How to draw this function?
I have a problem where I am given this function:
$z(x, y) = cos(x)cos(y)^{T}$
and now I am supposed to determine if a point is above or below this plane. But I am having trouble understanding what this plane actually looks like, so I am not sure how to solve the problem of the point position. Online plotting tools did not help, they don't give me any plot for this function. Also I am not sure if this $T$ does anything to $cos(y)$. So if anyone could explain this a bit, I would really appreciate it.
functions trigonometry graphing-functions plane-geometry
edited Dec 26 '18 at 21:26
Bernard
118k639112
asked Dec 26 '18 at 20:44
ivana14
425312
|
show 3 more comments
I have a problem where I am given this function:
$z(x, y) = cos(x)cos(y)^{T}$
and now I am supposed to determine if a point is above or below this plane. But I am having trouble understanding what this plane actually looks like, so I am not sure how to solve the problem of the point position. Online plotting tools did not help, they don't give me any plot for this function. Also I am not sure if this $T$ does anything to $cos(y)$. So if anyone could explain this a bit, I would really appreciate it.
functions trigonometry graphing-functions plane-geometry
edited Dec 26 '18 at 21:26
Bernard
118k639112
asked Dec 26 '18 at 20:44
ivana14
425312
Where did you get this problem & function from?
– John Omielan
Dec 26 '18 at 20:46
Where does the $T$ come from?
– saulspatz
Dec 26 '18 at 20:47
@saulspatz I thing that it means transpose, since we used that notation for matrices before, but I am not sure how to apply that here.
– ivana14
Dec 26 '18 at 20:49
1
The transpose doesn't do anything to a real number, so you can ignore it.
– saulspatz
Dec 26 '18 at 20:49
@JohnOmielan I am taking an AI class, and they have given us a problem to create a neural network that will learn how to classify points based on their position in regards to the plane of this function
– ivana14
Dec 26 '18 at 20:50
|
show 3 more comments
I have a problem where I am given this function:
$z(x, y) = cos(x)cos(y)^{T}$
and now I am supposed to determine if a point is above or below this plane. But I am having trouble understanding what this plane actually looks like, so I am not sure how to solve the problem of the point position. Online plotting tools did not help, they don't give me any plot for this function. Also I am not sure if this $T$ does anything to $cos(y)$. So if anyone could explain this a bit, I would really appreciate it.
functions trigonometry graphing-functions plane-geometry
edited Dec 26 '18 at 21:26
Bernard
118k639112
asked Dec 26 '18 at 20:44
ivana14
425312
I have a problem where I am given this function:
$z(x, y) = cos(x)cos(y)^{T}$
and now I am supposed to determine if a point is above or below this plane. But I am having trouble understanding what this plane actually looks like, so I am not sure how to solve the problem of the point position. Online plotting tools did not help, they don't give me any plot for this function. Also I am not sure if this $T$ does anything to $cos(y)$. So if anyone could explain this a bit, I would really appreciate it.
functions trigonometry graphing-functions plane-geometry
functions trigonometry graphing-functions plane-geometry
edited Dec 26 '18 at 21:26
Bernard
118k639112
asked Dec 26 '18 at 20:44
ivana14
425312
edited Dec 26 '18 at 21:26
Bernard
118k639112
asked Dec 26 '18 at 20:44
ivana14
425312
edited Dec 26 '18 at 21:26
Bernard
118k639112
edited Dec 26 '18 at 21:26
Bernard
118k639112
edited Dec 26 '18 at 21:26
Bernard
118k639112
118k639112
asked Dec 26 '18 at 20:44
ivana14
425312
asked Dec 26 '18 at 20:44
ivana14
425312
asked Dec 26 '18 at 20:44
ivana14
425312
425312
Where did you get this problem & function from?
– John Omielan
Dec 26 '18 at 20:46
Where does the $T$ come from?
– saulspatz
Dec 26 '18 at 20:47
@saulspatz I thing that it means transpose, since we used that notation for matrices before, but I am not sure how to apply that here.
– ivana14
Dec 26 '18 at 20:49
1
The transpose doesn't do anything to a real number, so you can ignore it.
– saulspatz
Dec 26 '18 at 20:49
@JohnOmielan I am taking an AI class, and they have given us a problem to create a neural network that will learn how to classify points based on their position in regards to the plane of this function
– ivana14
Dec 26 '18 at 20:50
|
show 3 more comments
Where did you get this problem & function from?
– John Omielan
Dec 26 '18 at 20:46
Where does the $T$ come from?
– saulspatz
Dec 26 '18 at 20:47
@saulspatz I thing that it means transpose, since we used that notation for matrices before, but I am not sure how to apply that here.
– ivana14
Dec 26 '18 at 20:49
1
The transpose doesn't do anything to a real number, so you can ignore it.
– saulspatz
Dec 26 '18 at 20:49
@JohnOmielan I am taking an AI class, and they have given us a problem to create a neural network that will learn how to classify points based on their position in regards to the plane of this function
– ivana14
Dec 26 '18 at 20:50
Where did you get this problem & function from?
– John Omielan
Dec 26 '18 at 20:46
Where did you get this problem & function from?
– John Omielan
Dec 26 '18 at 20:46
Where does the $T$ come from?
– saulspatz
Dec 26 '18 at 20:47
Where does the $T$ come from?
– saulspatz
Dec 26 '18 at 20:47
@saulspatz I thing that it means transpose, since we used that notation for matrices before, but I am not sure how to apply that here.
– ivana14
Dec 26 '18 at 20:49
@saulspatz I thing that it means transpose, since we used that notation for matrices before, but I am not sure how to apply that here.
– ivana14
Dec 26 '18 at 20:49
1
1
The transpose doesn't do anything to a real number, so you can ignore it.
– saulspatz
Dec 26 '18 at 20:49
The transpose doesn't do anything to a real number, so you can ignore it.
– saulspatz
Dec 26 '18 at 20:49
@JohnOmielan I am taking an AI class, and they have given us a problem to create a neural network that will learn how to classify points based on their position in regards to the plane of this function
– ivana14
Dec 26 '18 at 20:50
@JohnOmielan I am taking an AI class, and they have given us a problem to create a neural network that will learn how to classify points based on their position in regards to the plane of this function
– ivana14
Dec 26 '18 at 20:50
|
show 3 more comments
2 Answers
2
active
oldest
votes
This function forms a surface where a given point $left(x, y, zright)$ is "above" if the provided $z$ value is greater than your function value, i.e., $zleft(x,yright) = cosleft(xright)cosleft(yright)^T$, or "below" if it's less. In some very rare cases, the $z$ value may match exactly (due to computers of course having a limited accuracy for real number calculations), so you can decide how to classify those points.
As for what the surface looks like, there are many packages available to draw it for you, or you can even just simply try plugging in a few values yourself to get a feel for it. Also, as mentioned in the comments, the values have a $2 pi$ period due to the use of $cos$, so the surface will repeat itself every $2 pi$ units in both the $x$ and $y$ directions.
As for setting up the neural network, that obviously is something which this forum is not an appropriate place to discuss. Good luck with doing that.
answered Dec 26 '18 at 21:05
John Omielan
90418
Okay, it is a bit more clear to me now. Can you just please explain how I get the value of $z$?
– ivana14
Dec 26 '18 at 21:11
@ivana14 Note that "$z$" here actually refers to $2$ things. First, each point you are given, as it is in $3$-dimensional space, has a third co-ordinate that is usually called $z$. The surface you are checking the point against has a $z$ co-ordinate as well, determined by your equation of $zleft(x,yright)$ by using the $x$ and $y$, i.e., the first 2 co-ordinates, of the point you are given. I hope this answers your question and makes sense.
– John Omielan
Dec 26 '18 at 21:14
add a comment |
Take a look at the following "eggbox" representation of surface with equation :
$$z=cos(x)cos(y)$$
(no need for the transposition) with its associated level sets, of primary importance for the question you are asked.
Note that "peaks" are red and "pits" are blue, and that the same color convention is valid for level sets.
A way to understand this surface : If you fix for example $x=pi/3$, it means that when you cut this "cake" at this abscissa, as $cos(pi/3)=1/2$, the profile of the cut is a sine curve with equation $z=frac12 cos(y)$.
answered Dec 27 '18 at 18:34
Jean Marie
28.8k41949
add a comment |
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2 Answers
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2 Answers
2
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This function forms a surface where a given point $left(x, y, zright)$ is "above" if the provided $z$ value is greater than your function value, i.e., $zleft(x,yright) = cosleft(xright)cosleft(yright)^T$, or "below" if it's less. In some very rare cases, the $z$ value may match exactly (due to computers of course having a limited accuracy for real number calculations), so you can decide how to classify those points.
As for what the surface looks like, there are many packages available to draw it for you, or you can even just simply try plugging in a few values yourself to get a feel for it. Also, as mentioned in the comments, the values have a $2 pi$ period due to the use of $cos$, so the surface will repeat itself every $2 pi$ units in both the $x$ and $y$ directions.
As for setting up the neural network, that obviously is something which this forum is not an appropriate place to discuss. Good luck with doing that.
answered Dec 26 '18 at 21:05
John Omielan
90418
Okay, it is a bit more clear to me now. Can you just please explain how I get the value of $z$?
– ivana14
Dec 26 '18 at 21:11
@ivana14 Note that "$z$" here actually refers to $2$ things. First, each point you are given, as it is in $3$-dimensional space, has a third co-ordinate that is usually called $z$. The surface you are checking the point against has a $z$ co-ordinate as well, determined by your equation of $zleft(x,yright)$ by using the $x$ and $y$, i.e., the first 2 co-ordinates, of the point you are given. I hope this answers your question and makes sense.
– John Omielan
Dec 26 '18 at 21:14
add a comment |
This function forms a surface where a given point $left(x, y, zright)$ is "above" if the provided $z$ value is greater than your function value, i.e., $zleft(x,yright) = cosleft(xright)cosleft(yright)^T$, or "below" if it's less. In some very rare cases, the $z$ value may match exactly (due to computers of course having a limited accuracy for real number calculations), so you can decide how to classify those points.
As for what the surface looks like, there are many packages available to draw it for you, or you can even just simply try plugging in a few values yourself to get a feel for it. Also, as mentioned in the comments, the values have a $2 pi$ period due to the use of $cos$, so the surface will repeat itself every $2 pi$ units in both the $x$ and $y$ directions.
As for setting up the neural network, that obviously is something which this forum is not an appropriate place to discuss. Good luck with doing that.
answered Dec 26 '18 at 21:05
John Omielan
90418
Okay, it is a bit more clear to me now. Can you just please explain how I get the value of $z$?
– ivana14
Dec 26 '18 at 21:11
@ivana14 Note that "$z$" here actually refers to $2$ things. First, each point you are given, as it is in $3$-dimensional space, has a third co-ordinate that is usually called $z$. The surface you are checking the point against has a $z$ co-ordinate as well, determined by your equation of $zleft(x,yright)$ by using the $x$ and $y$, i.e., the first 2 co-ordinates, of the point you are given. I hope this answers your question and makes sense.
– John Omielan
Dec 26 '18 at 21:14
add a comment |
This function forms a surface where a given point $left(x, y, zright)$ is "above" if the provided $z$ value is greater than your function value, i.e., $zleft(x,yright) = cosleft(xright)cosleft(yright)^T$, or "below" if it's less. In some very rare cases, the $z$ value may match exactly (due to computers of course having a limited accuracy for real number calculations), so you can decide how to classify those points.
As for what the surface looks like, there are many packages available to draw it for you, or you can even just simply try plugging in a few values yourself to get a feel for it. Also, as mentioned in the comments, the values have a $2 pi$ period due to the use of $cos$, so the surface will repeat itself every $2 pi$ units in both the $x$ and $y$ directions.
As for setting up the neural network, that obviously is something which this forum is not an appropriate place to discuss. Good luck with doing that.
answered Dec 26 '18 at 21:05
John Omielan
90418
This function forms a surface where a given point $left(x, y, zright)$ is "above" if the provided $z$ value is greater than your function value, i.e., $zleft(x,yright) = cosleft(xright)cosleft(yright)^T$, or "below" if it's less. In some very rare cases, the $z$ value may match exactly (due to computers of course having a limited accuracy for real number calculations), so you can decide how to classify those points.
As for what the surface looks like, there are many packages available to draw it for you, or you can even just simply try plugging in a few values yourself to get a feel for it. Also, as mentioned in the comments, the values have a $2 pi$ period due to the use of $cos$, so the surface will repeat itself every $2 pi$ units in both the $x$ and $y$ directions.
As for setting up the neural network, that obviously is something which this forum is not an appropriate place to discuss. Good luck with doing that.
answered Dec 26 '18 at 21:05
John Omielan
90418
answered Dec 26 '18 at 21:05
John Omielan
90418
answered Dec 26 '18 at 21:05
John Omielan
90418
answered Dec 26 '18 at 21:05
John Omielan
90418
90418
Okay, it is a bit more clear to me now. Can you just please explain how I get the value of $z$?
– ivana14
Dec 26 '18 at 21:11
@ivana14 Note that "$z$" here actually refers to $2$ things. First, each point you are given, as it is in $3$-dimensional space, has a third co-ordinate that is usually called $z$. The surface you are checking the point against has a $z$ co-ordinate as well, determined by your equation of $zleft(x,yright)$ by using the $x$ and $y$, i.e., the first 2 co-ordinates, of the point you are given. I hope this answers your question and makes sense.
– John Omielan
Dec 26 '18 at 21:14
add a comment |
Okay, it is a bit more clear to me now. Can you just please explain how I get the value of $z$?
– ivana14
Dec 26 '18 at 21:11
@ivana14 Note that "$z$" here actually refers to $2$ things. First, each point you are given, as it is in $3$-dimensional space, has a third co-ordinate that is usually called $z$. The surface you are checking the point against has a $z$ co-ordinate as well, determined by your equation of $zleft(x,yright)$ by using the $x$ and $y$, i.e., the first 2 co-ordinates, of the point you are given. I hope this answers your question and makes sense.
– John Omielan
Dec 26 '18 at 21:14
Okay, it is a bit more clear to me now. Can you just please explain how I get the value of $z$?
– ivana14
Dec 26 '18 at 21:11
Okay, it is a bit more clear to me now. Can you just please explain how I get the value of $z$?
– ivana14
Dec 26 '18 at 21:11
@ivana14 Note that "$z$" here actually refers to $2$ things. First, each point you are given, as it is in $3$-dimensional space, has a third co-ordinate that is usually called $z$. The surface you are checking the point against has a $z$ co-ordinate as well, determined by your equation of $zleft(x,yright)$ by using the $x$ and $y$, i.e., the first 2 co-ordinates, of the point you are given. I hope this answers your question and makes sense.
– John Omielan
Dec 26 '18 at 21:14
@ivana14 Note that "$z$" here actually refers to $2$ things. First, each point you are given, as it is in $3$-dimensional space, has a third co-ordinate that is usually called $z$. The surface you are checking the point against has a $z$ co-ordinate as well, determined by your equation of $zleft(x,yright)$ by using the $x$ and $y$, i.e., the first 2 co-ordinates, of the point you are given. I hope this answers your question and makes sense.
– John Omielan
Dec 26 '18 at 21:14
add a comment |
Take a look at the following "eggbox" representation of surface with equation :
$$z=cos(x)cos(y)$$
(no need for the transposition) with its associated level sets, of primary importance for the question you are asked.
Note that "peaks" are red and "pits" are blue, and that the same color convention is valid for level sets.
A way to understand this surface : If you fix for example $x=pi/3$, it means that when you cut this "cake" at this abscissa, as $cos(pi/3)=1/2$, the profile of the cut is a sine curve with equation $z=frac12 cos(y)$.
answered Dec 27 '18 at 18:34
Jean Marie
28.8k41949
add a comment |
Take a look at the following "eggbox" representation of surface with equation :
$$z=cos(x)cos(y)$$
(no need for the transposition) with its associated level sets, of primary importance for the question you are asked.
Note that "peaks" are red and "pits" are blue, and that the same color convention is valid for level sets.
A way to understand this surface : If you fix for example $x=pi/3$, it means that when you cut this "cake" at this abscissa, as $cos(pi/3)=1/2$, the profile of the cut is a sine curve with equation $z=frac12 cos(y)$.
answered Dec 27 '18 at 18:34
Jean Marie
28.8k41949
add a comment |
Take a look at the following "eggbox" representation of surface with equation :
$$z=cos(x)cos(y)$$
(no need for the transposition) with its associated level sets, of primary importance for the question you are asked.
Note that "peaks" are red and "pits" are blue, and that the same color convention is valid for level sets.
A way to understand this surface : If you fix for example $x=pi/3$, it means that when you cut this "cake" at this abscissa, as $cos(pi/3)=1/2$, the profile of the cut is a sine curve with equation $z=frac12 cos(y)$.
answered Dec 27 '18 at 18:34
Jean Marie
28.8k41949
Take a look at the following "eggbox" representation of surface with equation :
$$z=cos(x)cos(y)$$
(no need for the transposition) with its associated level sets, of primary importance for the question you are asked.
Note that "peaks" are red and "pits" are blue, and that the same color convention is valid for level sets.
A way to understand this surface : If you fix for example $x=pi/3$, it means that when you cut this "cake" at this abscissa, as $cos(pi/3)=1/2$, the profile of the cut is a sine curve with equation $z=frac12 cos(y)$.
answered Dec 27 '18 at 18:34
Jean Marie
28.8k41949
answered Dec 27 '18 at 18:34
Jean Marie
28.8k41949
answered Dec 27 '18 at 18:34
Jean Marie
28.8k41949
answered Dec 27 '18 at 18:34
Jean Marie
28.8k41949
28.8k41949
add a comment |
add a comment |
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Where did you get this problem & function from?
– John Omielan
Dec 26 '18 at 20:46
Where does the $T$ come from?
– saulspatz
Dec 26 '18 at 20:47
@saulspatz I thing that it means transpose, since we used that notation for matrices before, but I am not sure how to apply that here.
– ivana14
Dec 26 '18 at 20:49
1
The transpose doesn't do anything to a real number, so you can ignore it.
– saulspatz
Dec 26 '18 at 20:49
@JohnOmielan I am taking an AI class, and they have given us a problem to create a neural network that will learn how to classify points based on their position in regards to the plane of this function
– ivana14
Dec 26 '18 at 20:50