Standard Cartesian Plane
$begingroup$
I have posted before.
But I had a question,
Is the **standard, ** used cartesian coordinate plane this:
Where the vertical axis represents $f(x) = y$ and the horizontal axis represents $x$?
I was wondering because in evaluation of
$$I = int_{-infty}^{infty} e^{-x^2} dx$$
We take:
$$I = int_{-infty}^{infty} e^{-y^2} dy$$
Then we take $I^2$.
Are you then using $f(y) = e^{-y^2}$ or just changing $x to y$ ?? So $y$ is the independent variable?
calculus functions
edited Jan 6 '15 at 18:02
Mark Fantini
4,86041936
asked Jan 6 '15 at 17:56
anonymousanonymous
12727
$endgroup$
|
show 1 more comment
$begingroup$
I have posted before.
But I had a question,
Is the **standard, ** used cartesian coordinate plane this:
Where the vertical axis represents $f(x) = y$ and the horizontal axis represents $x$?
I was wondering because in evaluation of
$$I = int_{-infty}^{infty} e^{-x^2} dx$$
We take:
$$I = int_{-infty}^{infty} e^{-y^2} dy$$
Then we take $I^2$.
Are you then using $f(y) = e^{-y^2}$ or just changing $x to y$ ?? So $y$ is the independent variable?
calculus functions
edited Jan 6 '15 at 18:02
Mark Fantini
4,86041936
asked Jan 6 '15 at 17:56
anonymousanonymous
12727
$endgroup$
1
$begingroup$
The vertical axis is the set of points $(x,y) in mathbb{R}^2$ such that $x=0$. The set of points such that $y=f(x)$ represents a curve in the $xy$ plane. Would it make you feel better if we wrote $$I = int_{- infty}^{+ infty} e^{-t^2} , dt$$ and $$I = int_{- infty}^{+ infty} e^{- w^2} , dw?$$ We use $x$ and $y$ because of inertia in switching to polar coordinates. In the example I gave we can write $t = r cos(theta), w = r sin(theta)$, but for some reason some people aren't comfortable with that and think only $x$ and $y$ can be written in polar coordinates.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:01
$begingroup$
But the problem is, when you combine it, when it becomes an iterated integral, then do you have a $3D$ plot of the curve? It becomes $e^{-(t^2 + w^2)}$ but you cant have a $3D$ plot because the nor $w$ nor $t$ are the vertical axis in the 2D graph.
$endgroup$
– anonymous
Jan 6 '15 at 18:04
$begingroup$
It's no longer a curve, it becomes a surface.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:07
$begingroup$
Exactly, how? Given: en.wikipedia.org/wiki/… Then one of the axes must be $y$ for example, and one of the axes must be $x$ for it to be a 3D system, and a surface.?
$endgroup$
– anonymous
Jan 6 '15 at 18:08
$begingroup$
@MarkFantini, do you see? In the 3D plane there is a seperate $xy$ plane, so if we have $e^{-(x^2 + y^2)}$ there then the issue would be that either $y$ is a dependent variable for $x$ or either $x$ is a dependent variable for $y$.
$endgroup$
– anonymous
Jan 6 '15 at 18:13
|
show 1 more comment
$begingroup$
I have posted before.
But I had a question,
Is the **standard, ** used cartesian coordinate plane this:
Where the vertical axis represents $f(x) = y$ and the horizontal axis represents $x$?
I was wondering because in evaluation of
$$I = int_{-infty}^{infty} e^{-x^2} dx$$
We take:
$$I = int_{-infty}^{infty} e^{-y^2} dy$$
Then we take $I^2$.
Are you then using $f(y) = e^{-y^2}$ or just changing $x to y$ ?? So $y$ is the independent variable?
calculus functions
edited Jan 6 '15 at 18:02
Mark Fantini
4,86041936
asked Jan 6 '15 at 17:56
anonymousanonymous
12727
$endgroup$
I have posted before.
But I had a question,
Is the **standard, ** used cartesian coordinate plane this:
Where the vertical axis represents $f(x) = y$ and the horizontal axis represents $x$?
I was wondering because in evaluation of
$$I = int_{-infty}^{infty} e^{-x^2} dx$$
We take:
$$I = int_{-infty}^{infty} e^{-y^2} dy$$
Then we take $I^2$.
Are you then using $f(y) = e^{-y^2}$ or just changing $x to y$ ?? So $y$ is the independent variable?
calculus functions
calculus functions
edited Jan 6 '15 at 18:02
Mark Fantini
4,86041936
asked Jan 6 '15 at 17:56
anonymousanonymous
12727
edited Jan 6 '15 at 18:02
Mark Fantini
4,86041936
asked Jan 6 '15 at 17:56
anonymousanonymous
12727
edited Jan 6 '15 at 18:02
Mark Fantini
4,86041936
edited Jan 6 '15 at 18:02
Mark Fantini
4,86041936
edited Jan 6 '15 at 18:02
Mark Fantini
4,86041936
4,86041936
asked Jan 6 '15 at 17:56
anonymousanonymous
12727
asked Jan 6 '15 at 17:56
anonymousanonymous
12727
asked Jan 6 '15 at 17:56
anonymousanonymous
12727
12727
1
$begingroup$
The vertical axis is the set of points $(x,y) in mathbb{R}^2$ such that $x=0$. The set of points such that $y=f(x)$ represents a curve in the $xy$ plane. Would it make you feel better if we wrote $$I = int_{- infty}^{+ infty} e^{-t^2} , dt$$ and $$I = int_{- infty}^{+ infty} e^{- w^2} , dw?$$ We use $x$ and $y$ because of inertia in switching to polar coordinates. In the example I gave we can write $t = r cos(theta), w = r sin(theta)$, but for some reason some people aren't comfortable with that and think only $x$ and $y$ can be written in polar coordinates.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:01
$begingroup$
But the problem is, when you combine it, when it becomes an iterated integral, then do you have a $3D$ plot of the curve? It becomes $e^{-(t^2 + w^2)}$ but you cant have a $3D$ plot because the nor $w$ nor $t$ are the vertical axis in the 2D graph.
$endgroup$
– anonymous
Jan 6 '15 at 18:04
$begingroup$
It's no longer a curve, it becomes a surface.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:07
$begingroup$
Exactly, how? Given: en.wikipedia.org/wiki/… Then one of the axes must be $y$ for example, and one of the axes must be $x$ for it to be a 3D system, and a surface.?
$endgroup$
– anonymous
Jan 6 '15 at 18:08
$begingroup$
@MarkFantini, do you see? In the 3D plane there is a seperate $xy$ plane, so if we have $e^{-(x^2 + y^2)}$ there then the issue would be that either $y$ is a dependent variable for $x$ or either $x$ is a dependent variable for $y$.
$endgroup$
– anonymous
Jan 6 '15 at 18:13
|
show 1 more comment
1
$begingroup$
The vertical axis is the set of points $(x,y) in mathbb{R}^2$ such that $x=0$. The set of points such that $y=f(x)$ represents a curve in the $xy$ plane. Would it make you feel better if we wrote $$I = int_{- infty}^{+ infty} e^{-t^2} , dt$$ and $$I = int_{- infty}^{+ infty} e^{- w^2} , dw?$$ We use $x$ and $y$ because of inertia in switching to polar coordinates. In the example I gave we can write $t = r cos(theta), w = r sin(theta)$, but for some reason some people aren't comfortable with that and think only $x$ and $y$ can be written in polar coordinates.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:01
$begingroup$
But the problem is, when you combine it, when it becomes an iterated integral, then do you have a $3D$ plot of the curve? It becomes $e^{-(t^2 + w^2)}$ but you cant have a $3D$ plot because the nor $w$ nor $t$ are the vertical axis in the 2D graph.
$endgroup$
– anonymous
Jan 6 '15 at 18:04
$begingroup$
It's no longer a curve, it becomes a surface.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:07
$begingroup$
Exactly, how? Given: en.wikipedia.org/wiki/… Then one of the axes must be $y$ for example, and one of the axes must be $x$ for it to be a 3D system, and a surface.?
$endgroup$
– anonymous
Jan 6 '15 at 18:08
$begingroup$
@MarkFantini, do you see? In the 3D plane there is a seperate $xy$ plane, so if we have $e^{-(x^2 + y^2)}$ there then the issue would be that either $y$ is a dependent variable for $x$ or either $x$ is a dependent variable for $y$.
$endgroup$
– anonymous
Jan 6 '15 at 18:13
1
1
$begingroup$
The vertical axis is the set of points $(x,y) in mathbb{R}^2$ such that $x=0$. The set of points such that $y=f(x)$ represents a curve in the $xy$ plane. Would it make you feel better if we wrote $$I = int_{- infty}^{+ infty} e^{-t^2} , dt$$ and $$I = int_{- infty}^{+ infty} e^{- w^2} , dw?$$ We use $x$ and $y$ because of inertia in switching to polar coordinates. In the example I gave we can write $t = r cos(theta), w = r sin(theta)$, but for some reason some people aren't comfortable with that and think only $x$ and $y$ can be written in polar coordinates.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:01
$begingroup$
The vertical axis is the set of points $(x,y) in mathbb{R}^2$ such that $x=0$. The set of points such that $y=f(x)$ represents a curve in the $xy$ plane. Would it make you feel better if we wrote $$I = int_{- infty}^{+ infty} e^{-t^2} , dt$$ and $$I = int_{- infty}^{+ infty} e^{- w^2} , dw?$$ We use $x$ and $y$ because of inertia in switching to polar coordinates. In the example I gave we can write $t = r cos(theta), w = r sin(theta)$, but for some reason some people aren't comfortable with that and think only $x$ and $y$ can be written in polar coordinates.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:01
$begingroup$
But the problem is, when you combine it, when it becomes an iterated integral, then do you have a $3D$ plot of the curve? It becomes $e^{-(t^2 + w^2)}$ but you cant have a $3D$ plot because the nor $w$ nor $t$ are the vertical axis in the 2D graph.
$endgroup$
– anonymous
Jan 6 '15 at 18:04
$begingroup$
But the problem is, when you combine it, when it becomes an iterated integral, then do you have a $3D$ plot of the curve? It becomes $e^{-(t^2 + w^2)}$ but you cant have a $3D$ plot because the nor $w$ nor $t$ are the vertical axis in the 2D graph.
$endgroup$
– anonymous
Jan 6 '15 at 18:04
$begingroup$
It's no longer a curve, it becomes a surface.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:07
$begingroup$
It's no longer a curve, it becomes a surface.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:07
$begingroup$
Exactly, how? Given: en.wikipedia.org/wiki/… Then one of the axes must be $y$ for example, and one of the axes must be $x$ for it to be a 3D system, and a surface.?
$endgroup$
– anonymous
Jan 6 '15 at 18:08
$begingroup$
Exactly, how? Given: en.wikipedia.org/wiki/… Then one of the axes must be $y$ for example, and one of the axes must be $x$ for it to be a 3D system, and a surface.?
$endgroup$
– anonymous
Jan 6 '15 at 18:08
$begingroup$
@MarkFantini, do you see? In the 3D plane there is a seperate $xy$ plane, so if we have $e^{-(x^2 + y^2)}$ there then the issue would be that either $y$ is a dependent variable for $x$ or either $x$ is a dependent variable for $y$.
$endgroup$
– anonymous
Jan 6 '15 at 18:13
$begingroup$
@MarkFantini, do you see? In the 3D plane there is a seperate $xy$ plane, so if we have $e^{-(x^2 + y^2)}$ there then the issue would be that either $y$ is a dependent variable for $x$ or either $x$ is a dependent variable for $y$.
$endgroup$
– anonymous
Jan 6 '15 at 18:13
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
What you appear to be looking at is
the usual way to evaluate $I=int_{-infty}^{infty} f(x);dx$
where $f(x)=e^{-x^2}.$
We show that
$$begin{eqnarray}
I^2 &=& left(int_{-infty}^{infty} f(x);dxright)
left(int_{-infty}^{infty} f(y);dyright)\
&=& int_{-infty}^{infty}int_{-infty}^{infty} f(x) f(y);dx;dy \
&=& int_{-infty}^{infty}int_{-infty}^{infty} g(x,y);dx;dy.
end{eqnarray}$$
That is, instead of doing a single integration of the single-variable
function $f$, we do a double integration of the two-variable function $g$,
where $g(x,y)=e^{-(x^2+y^2)}.$
Then we change variables to polar coordinates in order to put
the integral in a form that's much easier to evaluate.
While we can make a simple and reasonably complete graph of $y = f(x)$
in the $x,y$-plane consisting of one or more simple curves
that pass the "vertical line" test
(at least for reasonable functions $f$ such as the functions one uses for most practical purposes),
we cannot graph a multivariable function $g(x,y)$ that way,
because $g(x,y)$ assigns a value to every point in its domain,
in this case every point in the plane.
There are various other ways of plotting $g(x,y)$, but they involve
different techniques such as three-dimensional visualization,
contour lines, or other techniques to indicate that the function does
not just have a single value for any given $x$, but rather has a value for
every combination of $x$ and $y$.
I am reminded of this question, which is different from yours
but also hinges on the confusion that can occur if one tries to apply
principles of graphing single-variable functions
to a problem concerning a multiple-variable function.
Both single-variable and multiple-variable functions can be related
to the same $x,y$-coordinate plane, but they relate to that plane
in very different ways.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 7 '15 at 20:47
David KDavid K
54.4k343120
$endgroup$
$begingroup$
I do not understand some things here.How does $f(x)f(y) = g(x,y)$ $x$ and $y$ are collinear points on the axis.
$endgroup$
– anonymous
Jan 8 '15 at 13:21
$begingroup$
In $int_{-infty}^{infty} f(x);dx$, the symbol $x$ is a name for the variable of integration. It is not a point, though sometimes it is helpful to relate values of $x$ to labeled points on the $x$-axis of a graph. The notation $g(x,y)$ in the equations is there merely to remind us that when we set up the double integral, the integrand is a function of two variables instead of just one.
$endgroup$
– David K
Jan 8 '15 at 17:27
$begingroup$
It is a new function right? you didnt transform the single variable functions into one, you created a new function, in that form? Also, the formal reason why we can change is because these are BOUND variables right? By definition we are allowed to change the variables: en.wikipedia.org/wiki/Free_variables_and_bound_variables
$endgroup$
– anonymous
Jan 8 '15 at 17:44
$begingroup$
Yes, $g$ is a new function. And yes, it is important that in these integrals, $x$ and $y$ are bound variables.
$endgroup$
– David K
Jan 8 '15 at 18:24
$begingroup$
Okay, so how does $x^2+ y^2 = r^2$ when converting it into polar?
$endgroup$
– anonymous
Jan 9 '15 at 12:36
|
show 1 more comment
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you appear to be looking at is
the usual way to evaluate $I=int_{-infty}^{infty} f(x);dx$
where $f(x)=e^{-x^2}.$
We show that
$$begin{eqnarray}
I^2 &=& left(int_{-infty}^{infty} f(x);dxright)
left(int_{-infty}^{infty} f(y);dyright)\
&=& int_{-infty}^{infty}int_{-infty}^{infty} f(x) f(y);dx;dy \
&=& int_{-infty}^{infty}int_{-infty}^{infty} g(x,y);dx;dy.
end{eqnarray}$$
That is, instead of doing a single integration of the single-variable
function $f$, we do a double integration of the two-variable function $g$,
where $g(x,y)=e^{-(x^2+y^2)}.$
Then we change variables to polar coordinates in order to put
the integral in a form that's much easier to evaluate.
While we can make a simple and reasonably complete graph of $y = f(x)$
in the $x,y$-plane consisting of one or more simple curves
that pass the "vertical line" test
(at least for reasonable functions $f$ such as the functions one uses for most practical purposes),
we cannot graph a multivariable function $g(x,y)$ that way,
because $g(x,y)$ assigns a value to every point in its domain,
in this case every point in the plane.
There are various other ways of plotting $g(x,y)$, but they involve
different techniques such as three-dimensional visualization,
contour lines, or other techniques to indicate that the function does
not just have a single value for any given $x$, but rather has a value for
every combination of $x$ and $y$.
I am reminded of this question, which is different from yours
but also hinges on the confusion that can occur if one tries to apply
principles of graphing single-variable functions
to a problem concerning a multiple-variable function.
Both single-variable and multiple-variable functions can be related
to the same $x,y$-coordinate plane, but they relate to that plane
in very different ways.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 7 '15 at 20:47
David KDavid K
54.4k343120
$endgroup$
$begingroup$
I do not understand some things here.How does $f(x)f(y) = g(x,y)$ $x$ and $y$ are collinear points on the axis.
$endgroup$
– anonymous
Jan 8 '15 at 13:21
$begingroup$
In $int_{-infty}^{infty} f(x);dx$, the symbol $x$ is a name for the variable of integration. It is not a point, though sometimes it is helpful to relate values of $x$ to labeled points on the $x$-axis of a graph. The notation $g(x,y)$ in the equations is there merely to remind us that when we set up the double integral, the integrand is a function of two variables instead of just one.
$endgroup$
– David K
Jan 8 '15 at 17:27
$begingroup$
It is a new function right? you didnt transform the single variable functions into one, you created a new function, in that form? Also, the formal reason why we can change is because these are BOUND variables right? By definition we are allowed to change the variables: en.wikipedia.org/wiki/Free_variables_and_bound_variables
$endgroup$
– anonymous
Jan 8 '15 at 17:44
$begingroup$
Yes, $g$ is a new function. And yes, it is important that in these integrals, $x$ and $y$ are bound variables.
$endgroup$
– David K
Jan 8 '15 at 18:24
$begingroup$
Okay, so how does $x^2+ y^2 = r^2$ when converting it into polar?
$endgroup$
– anonymous
Jan 9 '15 at 12:36
|
show 1 more comment
$begingroup$
What you appear to be looking at is
the usual way to evaluate $I=int_{-infty}^{infty} f(x);dx$
where $f(x)=e^{-x^2}.$
We show that
$$begin{eqnarray}
I^2 &=& left(int_{-infty}^{infty} f(x);dxright)
left(int_{-infty}^{infty} f(y);dyright)\
&=& int_{-infty}^{infty}int_{-infty}^{infty} f(x) f(y);dx;dy \
&=& int_{-infty}^{infty}int_{-infty}^{infty} g(x,y);dx;dy.
end{eqnarray}$$
That is, instead of doing a single integration of the single-variable
function $f$, we do a double integration of the two-variable function $g$,
where $g(x,y)=e^{-(x^2+y^2)}.$
Then we change variables to polar coordinates in order to put
the integral in a form that's much easier to evaluate.
While we can make a simple and reasonably complete graph of $y = f(x)$
in the $x,y$-plane consisting of one or more simple curves
that pass the "vertical line" test
(at least for reasonable functions $f$ such as the functions one uses for most practical purposes),
we cannot graph a multivariable function $g(x,y)$ that way,
because $g(x,y)$ assigns a value to every point in its domain,
in this case every point in the plane.
There are various other ways of plotting $g(x,y)$, but they involve
different techniques such as three-dimensional visualization,
contour lines, or other techniques to indicate that the function does
not just have a single value for any given $x$, but rather has a value for
every combination of $x$ and $y$.
I am reminded of this question, which is different from yours
but also hinges on the confusion that can occur if one tries to apply
principles of graphing single-variable functions
to a problem concerning a multiple-variable function.
Both single-variable and multiple-variable functions can be related
to the same $x,y$-coordinate plane, but they relate to that plane
in very different ways.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 7 '15 at 20:47
David KDavid K
54.4k343120
$endgroup$
$begingroup$
I do not understand some things here.How does $f(x)f(y) = g(x,y)$ $x$ and $y$ are collinear points on the axis.
$endgroup$
– anonymous
Jan 8 '15 at 13:21
$begingroup$
In $int_{-infty}^{infty} f(x);dx$, the symbol $x$ is a name for the variable of integration. It is not a point, though sometimes it is helpful to relate values of $x$ to labeled points on the $x$-axis of a graph. The notation $g(x,y)$ in the equations is there merely to remind us that when we set up the double integral, the integrand is a function of two variables instead of just one.
$endgroup$
– David K
Jan 8 '15 at 17:27
$begingroup$
It is a new function right? you didnt transform the single variable functions into one, you created a new function, in that form? Also, the formal reason why we can change is because these are BOUND variables right? By definition we are allowed to change the variables: en.wikipedia.org/wiki/Free_variables_and_bound_variables
$endgroup$
– anonymous
Jan 8 '15 at 17:44
$begingroup$
Yes, $g$ is a new function. And yes, it is important that in these integrals, $x$ and $y$ are bound variables.
$endgroup$
– David K
Jan 8 '15 at 18:24
$begingroup$
Okay, so how does $x^2+ y^2 = r^2$ when converting it into polar?
$endgroup$
– anonymous
Jan 9 '15 at 12:36
|
show 1 more comment
$begingroup$
What you appear to be looking at is
the usual way to evaluate $I=int_{-infty}^{infty} f(x);dx$
where $f(x)=e^{-x^2}.$
We show that
$$begin{eqnarray}
I^2 &=& left(int_{-infty}^{infty} f(x);dxright)
left(int_{-infty}^{infty} f(y);dyright)\
&=& int_{-infty}^{infty}int_{-infty}^{infty} f(x) f(y);dx;dy \
&=& int_{-infty}^{infty}int_{-infty}^{infty} g(x,y);dx;dy.
end{eqnarray}$$
That is, instead of doing a single integration of the single-variable
function $f$, we do a double integration of the two-variable function $g$,
where $g(x,y)=e^{-(x^2+y^2)}.$
Then we change variables to polar coordinates in order to put
the integral in a form that's much easier to evaluate.
While we can make a simple and reasonably complete graph of $y = f(x)$
in the $x,y$-plane consisting of one or more simple curves
that pass the "vertical line" test
(at least for reasonable functions $f$ such as the functions one uses for most practical purposes),
we cannot graph a multivariable function $g(x,y)$ that way,
because $g(x,y)$ assigns a value to every point in its domain,
in this case every point in the plane.
There are various other ways of plotting $g(x,y)$, but they involve
different techniques such as three-dimensional visualization,
contour lines, or other techniques to indicate that the function does
not just have a single value for any given $x$, but rather has a value for
every combination of $x$ and $y$.
I am reminded of this question, which is different from yours
but also hinges on the confusion that can occur if one tries to apply
principles of graphing single-variable functions
to a problem concerning a multiple-variable function.
Both single-variable and multiple-variable functions can be related
to the same $x,y$-coordinate plane, but they relate to that plane
in very different ways.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 7 '15 at 20:47
David KDavid K
54.4k343120
$endgroup$
What you appear to be looking at is
the usual way to evaluate $I=int_{-infty}^{infty} f(x);dx$
where $f(x)=e^{-x^2}.$
We show that
$$begin{eqnarray}
I^2 &=& left(int_{-infty}^{infty} f(x);dxright)
left(int_{-infty}^{infty} f(y);dyright)\
&=& int_{-infty}^{infty}int_{-infty}^{infty} f(x) f(y);dx;dy \
&=& int_{-infty}^{infty}int_{-infty}^{infty} g(x,y);dx;dy.
end{eqnarray}$$
That is, instead of doing a single integration of the single-variable
function $f$, we do a double integration of the two-variable function $g$,
where $g(x,y)=e^{-(x^2+y^2)}.$
Then we change variables to polar coordinates in order to put
the integral in a form that's much easier to evaluate.
While we can make a simple and reasonably complete graph of $y = f(x)$
in the $x,y$-plane consisting of one or more simple curves
that pass the "vertical line" test
(at least for reasonable functions $f$ such as the functions one uses for most practical purposes),
we cannot graph a multivariable function $g(x,y)$ that way,
because $g(x,y)$ assigns a value to every point in its domain,
in this case every point in the plane.
There are various other ways of plotting $g(x,y)$, but they involve
different techniques such as three-dimensional visualization,
contour lines, or other techniques to indicate that the function does
not just have a single value for any given $x$, but rather has a value for
every combination of $x$ and $y$.
I am reminded of this question, which is different from yours
but also hinges on the confusion that can occur if one tries to apply
principles of graphing single-variable functions
to a problem concerning a multiple-variable function.
Both single-variable and multiple-variable functions can be related
to the same $x,y$-coordinate plane, but they relate to that plane
in very different ways.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 7 '15 at 20:47
David KDavid K
54.4k343120
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jan 7 '15 at 20:47
David KDavid K
54.4k343120
answered Jan 7 '15 at 20:47
David KDavid K
54.4k343120
answered Jan 7 '15 at 20:47
David KDavid K
54.4k343120
54.4k343120
$begingroup$
I do not understand some things here.How does $f(x)f(y) = g(x,y)$ $x$ and $y$ are collinear points on the axis.
$endgroup$
– anonymous
Jan 8 '15 at 13:21
$begingroup$
In $int_{-infty}^{infty} f(x);dx$, the symbol $x$ is a name for the variable of integration. It is not a point, though sometimes it is helpful to relate values of $x$ to labeled points on the $x$-axis of a graph. The notation $g(x,y)$ in the equations is there merely to remind us that when we set up the double integral, the integrand is a function of two variables instead of just one.
$endgroup$
– David K
Jan 8 '15 at 17:27
$begingroup$
It is a new function right? you didnt transform the single variable functions into one, you created a new function, in that form? Also, the formal reason why we can change is because these are BOUND variables right? By definition we are allowed to change the variables: en.wikipedia.org/wiki/Free_variables_and_bound_variables
$endgroup$
– anonymous
Jan 8 '15 at 17:44
$begingroup$
Yes, $g$ is a new function. And yes, it is important that in these integrals, $x$ and $y$ are bound variables.
$endgroup$
– David K
Jan 8 '15 at 18:24
$begingroup$
Okay, so how does $x^2+ y^2 = r^2$ when converting it into polar?
$endgroup$
– anonymous
Jan 9 '15 at 12:36
|
show 1 more comment
$begingroup$
I do not understand some things here.How does $f(x)f(y) = g(x,y)$ $x$ and $y$ are collinear points on the axis.
$endgroup$
– anonymous
Jan 8 '15 at 13:21
$begingroup$
In $int_{-infty}^{infty} f(x);dx$, the symbol $x$ is a name for the variable of integration. It is not a point, though sometimes it is helpful to relate values of $x$ to labeled points on the $x$-axis of a graph. The notation $g(x,y)$ in the equations is there merely to remind us that when we set up the double integral, the integrand is a function of two variables instead of just one.
$endgroup$
– David K
Jan 8 '15 at 17:27
$begingroup$
It is a new function right? you didnt transform the single variable functions into one, you created a new function, in that form? Also, the formal reason why we can change is because these are BOUND variables right? By definition we are allowed to change the variables: en.wikipedia.org/wiki/Free_variables_and_bound_variables
$endgroup$
– anonymous
Jan 8 '15 at 17:44
$begingroup$
Yes, $g$ is a new function. And yes, it is important that in these integrals, $x$ and $y$ are bound variables.
$endgroup$
– David K
Jan 8 '15 at 18:24
$begingroup$
Okay, so how does $x^2+ y^2 = r^2$ when converting it into polar?
$endgroup$
– anonymous
Jan 9 '15 at 12:36
$begingroup$
I do not understand some things here.How does $f(x)f(y) = g(x,y)$ $x$ and $y$ are collinear points on the axis.
$endgroup$
– anonymous
Jan 8 '15 at 13:21
$begingroup$
I do not understand some things here.How does $f(x)f(y) = g(x,y)$ $x$ and $y$ are collinear points on the axis.
$endgroup$
– anonymous
Jan 8 '15 at 13:21
$begingroup$
In $int_{-infty}^{infty} f(x);dx$, the symbol $x$ is a name for the variable of integration. It is not a point, though sometimes it is helpful to relate values of $x$ to labeled points on the $x$-axis of a graph. The notation $g(x,y)$ in the equations is there merely to remind us that when we set up the double integral, the integrand is a function of two variables instead of just one.
$endgroup$
– David K
Jan 8 '15 at 17:27
$begingroup$
In $int_{-infty}^{infty} f(x);dx$, the symbol $x$ is a name for the variable of integration. It is not a point, though sometimes it is helpful to relate values of $x$ to labeled points on the $x$-axis of a graph. The notation $g(x,y)$ in the equations is there merely to remind us that when we set up the double integral, the integrand is a function of two variables instead of just one.
$endgroup$
– David K
Jan 8 '15 at 17:27
$begingroup$
It is a new function right? you didnt transform the single variable functions into one, you created a new function, in that form? Also, the formal reason why we can change is because these are BOUND variables right? By definition we are allowed to change the variables: en.wikipedia.org/wiki/Free_variables_and_bound_variables
$endgroup$
– anonymous
Jan 8 '15 at 17:44
$begingroup$
It is a new function right? you didnt transform the single variable functions into one, you created a new function, in that form? Also, the formal reason why we can change is because these are BOUND variables right? By definition we are allowed to change the variables: en.wikipedia.org/wiki/Free_variables_and_bound_variables
$endgroup$
– anonymous
Jan 8 '15 at 17:44
$begingroup$
Yes, $g$ is a new function. And yes, it is important that in these integrals, $x$ and $y$ are bound variables.
$endgroup$
– David K
Jan 8 '15 at 18:24
$begingroup$
Yes, $g$ is a new function. And yes, it is important that in these integrals, $x$ and $y$ are bound variables.
$endgroup$
– David K
Jan 8 '15 at 18:24
$begingroup$
Okay, so how does $x^2+ y^2 = r^2$ when converting it into polar?
$endgroup$
– anonymous
Jan 9 '15 at 12:36
$begingroup$
Okay, so how does $x^2+ y^2 = r^2$ when converting it into polar?
$endgroup$
– anonymous
Jan 9 '15 at 12:36
|
show 1 more comment
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$begingroup$
The vertical axis is the set of points $(x,y) in mathbb{R}^2$ such that $x=0$. The set of points such that $y=f(x)$ represents a curve in the $xy$ plane. Would it make you feel better if we wrote $$I = int_{- infty}^{+ infty} e^{-t^2} , dt$$ and $$I = int_{- infty}^{+ infty} e^{- w^2} , dw?$$ We use $x$ and $y$ because of inertia in switching to polar coordinates. In the example I gave we can write $t = r cos(theta), w = r sin(theta)$, but for some reason some people aren't comfortable with that and think only $x$ and $y$ can be written in polar coordinates.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:01
$begingroup$
But the problem is, when you combine it, when it becomes an iterated integral, then do you have a $3D$ plot of the curve? It becomes $e^{-(t^2 + w^2)}$ but you cant have a $3D$ plot because the nor $w$ nor $t$ are the vertical axis in the 2D graph.
$endgroup$
– anonymous
Jan 6 '15 at 18:04
$begingroup$
It's no longer a curve, it becomes a surface.
$endgroup$
– Mark Fantini
Jan 6 '15 at 18:07
$begingroup$
Exactly, how? Given: en.wikipedia.org/wiki/… Then one of the axes must be $y$ for example, and one of the axes must be $x$ for it to be a 3D system, and a surface.?
$endgroup$
– anonymous
Jan 6 '15 at 18:08
$begingroup$
@MarkFantini, do you see? In the 3D plane there is a seperate $xy$ plane, so if we have $e^{-(x^2 + y^2)}$ there then the issue would be that either $y$ is a dependent variable for $x$ or either $x$ is a dependent variable for $y$.
$endgroup$
– anonymous
Jan 6 '15 at 18:13