) - 4xy = 0
This is trivial, and now for each x-value, we can get a y value (or many y-values) using a aproximation methode like Newton, which use derivates.
Daniel
P.S. see https://en.wikipedia.org/wiki/Newton%27s_method
$endgroup$
An aproach for this would be to take some values.
x arcsin( x/y) + sqrt( y^2 - x^2) = 4xy
For example x=5. We substitute this value. And later we try different values for y until we have a equality (left part is some as right side). Some mechanisms help to do this, looking on how quick we get a approximation. I would first convert this to a equation equal to 0 :
x arcsin( x/y) + sqrt( y^2 - x^2) - 4xy = 0
This is trivial, and now for each x-value, we can get a y value (or many y-values) using a aproximation methode like Newton, which use derivates.
Daniel
P.S. see https://en.wikipedia.org/wiki/Newton%27s_method
answered Sep 18 '17 at 6:25
Daniel PolDaniel Pol
33425
33425
add a comment |
add a comment |
$begingroup$
According from what you asked, let's see images below and take conclusions.
https://i.ytimg.com/vi/y7N0Pr3N1RY/maxresdefault.jpg
(points tracing)
https://i.stack.imgur.com/SAubS.png
(from the same equation in 3D graph)
To trace a XY graph which is non-function, it is actually calculate the equation with every X and Y in the graph in range.
And it is determined by intersection between XY plane.
$endgroup$
add a comment |
$begingroup$
According from what you asked, let's see images below and take conclusions.
https://i.ytimg.com/vi/y7N0Pr3N1RY/maxresdefault.jpg
(points tracing)
https://i.stack.imgur.com/SAubS.png
(from the same equation in 3D graph)
To trace a XY graph which is non-function, it is actually calculate the equation with every X and Y in the graph in range.
And it is determined by intersection between XY plane.
$endgroup$
add a comment |
$begingroup$
According from what you asked, let's see images below and take conclusions.
https://i.ytimg.com/vi/y7N0Pr3N1RY/maxresdefault.jpg
(points tracing)
https://i.stack.imgur.com/SAubS.png
(from the same equation in 3D graph)
To trace a XY graph which is non-function, it is actually calculate the equation with every X and Y in the graph in range.
And it is determined by intersection between XY plane.
$endgroup$
According from what you asked, let's see images below and take conclusions.
https://i.ytimg.com/vi/y7N0Pr3N1RY/maxresdefault.jpg
(points tracing)
https://i.stack.imgur.com/SAubS.png
(from the same equation in 3D graph)
To trace a XY graph which is non-function, it is actually calculate the equation with every X and Y in the graph in range.
And it is determined by intersection between XY plane.
edited Jan 5 '18 at 7:04
answered Jan 3 '18 at 9:58
A-KouZ1A-KouZ1
11
11
add a comment |
add a comment |
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