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How can a gradient be though of as a function?
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If a function $f(x,y)$ would output a value in a third dimension, $z=f(x,y)$ for example. How can we treat the gradient of f as a function in x and y, when the output of the gradient is a vector in the two dimensions $x$ and $y$ which are the dimensions of the inputs. I guess my question is, is it normal for a function to map to the dimensions of its inputs?
multivariable-calculus functions
edited 35 mins ago
Ethan Bolker
38.7k543102
asked 3 hours ago
Omar Hossam Ahmed
261
add a comment |
up vote
4
down vote
favorite
If a function $f(x,y)$ would output a value in a third dimension, $z=f(x,y)$ for example. How can we treat the gradient of f as a function in x and y, when the output of the gradient is a vector in the two dimensions $x$ and $y$ which are the dimensions of the inputs. I guess my question is, is it normal for a function to map to the dimensions of its inputs?
multivariable-calculus functions
edited 35 mins ago
Ethan Bolker
38.7k543102
asked 3 hours ago
Omar Hossam Ahmed
261
Problems in calculus often stem from physics and engineering, here we deal a lot with $mathbbR^3$ and $mathbbR^2$, so it makes sense that we visualise this by writing out a certain function in terms of $x,y,z$ we could define other basis vectors and write out some other functions for instance in terms of spherical polar coordinates, but the input is hard to visualise. We often transform to other coordinates to do integrations, for instance when we want to integrate over a sphere. The corresponding problem then transforms to some other graph in terms of these new coordinates.
– WesleyGroupshaveFeelingsToo
11 mins ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
If a function $f(x,y)$ would output a value in a third dimension, $z=f(x,y)$ for example. How can we treat the gradient of f as a function in x and y, when the output of the gradient is a vector in the two dimensions $x$ and $y$ which are the dimensions of the inputs. I guess my question is, is it normal for a function to map to the dimensions of its inputs?
multivariable-calculus functions
edited 35 mins ago
Ethan Bolker
38.7k543102
asked 3 hours ago
Omar Hossam Ahmed
261
If a function $f(x,y)$ would output a value in a third dimension, $z=f(x,y)$ for example. How can we treat the gradient of f as a function in x and y, when the output of the gradient is a vector in the two dimensions $x$ and $y$ which are the dimensions of the inputs. I guess my question is, is it normal for a function to map to the dimensions of its inputs?
multivariable-calculus functions
multivariable-calculus functions
edited 35 mins ago
Ethan Bolker
38.7k543102
asked 3 hours ago
Omar Hossam Ahmed
261
edited 35 mins ago
Ethan Bolker
38.7k543102
asked 3 hours ago
Omar Hossam Ahmed
261
edited 35 mins ago
Ethan Bolker
38.7k543102
edited 35 mins ago
Ethan Bolker
38.7k543102
edited 35 mins ago
Ethan Bolker
38.7k543102
38.7k543102
asked 3 hours ago
Omar Hossam Ahmed
261
asked 3 hours ago
Omar Hossam Ahmed
261
asked 3 hours ago
Omar Hossam Ahmed
261
261
Problems in calculus often stem from physics and engineering, here we deal a lot with $mathbbR^3$ and $mathbbR^2$, so it makes sense that we visualise this by writing out a certain function in terms of $x,y,z$ we could define other basis vectors and write out some other functions for instance in terms of spherical polar coordinates, but the input is hard to visualise. We often transform to other coordinates to do integrations, for instance when we want to integrate over a sphere. The corresponding problem then transforms to some other graph in terms of these new coordinates.
– WesleyGroupshaveFeelingsToo
11 mins ago
add a comment |
Problems in calculus often stem from physics and engineering, here we deal a lot with $mathbbR^3$ and $mathbbR^2$, so it makes sense that we visualise this by writing out a certain function in terms of $x,y,z$ we could define other basis vectors and write out some other functions for instance in terms of spherical polar coordinates, but the input is hard to visualise. We often transform to other coordinates to do integrations, for instance when we want to integrate over a sphere. The corresponding problem then transforms to some other graph in terms of these new coordinates.
– WesleyGroupshaveFeelingsToo
11 mins ago
Problems in calculus often stem from physics and engineering, here we deal a lot with $mathbbR^3$ and $mathbbR^2$, so it makes sense that we visualise this by writing out a certain function in terms of $x,y,z$ we could define other basis vectors and write out some other functions for instance in terms of spherical polar coordinates, but the input is hard to visualise. We often transform to other coordinates to do integrations, for instance when we want to integrate over a sphere. The corresponding problem then transforms to some other graph in terms of these new coordinates.
– WesleyGroupshaveFeelingsToo
11 mins ago
Problems in calculus often stem from physics and engineering, here we deal a lot with $mathbbR^3$ and $mathbbR^2$, so it makes sense that we visualise this by writing out a certain function in terms of $x,y,z$ we could define other basis vectors and write out some other functions for instance in terms of spherical polar coordinates, but the input is hard to visualise. We often transform to other coordinates to do integrations, for instance when we want to integrate over a sphere. The corresponding problem then transforms to some other graph in terms of these new coordinates.
– WesleyGroupshaveFeelingsToo
11 mins ago
add a comment |
2 Answers
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A function $F$ can be thought of as a machine that takes in inputs from one set, say $X$, and outputs elements in another set, say $Y$. A function outputs a single element for every input, and we write it $F : X rightarrow Y$.
In your case, the gradient of $f$ is just a function $nabla f : mathbbR^2 rightarrow mathbbR^2$, where for each input $(x,y)$, it outputs the element $left( fracpartial fpartial x(x,y), fracpartial fpartial y(x,y) right)$.
answered 3 hours ago
Sambo
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up vote
2
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$(1)$ The gradient is a vector-valued function, this maps pairs of numbers $(x,y)$ to some other pair of numbers $(x',y')$. These pairs of numbers we call vectors and they have a very geometric interpretation: they have a length and a direction. Specifically the gradient corresponds to the direction and magnitude of steepest ascent.
Also see:
https://en.wikipedia.org/wiki/Vector-valued_function
This Khan Academy link I found as well is very useful as he also thought of the same example as I did:
https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/gradient-and-directional-derivatives/v/gradient-and-graphs
$(2)$ On the contrary, when you plot a function that maps from $mathbbR^2 rightarrow mathbbR$ like $f(x,y)=x^2 +y^2$, when you want to plot this you often define a third variable $z=f(x,y)$ and you let the value of this variable be equal to the function value. I have plotted $x^2 +y^2$ in this way below:
Without introducing another axis, we can also just give different function value ranges a different colour, so a large value could be very dark and a low value could be very light or vice versa. We recognise the same function:
The difference between $(1)$ and $(2)$ is the notion of direction. The function you describe is usually a "scalar field", we only have the notion of magnitude or "value" but not of direction. Gradients will give you a so-called "vector field" as physicists often call it, we usually visualise this using VectorPlots. Below you find such a method, I've plotted the vector field $f(x,y)=(x^2+y,y^2+x )$
which has gradient $grad(f)(x,y)=(2x,2y)$
answered 37 mins ago
WesleyGroupshaveFeelingsToo
564216
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
A function $F$ can be thought of as a machine that takes in inputs from one set, say $X$, and outputs elements in another set, say $Y$. A function outputs a single element for every input, and we write it $F : X rightarrow Y$.
In your case, the gradient of $f$ is just a function $nabla f : mathbbR^2 rightarrow mathbbR^2$, where for each input $(x,y)$, it outputs the element $left( fracpartial fpartial x(x,y), fracpartial fpartial y(x,y) right)$.
answered 3 hours ago
Sambo
1,8652428
add a comment |
up vote
6
down vote
A function $F$ can be thought of as a machine that takes in inputs from one set, say $X$, and outputs elements in another set, say $Y$. A function outputs a single element for every input, and we write it $F : X rightarrow Y$.
In your case, the gradient of $f$ is just a function $nabla f : mathbbR^2 rightarrow mathbbR^2$, where for each input $(x,y)$, it outputs the element $left( fracpartial fpartial x(x,y), fracpartial fpartial y(x,y) right)$.
answered 3 hours ago
Sambo
1,8652428
add a comment |
up vote
6
down vote
up vote
6
down vote
A function $F$ can be thought of as a machine that takes in inputs from one set, say $X$, and outputs elements in another set, say $Y$. A function outputs a single element for every input, and we write it $F : X rightarrow Y$.
In your case, the gradient of $f$ is just a function $nabla f : mathbbR^2 rightarrow mathbbR^2$, where for each input $(x,y)$, it outputs the element $left( fracpartial fpartial x(x,y), fracpartial fpartial y(x,y) right)$.
answered 3 hours ago
Sambo
1,8652428
A function $F$ can be thought of as a machine that takes in inputs from one set, say $X$, and outputs elements in another set, say $Y$. A function outputs a single element for every input, and we write it $F : X rightarrow Y$.
In your case, the gradient of $f$ is just a function $nabla f : mathbbR^2 rightarrow mathbbR^2$, where for each input $(x,y)$, it outputs the element $left( fracpartial fpartial x(x,y), fracpartial fpartial y(x,y) right)$.
answered 3 hours ago
Sambo
1,8652428
answered 3 hours ago
Sambo
1,8652428
answered 3 hours ago
Sambo
1,8652428
answered 3 hours ago
Sambo
1,8652428
1,8652428
add a comment |
add a comment |
up vote
2
down vote
$(1)$ The gradient is a vector-valued function, this maps pairs of numbers $(x,y)$ to some other pair of numbers $(x',y')$. These pairs of numbers we call vectors and they have a very geometric interpretation: they have a length and a direction. Specifically the gradient corresponds to the direction and magnitude of steepest ascent.
Also see:
https://en.wikipedia.org/wiki/Vector-valued_function
This Khan Academy link I found as well is very useful as he also thought of the same example as I did:
https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/gradient-and-directional-derivatives/v/gradient-and-graphs
$(2)$ On the contrary, when you plot a function that maps from $mathbbR^2 rightarrow mathbbR$ like $f(x,y)=x^2 +y^2$, when you want to plot this you often define a third variable $z=f(x,y)$ and you let the value of this variable be equal to the function value. I have plotted $x^2 +y^2$ in this way below:
Without introducing another axis, we can also just give different function value ranges a different colour, so a large value could be very dark and a low value could be very light or vice versa. We recognise the same function:
The difference between $(1)$ and $(2)$ is the notion of direction. The function you describe is usually a "scalar field", we only have the notion of magnitude or "value" but not of direction. Gradients will give you a so-called "vector field" as physicists often call it, we usually visualise this using VectorPlots. Below you find such a method, I've plotted the vector field $f(x,y)=(x^2+y,y^2+x )$
which has gradient $grad(f)(x,y)=(2x,2y)$
answered 37 mins ago
WesleyGroupshaveFeelingsToo
564216
add a comment |
up vote
2
down vote
$(1)$ The gradient is a vector-valued function, this maps pairs of numbers $(x,y)$ to some other pair of numbers $(x',y')$. These pairs of numbers we call vectors and they have a very geometric interpretation: they have a length and a direction. Specifically the gradient corresponds to the direction and magnitude of steepest ascent.
Also see:
https://en.wikipedia.org/wiki/Vector-valued_function
This Khan Academy link I found as well is very useful as he also thought of the same example as I did:
https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/gradient-and-directional-derivatives/v/gradient-and-graphs
$(2)$ On the contrary, when you plot a function that maps from $mathbbR^2 rightarrow mathbbR$ like $f(x,y)=x^2 +y^2$, when you want to plot this you often define a third variable $z=f(x,y)$ and you let the value of this variable be equal to the function value. I have plotted $x^2 +y^2$ in this way below:
Without introducing another axis, we can also just give different function value ranges a different colour, so a large value could be very dark and a low value could be very light or vice versa. We recognise the same function:
The difference between $(1)$ and $(2)$ is the notion of direction. The function you describe is usually a "scalar field", we only have the notion of magnitude or "value" but not of direction. Gradients will give you a so-called "vector field" as physicists often call it, we usually visualise this using VectorPlots. Below you find such a method, I've plotted the vector field $f(x,y)=(x^2+y,y^2+x )$
which has gradient $grad(f)(x,y)=(2x,2y)$
answered 37 mins ago
WesleyGroupshaveFeelingsToo
564216
add a comment |
up vote
2
down vote
up vote
2
down vote
$(1)$ The gradient is a vector-valued function, this maps pairs of numbers $(x,y)$ to some other pair of numbers $(x',y')$. These pairs of numbers we call vectors and they have a very geometric interpretation: they have a length and a direction. Specifically the gradient corresponds to the direction and magnitude of steepest ascent.
Also see:
https://en.wikipedia.org/wiki/Vector-valued_function
This Khan Academy link I found as well is very useful as he also thought of the same example as I did:
https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/gradient-and-directional-derivatives/v/gradient-and-graphs
$(2)$ On the contrary, when you plot a function that maps from $mathbbR^2 rightarrow mathbbR$ like $f(x,y)=x^2 +y^2$, when you want to plot this you often define a third variable $z=f(x,y)$ and you let the value of this variable be equal to the function value. I have plotted $x^2 +y^2$ in this way below:
Without introducing another axis, we can also just give different function value ranges a different colour, so a large value could be very dark and a low value could be very light or vice versa. We recognise the same function:
The difference between $(1)$ and $(2)$ is the notion of direction. The function you describe is usually a "scalar field", we only have the notion of magnitude or "value" but not of direction. Gradients will give you a so-called "vector field" as physicists often call it, we usually visualise this using VectorPlots. Below you find such a method, I've plotted the vector field $f(x,y)=(x^2+y,y^2+x )$
which has gradient $grad(f)(x,y)=(2x,2y)$
answered 37 mins ago
WesleyGroupshaveFeelingsToo
564216
$(1)$ The gradient is a vector-valued function, this maps pairs of numbers $(x,y)$ to some other pair of numbers $(x',y')$. These pairs of numbers we call vectors and they have a very geometric interpretation: they have a length and a direction. Specifically the gradient corresponds to the direction and magnitude of steepest ascent.
Also see:
https://en.wikipedia.org/wiki/Vector-valued_function
This Khan Academy link I found as well is very useful as he also thought of the same example as I did:
https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/gradient-and-directional-derivatives/v/gradient-and-graphs
$(2)$ On the contrary, when you plot a function that maps from $mathbbR^2 rightarrow mathbbR$ like $f(x,y)=x^2 +y^2$, when you want to plot this you often define a third variable $z=f(x,y)$ and you let the value of this variable be equal to the function value. I have plotted $x^2 +y^2$ in this way below:
Without introducing another axis, we can also just give different function value ranges a different colour, so a large value could be very dark and a low value could be very light or vice versa. We recognise the same function:
The difference between $(1)$ and $(2)$ is the notion of direction. The function you describe is usually a "scalar field", we only have the notion of magnitude or "value" but not of direction. Gradients will give you a so-called "vector field" as physicists often call it, we usually visualise this using VectorPlots. Below you find such a method, I've plotted the vector field $f(x,y)=(x^2+y,y^2+x )$
which has gradient $grad(f)(x,y)=(2x,2y)$
answered 37 mins ago
WesleyGroupshaveFeelingsToo
564216
edited 4 mins ago
answered 37 mins ago
WesleyGroupshaveFeelingsToo
564216
answered 37 mins ago
WesleyGroupshaveFeelingsToo
564216
answered 37 mins ago
WesleyGroupshaveFeelingsToo
564216
564216
add a comment |
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Problems in calculus often stem from physics and engineering, here we deal a lot with $mathbbR^3$ and $mathbbR^2$, so it makes sense that we visualise this by writing out a certain function in terms of $x,y,z$ we could define other basis vectors and write out some other functions for instance in terms of spherical polar coordinates, but the input is hard to visualise. We often transform to other coordinates to do integrations, for instance when we want to integrate over a sphere. The corresponding problem then transforms to some other graph in terms of these new coordinates.
– WesleyGroupshaveFeelingsToo
11 mins ago