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Maximal open set for implicit function theorem
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We consider the equation $cos (xy) + sin(xy) = y$ with $(x,y)inmathbb{R}^2$.
From the implicit function theorem, one has the existence of an open neighborhood of $0$ on which there exists a unique solution $y : xmapsto y(x)$ to the equation.
I would like to know how to determine a maximal open set on which the map $y$ is defined. The hint proposes the use a drawing. But I do not know how to draw the implicit plot of the equation.
analysis graphing-functions implicit-function-theorem
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hecho
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up vote
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We consider the equation $cos (xy) + sin(xy) = y$ with $(x,y)inmathbb{R}^2$.
From the implicit function theorem, one has the existence of an open neighborhood of $0$ on which there exists a unique solution $y : xmapsto y(x)$ to the equation.
I would like to know how to determine a maximal open set on which the map $y$ is defined. The hint proposes the use a drawing. But I do not know how to draw the implicit plot of the equation.
analysis graphing-functions implicit-function-theorem
asked yesterday
hecho
112
New contributor
hecho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We consider the equation $cos (xy) + sin(xy) = y$ with $(x,y)inmathbb{R}^2$.
From the implicit function theorem, one has the existence of an open neighborhood of $0$ on which there exists a unique solution $y : xmapsto y(x)$ to the equation.
I would like to know how to determine a maximal open set on which the map $y$ is defined. The hint proposes the use a drawing. But I do not know how to draw the implicit plot of the equation.
analysis graphing-functions implicit-function-theorem
asked yesterday
hecho
112
New contributor
hecho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We consider the equation $cos (xy) + sin(xy) = y$ with $(x,y)inmathbb{R}^2$.
From the implicit function theorem, one has the existence of an open neighborhood of $0$ on which there exists a unique solution $y : xmapsto y(x)$ to the equation.
I would like to know how to determine a maximal open set on which the map $y$ is defined. The hint proposes the use a drawing. But I do not know how to draw the implicit plot of the equation.
analysis graphing-functions implicit-function-theorem
analysis graphing-functions implicit-function-theorem
asked yesterday
hecho
112
New contributor
hecho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
hecho
112
New contributor
hecho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
hecho
112
New contributor
hecho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
hecho
112
asked yesterday
hecho
112
112
New contributor
hecho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
hecho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
hecho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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1
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We are told to study the equation
$$f(x,y):=cos(xy)+sin(xy)-y=0tag{1}$$
in the neighborhood of an initial point $(0,y)$ satisfying the equation.
Since $f(0,y)=0$ implies $y=1$ there is exactly one such point, namely ${bf z}_0:=(0,1)$. The point ${bf z}_0$ is marked in the following contour plot of $f$.
One computes
$$f_x=ybigl(-sin(xy)+cos(xy)bigr),quad f_y=xbigl(-sin(xy)+cos(xy)bigr)-1 .$$
As $f_y({bf z}_0)=-1ne0$ the implicit function theorem allows to conclude the following: There are a window $W=[-a,a]times[1-b,1+b]$ centered at ${bf z}_0$ and a $C^1$-function
$$phi:quad [-a,a]to[1-b,1+b],qquad xmapsto y=phi(x) ,$$
such that within $W$ the equation $(1)$ is equivalent with $y=phi(x)$, in particular $phi(0)=1$. Furthermore one has
$$phi'(0)=-{f_x({bf z}_0)over f_y({bf z}_0)}=1 .$$
The graph of this $phi$ is shown in red in the above figure.
Now this implicit function statement does not tell us anything about the size of this window $W$, nor about what's happening with $phi$ outside of $W$. In order to gain global information about $phi$ we have to invoke other tools. From the figure we conjecture that $phi$ is actually defined for $-infty<x<infty$. But note that the full level set $N_0:=f^{-1}(0)$ has a very complicated structure; see the following figure, where $N_0$ is the boundary between the blue and the white parts of the figure. One obtains the impression that the red curve of the first figure is special, in so far as it is a graph over all of ${mathbb R}$.
In order to get further hold on this $phi$ we write the equation $(1)$ (with the help of a well known trigonometric formula) in the form
$$sqrt{2}sinleft(xy+{piover4}right)-y=0 .tag{2}$$
We can solve $(2)$ "formally" explicitly for $x$ and obtain something like
$$x=phi^{-1}(y)={1over y}left(arcsin{yoversqrt{2}}-{piover4}right) .$$
This has to be taken with a grain of salt. In fact you should obtain two such expressions, one of them giving the part of the red graph to the left of the maximum at $bigl({piover4sqrt{2}},sqrt{2}bigr)$, and the other expression the part of the red graph to the right of this point.
I stop here.
answered yesterday
Christian Blatter
170k7111324
Thanks for you answer. How did you get equation $(1)$? Also, how I am suppose to draw $N_0$ without a computer?
– hecho
11 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We are told to study the equation
$$f(x,y):=cos(xy)+sin(xy)-y=0tag{1}$$
in the neighborhood of an initial point $(0,y)$ satisfying the equation.
Since $f(0,y)=0$ implies $y=1$ there is exactly one such point, namely ${bf z}_0:=(0,1)$. The point ${bf z}_0$ is marked in the following contour plot of $f$.
One computes
$$f_x=ybigl(-sin(xy)+cos(xy)bigr),quad f_y=xbigl(-sin(xy)+cos(xy)bigr)-1 .$$
As $f_y({bf z}_0)=-1ne0$ the implicit function theorem allows to conclude the following: There are a window $W=[-a,a]times[1-b,1+b]$ centered at ${bf z}_0$ and a $C^1$-function
$$phi:quad [-a,a]to[1-b,1+b],qquad xmapsto y=phi(x) ,$$
such that within $W$ the equation $(1)$ is equivalent with $y=phi(x)$, in particular $phi(0)=1$. Furthermore one has
$$phi'(0)=-{f_x({bf z}_0)over f_y({bf z}_0)}=1 .$$
The graph of this $phi$ is shown in red in the above figure.
Now this implicit function statement does not tell us anything about the size of this window $W$, nor about what's happening with $phi$ outside of $W$. In order to gain global information about $phi$ we have to invoke other tools. From the figure we conjecture that $phi$ is actually defined for $-infty<x<infty$. But note that the full level set $N_0:=f^{-1}(0)$ has a very complicated structure; see the following figure, where $N_0$ is the boundary between the blue and the white parts of the figure. One obtains the impression that the red curve of the first figure is special, in so far as it is a graph over all of ${mathbb R}$.
In order to get further hold on this $phi$ we write the equation $(1)$ (with the help of a well known trigonometric formula) in the form
$$sqrt{2}sinleft(xy+{piover4}right)-y=0 .tag{2}$$
We can solve $(2)$ "formally" explicitly for $x$ and obtain something like
$$x=phi^{-1}(y)={1over y}left(arcsin{yoversqrt{2}}-{piover4}right) .$$
This has to be taken with a grain of salt. In fact you should obtain two such expressions, one of them giving the part of the red graph to the left of the maximum at $bigl({piover4sqrt{2}},sqrt{2}bigr)$, and the other expression the part of the red graph to the right of this point.
I stop here.
answered yesterday
Christian Blatter
170k7111324
Thanks for you answer. How did you get equation $(1)$? Also, how I am suppose to draw $N_0$ without a computer?
– hecho
11 hours ago
add a comment |
up vote
1
down vote
We are told to study the equation
$$f(x,y):=cos(xy)+sin(xy)-y=0tag{1}$$
in the neighborhood of an initial point $(0,y)$ satisfying the equation.
Since $f(0,y)=0$ implies $y=1$ there is exactly one such point, namely ${bf z}_0:=(0,1)$. The point ${bf z}_0$ is marked in the following contour plot of $f$.
One computes
$$f_x=ybigl(-sin(xy)+cos(xy)bigr),quad f_y=xbigl(-sin(xy)+cos(xy)bigr)-1 .$$
As $f_y({bf z}_0)=-1ne0$ the implicit function theorem allows to conclude the following: There are a window $W=[-a,a]times[1-b,1+b]$ centered at ${bf z}_0$ and a $C^1$-function
$$phi:quad [-a,a]to[1-b,1+b],qquad xmapsto y=phi(x) ,$$
such that within $W$ the equation $(1)$ is equivalent with $y=phi(x)$, in particular $phi(0)=1$. Furthermore one has
$$phi'(0)=-{f_x({bf z}_0)over f_y({bf z}_0)}=1 .$$
The graph of this $phi$ is shown in red in the above figure.
Now this implicit function statement does not tell us anything about the size of this window $W$, nor about what's happening with $phi$ outside of $W$. In order to gain global information about $phi$ we have to invoke other tools. From the figure we conjecture that $phi$ is actually defined for $-infty<x<infty$. But note that the full level set $N_0:=f^{-1}(0)$ has a very complicated structure; see the following figure, where $N_0$ is the boundary between the blue and the white parts of the figure. One obtains the impression that the red curve of the first figure is special, in so far as it is a graph over all of ${mathbb R}$.
In order to get further hold on this $phi$ we write the equation $(1)$ (with the help of a well known trigonometric formula) in the form
$$sqrt{2}sinleft(xy+{piover4}right)-y=0 .tag{2}$$
We can solve $(2)$ "formally" explicitly for $x$ and obtain something like
$$x=phi^{-1}(y)={1over y}left(arcsin{yoversqrt{2}}-{piover4}right) .$$
This has to be taken with a grain of salt. In fact you should obtain two such expressions, one of them giving the part of the red graph to the left of the maximum at $bigl({piover4sqrt{2}},sqrt{2}bigr)$, and the other expression the part of the red graph to the right of this point.
I stop here.
answered yesterday
Christian Blatter
170k7111324
Thanks for you answer. How did you get equation $(1)$? Also, how I am suppose to draw $N_0$ without a computer?
– hecho
11 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
We are told to study the equation
$$f(x,y):=cos(xy)+sin(xy)-y=0tag{1}$$
in the neighborhood of an initial point $(0,y)$ satisfying the equation.
Since $f(0,y)=0$ implies $y=1$ there is exactly one such point, namely ${bf z}_0:=(0,1)$. The point ${bf z}_0$ is marked in the following contour plot of $f$.
One computes
$$f_x=ybigl(-sin(xy)+cos(xy)bigr),quad f_y=xbigl(-sin(xy)+cos(xy)bigr)-1 .$$
As $f_y({bf z}_0)=-1ne0$ the implicit function theorem allows to conclude the following: There are a window $W=[-a,a]times[1-b,1+b]$ centered at ${bf z}_0$ and a $C^1$-function
$$phi:quad [-a,a]to[1-b,1+b],qquad xmapsto y=phi(x) ,$$
such that within $W$ the equation $(1)$ is equivalent with $y=phi(x)$, in particular $phi(0)=1$. Furthermore one has
$$phi'(0)=-{f_x({bf z}_0)over f_y({bf z}_0)}=1 .$$
The graph of this $phi$ is shown in red in the above figure.
Now this implicit function statement does not tell us anything about the size of this window $W$, nor about what's happening with $phi$ outside of $W$. In order to gain global information about $phi$ we have to invoke other tools. From the figure we conjecture that $phi$ is actually defined for $-infty<x<infty$. But note that the full level set $N_0:=f^{-1}(0)$ has a very complicated structure; see the following figure, where $N_0$ is the boundary between the blue and the white parts of the figure. One obtains the impression that the red curve of the first figure is special, in so far as it is a graph over all of ${mathbb R}$.
In order to get further hold on this $phi$ we write the equation $(1)$ (with the help of a well known trigonometric formula) in the form
$$sqrt{2}sinleft(xy+{piover4}right)-y=0 .tag{2}$$
We can solve $(2)$ "formally" explicitly for $x$ and obtain something like
$$x=phi^{-1}(y)={1over y}left(arcsin{yoversqrt{2}}-{piover4}right) .$$
This has to be taken with a grain of salt. In fact you should obtain two such expressions, one of them giving the part of the red graph to the left of the maximum at $bigl({piover4sqrt{2}},sqrt{2}bigr)$, and the other expression the part of the red graph to the right of this point.
I stop here.
answered yesterday
Christian Blatter
170k7111324
We are told to study the equation
$$f(x,y):=cos(xy)+sin(xy)-y=0tag{1}$$
in the neighborhood of an initial point $(0,y)$ satisfying the equation.
Since $f(0,y)=0$ implies $y=1$ there is exactly one such point, namely ${bf z}_0:=(0,1)$. The point ${bf z}_0$ is marked in the following contour plot of $f$.
One computes
$$f_x=ybigl(-sin(xy)+cos(xy)bigr),quad f_y=xbigl(-sin(xy)+cos(xy)bigr)-1 .$$
As $f_y({bf z}_0)=-1ne0$ the implicit function theorem allows to conclude the following: There are a window $W=[-a,a]times[1-b,1+b]$ centered at ${bf z}_0$ and a $C^1$-function
$$phi:quad [-a,a]to[1-b,1+b],qquad xmapsto y=phi(x) ,$$
such that within $W$ the equation $(1)$ is equivalent with $y=phi(x)$, in particular $phi(0)=1$. Furthermore one has
$$phi'(0)=-{f_x({bf z}_0)over f_y({bf z}_0)}=1 .$$
The graph of this $phi$ is shown in red in the above figure.
Now this implicit function statement does not tell us anything about the size of this window $W$, nor about what's happening with $phi$ outside of $W$. In order to gain global information about $phi$ we have to invoke other tools. From the figure we conjecture that $phi$ is actually defined for $-infty<x<infty$. But note that the full level set $N_0:=f^{-1}(0)$ has a very complicated structure; see the following figure, where $N_0$ is the boundary between the blue and the white parts of the figure. One obtains the impression that the red curve of the first figure is special, in so far as it is a graph over all of ${mathbb R}$.
In order to get further hold on this $phi$ we write the equation $(1)$ (with the help of a well known trigonometric formula) in the form
$$sqrt{2}sinleft(xy+{piover4}right)-y=0 .tag{2}$$
We can solve $(2)$ "formally" explicitly for $x$ and obtain something like
$$x=phi^{-1}(y)={1over y}left(arcsin{yoversqrt{2}}-{piover4}right) .$$
This has to be taken with a grain of salt. In fact you should obtain two such expressions, one of them giving the part of the red graph to the left of the maximum at $bigl({piover4sqrt{2}},sqrt{2}bigr)$, and the other expression the part of the red graph to the right of this point.
I stop here.
answered yesterday
Christian Blatter
170k7111324
edited 6 hours ago
answered yesterday
Christian Blatter
170k7111324
answered yesterday
Christian Blatter
170k7111324
answered yesterday
Christian Blatter
170k7111324
170k7111324
Thanks for you answer. How did you get equation $(1)$? Also, how I am suppose to draw $N_0$ without a computer?
– hecho
11 hours ago
add a comment |
Thanks for you answer. How did you get equation $(1)$? Also, how I am suppose to draw $N_0$ without a computer?
– hecho
11 hours ago
Thanks for you answer. How did you get equation $(1)$? Also, how I am suppose to draw $N_0$ without a computer?
– hecho
11 hours ago
Thanks for you answer. How did you get equation $(1)$? Also, how I am suppose to draw $N_0$ without a computer?
– hecho
11 hours ago
add a comment |
hecho is a new contributor. Be nice, and check out our Code of Conduct.
hecho is a new contributor. Be nice, and check out our Code of Conduct.
hecho is a new contributor. Be nice, and check out our Code of Conduct.
hecho is a new contributor. Be nice, and check out our Code of Conduct.
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