Mixing
Does a change of variable affect the function
$begingroup$
If I have a function $f: mathbb{R}^2 to mathbb{R}$ then I can get $f$ in polar coordinates by doing : $f circ g(r, theta)$ where $g(r, theta) = (r cos theta, r sin theta)$.
Now my question is the following :
Does the shape of the graph of $f$ in the cartesian plane will look exactly the same as the shape of $f circ g(r, theta)$ drawn in the polar coordinates plane ?
I am saking this question because for example if I take a function $h : mathbb{R} to mathbb{R}$ then the shape of $h(x)$ doesn't look like the sape of $h(2x)$.
So for me it's not clear that if I transform $f$ to get the function in polar coordinates then in the cartesian plane we calculate : $f circ g(sqrt{x^2+y^2}, Arctan(y/x))$ or we calculate $f circ g (x, y)$. Because in the second interpretation the functions aren't the same.
For example $h(x)$ and $h(2x)$ aren't the same functions.
Thank you !
real-analysis calculus multivariable-calculus polar-coordinates
asked Jan 27 at 14:48
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
If I have a function $f: mathbb{R}^2 to mathbb{R}$ then I can get $f$ in polar coordinates by doing : $f circ g(r, theta)$ where $g(r, theta) = (r cos theta, r sin theta)$.
Now my question is the following :
Does the shape of the graph of $f$ in the cartesian plane will look exactly the same as the shape of $f circ g(r, theta)$ drawn in the polar coordinates plane ?
I am saking this question because for example if I take a function $h : mathbb{R} to mathbb{R}$ then the shape of $h(x)$ doesn't look like the sape of $h(2x)$.
So for me it's not clear that if I transform $f$ to get the function in polar coordinates then in the cartesian plane we calculate : $f circ g(sqrt{x^2+y^2}, Arctan(y/x))$ or we calculate $f circ g (x, y)$. Because in the second interpretation the functions aren't the same.
For example $h(x)$ and $h(2x)$ aren't the same functions.
Thank you !
real-analysis calculus multivariable-calculus polar-coordinates
asked Jan 27 at 14:48
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
$begingroup$
The graphs could be the same or they could be different. I can't say for sure because turning a function into a picture is different for different people. The visualization of a function is secondary. Regardless of what it looks like, a function must be well defined, which isn't subject to interpretation. We can both turn $x^2$ into a picture and our pictures might be different. But we have the same function. With composite functions, when the inside function is invertible (bijective), there really is "not much" of a difference between $fcirc g$ and $f$. So picture form is subjective
$endgroup$
– DWade64
Jan 27 at 19:40
$begingroup$
To try to help further: if $g$ is a function of $r$ and $theta$, then you can't really write $g(x,y)$. What you did at the end was say $g$ is invertible so there is some $g^{-1}$ taking an (x,y) back to an (r, theta). $gcirc g^{-1} = (x,y)$ and you wrote $f circ g circ g^{-1}$ or just $f(x,y)$
$endgroup$
– DWade64
Jan 27 at 20:01
add a comment |
$begingroup$
If I have a function $f: mathbb{R}^2 to mathbb{R}$ then I can get $f$ in polar coordinates by doing : $f circ g(r, theta)$ where $g(r, theta) = (r cos theta, r sin theta)$.
Now my question is the following :
Does the shape of the graph of $f$ in the cartesian plane will look exactly the same as the shape of $f circ g(r, theta)$ drawn in the polar coordinates plane ?
I am saking this question because for example if I take a function $h : mathbb{R} to mathbb{R}$ then the shape of $h(x)$ doesn't look like the sape of $h(2x)$.
So for me it's not clear that if I transform $f$ to get the function in polar coordinates then in the cartesian plane we calculate : $f circ g(sqrt{x^2+y^2}, Arctan(y/x))$ or we calculate $f circ g (x, y)$. Because in the second interpretation the functions aren't the same.
For example $h(x)$ and $h(2x)$ aren't the same functions.
Thank you !
real-analysis calculus multivariable-calculus polar-coordinates
asked Jan 27 at 14:48
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
If I have a function $f: mathbb{R}^2 to mathbb{R}$ then I can get $f$ in polar coordinates by doing : $f circ g(r, theta)$ where $g(r, theta) = (r cos theta, r sin theta)$.
Now my question is the following :
Does the shape of the graph of $f$ in the cartesian plane will look exactly the same as the shape of $f circ g(r, theta)$ drawn in the polar coordinates plane ?
I am saking this question because for example if I take a function $h : mathbb{R} to mathbb{R}$ then the shape of $h(x)$ doesn't look like the sape of $h(2x)$.
So for me it's not clear that if I transform $f$ to get the function in polar coordinates then in the cartesian plane we calculate : $f circ g(sqrt{x^2+y^2}, Arctan(y/x))$ or we calculate $f circ g (x, y)$. Because in the second interpretation the functions aren't the same.
For example $h(x)$ and $h(2x)$ aren't the same functions.
Thank you !
real-analysis calculus multivariable-calculus polar-coordinates
real-analysis calculus multivariable-calculus polar-coordinates
asked Jan 27 at 14:48
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 14:48
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 14:48
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 14:48
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 14:48
dghkgfzyukzdghkgfzyukz
16612
16612
$begingroup$
The graphs could be the same or they could be different. I can't say for sure because turning a function into a picture is different for different people. The visualization of a function is secondary. Regardless of what it looks like, a function must be well defined, which isn't subject to interpretation. We can both turn $x^2$ into a picture and our pictures might be different. But we have the same function. With composite functions, when the inside function is invertible (bijective), there really is "not much" of a difference between $fcirc g$ and $f$. So picture form is subjective
$endgroup$
– DWade64
Jan 27 at 19:40
$begingroup$
To try to help further: if $g$ is a function of $r$ and $theta$, then you can't really write $g(x,y)$. What you did at the end was say $g$ is invertible so there is some $g^{-1}$ taking an (x,y) back to an (r, theta). $gcirc g^{-1} = (x,y)$ and you wrote $f circ g circ g^{-1}$ or just $f(x,y)$
$endgroup$
– DWade64
Jan 27 at 20:01
add a comment |
$begingroup$
The graphs could be the same or they could be different. I can't say for sure because turning a function into a picture is different for different people. The visualization of a function is secondary. Regardless of what it looks like, a function must be well defined, which isn't subject to interpretation. We can both turn $x^2$ into a picture and our pictures might be different. But we have the same function. With composite functions, when the inside function is invertible (bijective), there really is "not much" of a difference between $fcirc g$ and $f$. So picture form is subjective
$endgroup$
– DWade64
Jan 27 at 19:40
$begingroup$
To try to help further: if $g$ is a function of $r$ and $theta$, then you can't really write $g(x,y)$. What you did at the end was say $g$ is invertible so there is some $g^{-1}$ taking an (x,y) back to an (r, theta). $gcirc g^{-1} = (x,y)$ and you wrote $f circ g circ g^{-1}$ or just $f(x,y)$
$endgroup$
– DWade64
Jan 27 at 20:01
$begingroup$
The graphs could be the same or they could be different. I can't say for sure because turning a function into a picture is different for different people. The visualization of a function is secondary. Regardless of what it looks like, a function must be well defined, which isn't subject to interpretation. We can both turn $x^2$ into a picture and our pictures might be different. But we have the same function. With composite functions, when the inside function is invertible (bijective), there really is "not much" of a difference between $fcirc g$ and $f$. So picture form is subjective
$endgroup$
– DWade64
Jan 27 at 19:40
$begingroup$
The graphs could be the same or they could be different. I can't say for sure because turning a function into a picture is different for different people. The visualization of a function is secondary. Regardless of what it looks like, a function must be well defined, which isn't subject to interpretation. We can both turn $x^2$ into a picture and our pictures might be different. But we have the same function. With composite functions, when the inside function is invertible (bijective), there really is "not much" of a difference between $fcirc g$ and $f$. So picture form is subjective
$endgroup$
– DWade64
Jan 27 at 19:40
$begingroup$
To try to help further: if $g$ is a function of $r$ and $theta$, then you can't really write $g(x,y)$. What you did at the end was say $g$ is invertible so there is some $g^{-1}$ taking an (x,y) back to an (r, theta). $gcirc g^{-1} = (x,y)$ and you wrote $f circ g circ g^{-1}$ or just $f(x,y)$
$endgroup$
– DWade64
Jan 27 at 20:01
$begingroup$
To try to help further: if $g$ is a function of $r$ and $theta$, then you can't really write $g(x,y)$. What you did at the end was say $g$ is invertible so there is some $g^{-1}$ taking an (x,y) back to an (r, theta). $gcirc g^{-1} = (x,y)$ and you wrote $f circ g circ g^{-1}$ or just $f(x,y)$
$endgroup$
– DWade64
Jan 27 at 20:01
add a comment |
1 Answer
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$begingroup$
It doesn't really matter what the function looks like pictorially. Let us both graph the function $g(x) := x^2$. You might graph the function in the center of a sheet of paper. I might graph mine in a corner, at some angle. My tick marks might be extremely far (miles) apart, and so on. But we both graphed the function correctly.
Given your outside function $f$, what does the $(x,y)$ part of $f(x,y)$ mean to you? Strictly speaking, $(x,y)$ just means an element in an abstract 2-dimensional domain (an element of the domain could be (blue, tree) for instance). But in math, $x,y$ are continuous numbers. From algebra class, we visualize a pair of numbers $(x,y)$ as a point in space. A coordinate system is just a mapping between a physical point in space to a unique number. When you label a point in space with $(x,y)$, most people (including me) take this to mean that you are filling space with perpendicular, evenly-space, grid lines (Cartesian grid lines). But this is just a chosen, go-to, standard. $(x,y)$ just means you have some labeling system set up, where every point in space has a unique number associated with it. So even though your visualization of $g(x) := x^2$ looks different than mine, that's ok because it's just a visualization. You chose some labeling system, and I chose some labeling system. Regardless of what the numbers are, if you physically point with your finger to a position in space, I will be able to look at that exact same position. And as long as you and I both have a valid labeling system, there will be a unique (one-to-one) mapping between your numbers and my numbers. Such a mapping between our labeling systems may not have an nice analytical form, but we could write everything down: "ok your $(3,4)$ goes to my $(-1,3)$, and your $(3,5)$ goes to my etc..."
So when you write $f(x,y)$ lets just assume, as anyone would, you are filling space with perpendicular, evenly-spaced, grid lines. You number the vertical grid lines $1, 2, 3,...$ and the horizontal grid lines $1,2,3,...$ You point to a location in space corresponding to $(3,4)$. At that location in space, your function outputs say $10$. In coordinate system number 2, that same point has the label $(5, arctan(4/3)approx 53)$. If you wanted to graph $fcirc g(r,theta)$, which is a composite function with respect to variables $(r,theta)$. I would fill space again with perpendicular, evenly-space, grid lines again with the horizontal direction $r$ and the vertical direction $theta$. Since that same point is now called $(5, )$ in the horizontal direction, the new grid lines are finer than before. And since that same point is now called $(;,53)$ in the vertical direction, the new vertical direction grid lines are much finer than before. In this (chosen visualization, emphasis on chosen), the graphs would actually look the same. But if I kept the same spacing of grid lines as system 1, then the graphs would look much different because I need to now go 5 tick marks instead of 3 in the horizontal direction and 53 tick marks instead of 4 in the vertical direction (this is why our two graphs of $g(x) := x^2$ were different - but it doesn't matter because it's just a visualization). It's a question of 1) do we keep the physical points the same and change the grid lines or 2) do we keep the grid lines the same and change the physical points ($(3,4)$ now "expands" or gets dragged out to $(5,53)$ - I suspect in general relativity, when the say "space is expanding," this can be seen in the math with a "change of coordinates."
If you decide to fill space with a bunch of circular grid lines and ray-like grid lines for $r$ and $theta$ (exactly matching the Cartesian grid lines, so the circle $r = 5$ intersects perfectly where $x = 3$ and $y = 4$ vertical and horizontal grid lines intersected and so on), then the graphs would be exactly the same again. This is what I understand "polar coordinate grid lines" to mean.
You're right that if you have an outside $f$ and inside $g$ function, then the composition $fcirc g neq f$. But if you know $g$ and $g$ is an invertible function, then you don't actually "lose any information about $f$ when you form the composition $fcirc g$." Because if the person who has the function $f circ g$ inputs $(r, theta) = (5, 53)$, the person who has $f$ would understand "oh you were inputting my $(3,4)$." This is why you can do double integrals or triple integrals and whatnot in different coordinate systems. You don't lose any information with an invertible inside function. You just have to worry about conversion factors (which is the Jacobian).
answered Jan 27 at 19:48
DWade64DWade64
6282613
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add a comment |
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$begingroup$
It doesn't really matter what the function looks like pictorially. Let us both graph the function $g(x) := x^2$. You might graph the function in the center of a sheet of paper. I might graph mine in a corner, at some angle. My tick marks might be extremely far (miles) apart, and so on. But we both graphed the function correctly.
Given your outside function $f$, what does the $(x,y)$ part of $f(x,y)$ mean to you? Strictly speaking, $(x,y)$ just means an element in an abstract 2-dimensional domain (an element of the domain could be (blue, tree) for instance). But in math, $x,y$ are continuous numbers. From algebra class, we visualize a pair of numbers $(x,y)$ as a point in space. A coordinate system is just a mapping between a physical point in space to a unique number. When you label a point in space with $(x,y)$, most people (including me) take this to mean that you are filling space with perpendicular, evenly-space, grid lines (Cartesian grid lines). But this is just a chosen, go-to, standard. $(x,y)$ just means you have some labeling system set up, where every point in space has a unique number associated with it. So even though your visualization of $g(x) := x^2$ looks different than mine, that's ok because it's just a visualization. You chose some labeling system, and I chose some labeling system. Regardless of what the numbers are, if you physically point with your finger to a position in space, I will be able to look at that exact same position. And as long as you and I both have a valid labeling system, there will be a unique (one-to-one) mapping between your numbers and my numbers. Such a mapping between our labeling systems may not have an nice analytical form, but we could write everything down: "ok your $(3,4)$ goes to my $(-1,3)$, and your $(3,5)$ goes to my etc..."
So when you write $f(x,y)$ lets just assume, as anyone would, you are filling space with perpendicular, evenly-spaced, grid lines. You number the vertical grid lines $1, 2, 3,...$ and the horizontal grid lines $1,2,3,...$ You point to a location in space corresponding to $(3,4)$. At that location in space, your function outputs say $10$. In coordinate system number 2, that same point has the label $(5, arctan(4/3)approx 53)$. If you wanted to graph $fcirc g(r,theta)$, which is a composite function with respect to variables $(r,theta)$. I would fill space again with perpendicular, evenly-space, grid lines again with the horizontal direction $r$ and the vertical direction $theta$. Since that same point is now called $(5, )$ in the horizontal direction, the new grid lines are finer than before. And since that same point is now called $(;,53)$ in the vertical direction, the new vertical direction grid lines are much finer than before. In this (chosen visualization, emphasis on chosen), the graphs would actually look the same. But if I kept the same spacing of grid lines as system 1, then the graphs would look much different because I need to now go 5 tick marks instead of 3 in the horizontal direction and 53 tick marks instead of 4 in the vertical direction (this is why our two graphs of $g(x) := x^2$ were different - but it doesn't matter because it's just a visualization). It's a question of 1) do we keep the physical points the same and change the grid lines or 2) do we keep the grid lines the same and change the physical points ($(3,4)$ now "expands" or gets dragged out to $(5,53)$ - I suspect in general relativity, when the say "space is expanding," this can be seen in the math with a "change of coordinates."
If you decide to fill space with a bunch of circular grid lines and ray-like grid lines for $r$ and $theta$ (exactly matching the Cartesian grid lines, so the circle $r = 5$ intersects perfectly where $x = 3$ and $y = 4$ vertical and horizontal grid lines intersected and so on), then the graphs would be exactly the same again. This is what I understand "polar coordinate grid lines" to mean.
You're right that if you have an outside $f$ and inside $g$ function, then the composition $fcirc g neq f$. But if you know $g$ and $g$ is an invertible function, then you don't actually "lose any information about $f$ when you form the composition $fcirc g$." Because if the person who has the function $f circ g$ inputs $(r, theta) = (5, 53)$, the person who has $f$ would understand "oh you were inputting my $(3,4)$." This is why you can do double integrals or triple integrals and whatnot in different coordinate systems. You don't lose any information with an invertible inside function. You just have to worry about conversion factors (which is the Jacobian).
answered Jan 27 at 19:48
DWade64DWade64
6282613
$endgroup$
add a comment |
$begingroup$
It doesn't really matter what the function looks like pictorially. Let us both graph the function $g(x) := x^2$. You might graph the function in the center of a sheet of paper. I might graph mine in a corner, at some angle. My tick marks might be extremely far (miles) apart, and so on. But we both graphed the function correctly.
Given your outside function $f$, what does the $(x,y)$ part of $f(x,y)$ mean to you? Strictly speaking, $(x,y)$ just means an element in an abstract 2-dimensional domain (an element of the domain could be (blue, tree) for instance). But in math, $x,y$ are continuous numbers. From algebra class, we visualize a pair of numbers $(x,y)$ as a point in space. A coordinate system is just a mapping between a physical point in space to a unique number. When you label a point in space with $(x,y)$, most people (including me) take this to mean that you are filling space with perpendicular, evenly-space, grid lines (Cartesian grid lines). But this is just a chosen, go-to, standard. $(x,y)$ just means you have some labeling system set up, where every point in space has a unique number associated with it. So even though your visualization of $g(x) := x^2$ looks different than mine, that's ok because it's just a visualization. You chose some labeling system, and I chose some labeling system. Regardless of what the numbers are, if you physically point with your finger to a position in space, I will be able to look at that exact same position. And as long as you and I both have a valid labeling system, there will be a unique (one-to-one) mapping between your numbers and my numbers. Such a mapping between our labeling systems may not have an nice analytical form, but we could write everything down: "ok your $(3,4)$ goes to my $(-1,3)$, and your $(3,5)$ goes to my etc..."
So when you write $f(x,y)$ lets just assume, as anyone would, you are filling space with perpendicular, evenly-spaced, grid lines. You number the vertical grid lines $1, 2, 3,...$ and the horizontal grid lines $1,2,3,...$ You point to a location in space corresponding to $(3,4)$. At that location in space, your function outputs say $10$. In coordinate system number 2, that same point has the label $(5, arctan(4/3)approx 53)$. If you wanted to graph $fcirc g(r,theta)$, which is a composite function with respect to variables $(r,theta)$. I would fill space again with perpendicular, evenly-space, grid lines again with the horizontal direction $r$ and the vertical direction $theta$. Since that same point is now called $(5, )$ in the horizontal direction, the new grid lines are finer than before. And since that same point is now called $(;,53)$ in the vertical direction, the new vertical direction grid lines are much finer than before. In this (chosen visualization, emphasis on chosen), the graphs would actually look the same. But if I kept the same spacing of grid lines as system 1, then the graphs would look much different because I need to now go 5 tick marks instead of 3 in the horizontal direction and 53 tick marks instead of 4 in the vertical direction (this is why our two graphs of $g(x) := x^2$ were different - but it doesn't matter because it's just a visualization). It's a question of 1) do we keep the physical points the same and change the grid lines or 2) do we keep the grid lines the same and change the physical points ($(3,4)$ now "expands" or gets dragged out to $(5,53)$ - I suspect in general relativity, when the say "space is expanding," this can be seen in the math with a "change of coordinates."
If you decide to fill space with a bunch of circular grid lines and ray-like grid lines for $r$ and $theta$ (exactly matching the Cartesian grid lines, so the circle $r = 5$ intersects perfectly where $x = 3$ and $y = 4$ vertical and horizontal grid lines intersected and so on), then the graphs would be exactly the same again. This is what I understand "polar coordinate grid lines" to mean.
You're right that if you have an outside $f$ and inside $g$ function, then the composition $fcirc g neq f$. But if you know $g$ and $g$ is an invertible function, then you don't actually "lose any information about $f$ when you form the composition $fcirc g$." Because if the person who has the function $f circ g$ inputs $(r, theta) = (5, 53)$, the person who has $f$ would understand "oh you were inputting my $(3,4)$." This is why you can do double integrals or triple integrals and whatnot in different coordinate systems. You don't lose any information with an invertible inside function. You just have to worry about conversion factors (which is the Jacobian).
answered Jan 27 at 19:48
DWade64DWade64
6282613
$endgroup$
add a comment |
$begingroup$
It doesn't really matter what the function looks like pictorially. Let us both graph the function $g(x) := x^2$. You might graph the function in the center of a sheet of paper. I might graph mine in a corner, at some angle. My tick marks might be extremely far (miles) apart, and so on. But we both graphed the function correctly.
Given your outside function $f$, what does the $(x,y)$ part of $f(x,y)$ mean to you? Strictly speaking, $(x,y)$ just means an element in an abstract 2-dimensional domain (an element of the domain could be (blue, tree) for instance). But in math, $x,y$ are continuous numbers. From algebra class, we visualize a pair of numbers $(x,y)$ as a point in space. A coordinate system is just a mapping between a physical point in space to a unique number. When you label a point in space with $(x,y)$, most people (including me) take this to mean that you are filling space with perpendicular, evenly-space, grid lines (Cartesian grid lines). But this is just a chosen, go-to, standard. $(x,y)$ just means you have some labeling system set up, where every point in space has a unique number associated with it. So even though your visualization of $g(x) := x^2$ looks different than mine, that's ok because it's just a visualization. You chose some labeling system, and I chose some labeling system. Regardless of what the numbers are, if you physically point with your finger to a position in space, I will be able to look at that exact same position. And as long as you and I both have a valid labeling system, there will be a unique (one-to-one) mapping between your numbers and my numbers. Such a mapping between our labeling systems may not have an nice analytical form, but we could write everything down: "ok your $(3,4)$ goes to my $(-1,3)$, and your $(3,5)$ goes to my etc..."
So when you write $f(x,y)$ lets just assume, as anyone would, you are filling space with perpendicular, evenly-spaced, grid lines. You number the vertical grid lines $1, 2, 3,...$ and the horizontal grid lines $1,2,3,...$ You point to a location in space corresponding to $(3,4)$. At that location in space, your function outputs say $10$. In coordinate system number 2, that same point has the label $(5, arctan(4/3)approx 53)$. If you wanted to graph $fcirc g(r,theta)$, which is a composite function with respect to variables $(r,theta)$. I would fill space again with perpendicular, evenly-space, grid lines again with the horizontal direction $r$ and the vertical direction $theta$. Since that same point is now called $(5, )$ in the horizontal direction, the new grid lines are finer than before. And since that same point is now called $(;,53)$ in the vertical direction, the new vertical direction grid lines are much finer than before. In this (chosen visualization, emphasis on chosen), the graphs would actually look the same. But if I kept the same spacing of grid lines as system 1, then the graphs would look much different because I need to now go 5 tick marks instead of 3 in the horizontal direction and 53 tick marks instead of 4 in the vertical direction (this is why our two graphs of $g(x) := x^2$ were different - but it doesn't matter because it's just a visualization). It's a question of 1) do we keep the physical points the same and change the grid lines or 2) do we keep the grid lines the same and change the physical points ($(3,4)$ now "expands" or gets dragged out to $(5,53)$ - I suspect in general relativity, when the say "space is expanding," this can be seen in the math with a "change of coordinates."
If you decide to fill space with a bunch of circular grid lines and ray-like grid lines for $r$ and $theta$ (exactly matching the Cartesian grid lines, so the circle $r = 5$ intersects perfectly where $x = 3$ and $y = 4$ vertical and horizontal grid lines intersected and so on), then the graphs would be exactly the same again. This is what I understand "polar coordinate grid lines" to mean.
You're right that if you have an outside $f$ and inside $g$ function, then the composition $fcirc g neq f$. But if you know $g$ and $g$ is an invertible function, then you don't actually "lose any information about $f$ when you form the composition $fcirc g$." Because if the person who has the function $f circ g$ inputs $(r, theta) = (5, 53)$, the person who has $f$ would understand "oh you were inputting my $(3,4)$." This is why you can do double integrals or triple integrals and whatnot in different coordinate systems. You don't lose any information with an invertible inside function. You just have to worry about conversion factors (which is the Jacobian).
answered Jan 27 at 19:48
DWade64DWade64
6282613
$endgroup$
It doesn't really matter what the function looks like pictorially. Let us both graph the function $g(x) := x^2$. You might graph the function in the center of a sheet of paper. I might graph mine in a corner, at some angle. My tick marks might be extremely far (miles) apart, and so on. But we both graphed the function correctly.
Given your outside function $f$, what does the $(x,y)$ part of $f(x,y)$ mean to you? Strictly speaking, $(x,y)$ just means an element in an abstract 2-dimensional domain (an element of the domain could be (blue, tree) for instance). But in math, $x,y$ are continuous numbers. From algebra class, we visualize a pair of numbers $(x,y)$ as a point in space. A coordinate system is just a mapping between a physical point in space to a unique number. When you label a point in space with $(x,y)$, most people (including me) take this to mean that you are filling space with perpendicular, evenly-space, grid lines (Cartesian grid lines). But this is just a chosen, go-to, standard. $(x,y)$ just means you have some labeling system set up, where every point in space has a unique number associated with it. So even though your visualization of $g(x) := x^2$ looks different than mine, that's ok because it's just a visualization. You chose some labeling system, and I chose some labeling system. Regardless of what the numbers are, if you physically point with your finger to a position in space, I will be able to look at that exact same position. And as long as you and I both have a valid labeling system, there will be a unique (one-to-one) mapping between your numbers and my numbers. Such a mapping between our labeling systems may not have an nice analytical form, but we could write everything down: "ok your $(3,4)$ goes to my $(-1,3)$, and your $(3,5)$ goes to my etc..."
So when you write $f(x,y)$ lets just assume, as anyone would, you are filling space with perpendicular, evenly-spaced, grid lines. You number the vertical grid lines $1, 2, 3,...$ and the horizontal grid lines $1,2,3,...$ You point to a location in space corresponding to $(3,4)$. At that location in space, your function outputs say $10$. In coordinate system number 2, that same point has the label $(5, arctan(4/3)approx 53)$. If you wanted to graph $fcirc g(r,theta)$, which is a composite function with respect to variables $(r,theta)$. I would fill space again with perpendicular, evenly-space, grid lines again with the horizontal direction $r$ and the vertical direction $theta$. Since that same point is now called $(5, )$ in the horizontal direction, the new grid lines are finer than before. And since that same point is now called $(;,53)$ in the vertical direction, the new vertical direction grid lines are much finer than before. In this (chosen visualization, emphasis on chosen), the graphs would actually look the same. But if I kept the same spacing of grid lines as system 1, then the graphs would look much different because I need to now go 5 tick marks instead of 3 in the horizontal direction and 53 tick marks instead of 4 in the vertical direction (this is why our two graphs of $g(x) := x^2$ were different - but it doesn't matter because it's just a visualization). It's a question of 1) do we keep the physical points the same and change the grid lines or 2) do we keep the grid lines the same and change the physical points ($(3,4)$ now "expands" or gets dragged out to $(5,53)$ - I suspect in general relativity, when the say "space is expanding," this can be seen in the math with a "change of coordinates."
If you decide to fill space with a bunch of circular grid lines and ray-like grid lines for $r$ and $theta$ (exactly matching the Cartesian grid lines, so the circle $r = 5$ intersects perfectly where $x = 3$ and $y = 4$ vertical and horizontal grid lines intersected and so on), then the graphs would be exactly the same again. This is what I understand "polar coordinate grid lines" to mean.
You're right that if you have an outside $f$ and inside $g$ function, then the composition $fcirc g neq f$. But if you know $g$ and $g$ is an invertible function, then you don't actually "lose any information about $f$ when you form the composition $fcirc g$." Because if the person who has the function $f circ g$ inputs $(r, theta) = (5, 53)$, the person who has $f$ would understand "oh you were inputting my $(3,4)$." This is why you can do double integrals or triple integrals and whatnot in different coordinate systems. You don't lose any information with an invertible inside function. You just have to worry about conversion factors (which is the Jacobian).
answered Jan 27 at 19:48
DWade64DWade64
6282613
answered Jan 27 at 19:48
DWade64DWade64
6282613
answered Jan 27 at 19:48
DWade64DWade64
6282613
answered Jan 27 at 19:48
DWade64DWade64
6282613
6282613
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$begingroup$
The graphs could be the same or they could be different. I can't say for sure because turning a function into a picture is different for different people. The visualization of a function is secondary. Regardless of what it looks like, a function must be well defined, which isn't subject to interpretation. We can both turn $x^2$ into a picture and our pictures might be different. But we have the same function. With composite functions, when the inside function is invertible (bijective), there really is "not much" of a difference between $fcirc g$ and $f$. So picture form is subjective
$endgroup$
– DWade64
Jan 27 at 19:40
$begingroup$
To try to help further: if $g$ is a function of $r$ and $theta$, then you can't really write $g(x,y)$. What you did at the end was say $g$ is invertible so there is some $g^{-1}$ taking an (x,y) back to an (r, theta). $gcirc g^{-1} = (x,y)$ and you wrote $f circ g circ g^{-1}$ or just $f(x,y)$
$endgroup$
– DWade64
Jan 27 at 20:01