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Why is $sin(x+y)=c$ not a smooth curve for $c = pm 1$?
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I understand that, for $c= 1$ and $c=-1$, the derivative of the function $f(x,y)=sin(x+y)-c$ is a zero vector, and hence the locus which is given by $sin(x+y)-c=0$ fails to satisfy the definition of the smooth curve.
However, I have the following doubt:
Suppose $c = 1$.
Then,
$$ sin(x+y) = 1 implies y = -x + arcsin(1)$$
So, $$y = -x + (4n+1)pi/2, quadforall ninmathbbZ$$
satisfies the equation. These are a set of parallel lines.
But, why is it failing to satisfy the definition of a smooth curve?
I tried to plot the function for $c=1$ in WolframAlpha, but the plot did not contain any points. Most likely, I am doing some silly mistake in the above calculations. Can anyone help in identifying the flaw in my thinking?
algebra-precalculus multivariable-calculus trigonometry
asked Jul 25 at 4:32
Suhan Shetty
835
 |Â
show 2 more comments
up vote
2
down vote
favorite
I understand that, for $c= 1$ and $c=-1$, the derivative of the function $f(x,y)=sin(x+y)-c$ is a zero vector, and hence the locus which is given by $sin(x+y)-c=0$ fails to satisfy the definition of the smooth curve.
However, I have the following doubt:
Suppose $c = 1$.
Then,
$$ sin(x+y) = 1 implies y = -x + arcsin(1)$$
So, $$y = -x + (4n+1)pi/2, quadforall ninmathbbZ$$
satisfies the equation. These are a set of parallel lines.
But, why is it failing to satisfy the definition of a smooth curve?
I tried to plot the function for $c=1$ in WolframAlpha, but the plot did not contain any points. Most likely, I am doing some silly mistake in the above calculations. Can anyone help in identifying the flaw in my thinking?
algebra-precalculus multivariable-calculus trigonometry
asked Jul 25 at 4:32
Suhan Shetty
835
Derivative at where and of what kind?
– Parcly Taxel
Jul 25 at 4:36
1
A curve, which is a vector function $mathbfr(t)$ of a parameter $t$, is considered not smooth if $dmathbfr/dt$ is ever the zero vector $bf 0$. However, that doesn't mean the gradient $nabla f$ of a scalar function $f$ being the zero vector implies the level curve of $f$ is not smooth.
– anon
Jul 25 at 5:07
Yeah, as anon pointed out in a comment to my failed attempt at this (need more coffee, sorry), the problem seems to be that the usual two families of parallel lines coincide for these values of $c$. This makes IFT fail to diagnose the smoothness because all the solution points are, in a sense, double zeros.
– Jyrki Lahtonen
Jul 25 at 5:10
I do not see why $c = pm1$ plays a special role. The derivative $Df$ of $f$ does not depend on $c$. Moreover, the set $N_c = u in mathbbR mid sin(u) = c $ always consists of countably many isolated points $u_n$ so that the solution set decomposes into the parallel lines $y = -x + u_n$.
– Paul Frost
Jul 25 at 7:57
@PaulFrost Let $ f = sin(x+y) - c$ . Then, $Df = [cos(x+y), cos(x+y)]$. So, $Df$ will be a zero vector if $cos(x+y)=0$. But, $x+y = arcsin(c)$. So, $Df$ is a zero vector at $c=pm1$.
– Suhan Shetty
Jul 26 at 4:16
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I understand that, for $c= 1$ and $c=-1$, the derivative of the function $f(x,y)=sin(x+y)-c$ is a zero vector, and hence the locus which is given by $sin(x+y)-c=0$ fails to satisfy the definition of the smooth curve.
However, I have the following doubt:
Suppose $c = 1$.
Then,
$$ sin(x+y) = 1 implies y = -x + arcsin(1)$$
So, $$y = -x + (4n+1)pi/2, quadforall ninmathbbZ$$
satisfies the equation. These are a set of parallel lines.
But, why is it failing to satisfy the definition of a smooth curve?
I tried to plot the function for $c=1$ in WolframAlpha, but the plot did not contain any points. Most likely, I am doing some silly mistake in the above calculations. Can anyone help in identifying the flaw in my thinking?
algebra-precalculus multivariable-calculus trigonometry
asked Jul 25 at 4:32
Suhan Shetty
835
I understand that, for $c= 1$ and $c=-1$, the derivative of the function $f(x,y)=sin(x+y)-c$ is a zero vector, and hence the locus which is given by $sin(x+y)-c=0$ fails to satisfy the definition of the smooth curve.
However, I have the following doubt:
Suppose $c = 1$.
Then,
$$ sin(x+y) = 1 implies y = -x + arcsin(1)$$
So, $$y = -x + (4n+1)pi/2, quadforall ninmathbbZ$$
satisfies the equation. These are a set of parallel lines.
But, why is it failing to satisfy the definition of a smooth curve?
I tried to plot the function for $c=1$ in WolframAlpha, but the plot did not contain any points. Most likely, I am doing some silly mistake in the above calculations. Can anyone help in identifying the flaw in my thinking?
algebra-precalculus multivariable-calculus trigonometry
asked Jul 25 at 4:32
Suhan Shetty
835
edited Jul 26 at 16:01
asked Jul 25 at 4:32
Suhan Shetty
835
asked Jul 25 at 4:32
Suhan Shetty
835
asked Jul 25 at 4:32
Suhan Shetty
835
835
Derivative at where and of what kind?
– Parcly Taxel
Jul 25 at 4:36
1
A curve, which is a vector function $mathbfr(t)$ of a parameter $t$, is considered not smooth if $dmathbfr/dt$ is ever the zero vector $bf 0$. However, that doesn't mean the gradient $nabla f$ of a scalar function $f$ being the zero vector implies the level curve of $f$ is not smooth.
– anon
Jul 25 at 5:07
Yeah, as anon pointed out in a comment to my failed attempt at this (need more coffee, sorry), the problem seems to be that the usual two families of parallel lines coincide for these values of $c$. This makes IFT fail to diagnose the smoothness because all the solution points are, in a sense, double zeros.
– Jyrki Lahtonen
Jul 25 at 5:10
I do not see why $c = pm1$ plays a special role. The derivative $Df$ of $f$ does not depend on $c$. Moreover, the set $N_c = u in mathbbR mid sin(u) = c $ always consists of countably many isolated points $u_n$ so that the solution set decomposes into the parallel lines $y = -x + u_n$.
– Paul Frost
Jul 25 at 7:57
@PaulFrost Let $ f = sin(x+y) - c$ . Then, $Df = [cos(x+y), cos(x+y)]$. So, $Df$ will be a zero vector if $cos(x+y)=0$. But, $x+y = arcsin(c)$. So, $Df$ is a zero vector at $c=pm1$.
– Suhan Shetty
Jul 26 at 4:16
 |Â
show 2 more comments
Derivative at where and of what kind?
– Parcly Taxel
Jul 25 at 4:36
1
A curve, which is a vector function $mathbfr(t)$ of a parameter $t$, is considered not smooth if $dmathbfr/dt$ is ever the zero vector $bf 0$. However, that doesn't mean the gradient $nabla f$ of a scalar function $f$ being the zero vector implies the level curve of $f$ is not smooth.
– anon
Jul 25 at 5:07
Yeah, as anon pointed out in a comment to my failed attempt at this (need more coffee, sorry), the problem seems to be that the usual two families of parallel lines coincide for these values of $c$. This makes IFT fail to diagnose the smoothness because all the solution points are, in a sense, double zeros.
– Jyrki Lahtonen
Jul 25 at 5:10
I do not see why $c = pm1$ plays a special role. The derivative $Df$ of $f$ does not depend on $c$. Moreover, the set $N_c = u in mathbbR mid sin(u) = c $ always consists of countably many isolated points $u_n$ so that the solution set decomposes into the parallel lines $y = -x + u_n$.
– Paul Frost
Jul 25 at 7:57
@PaulFrost Let $ f = sin(x+y) - c$ . Then, $Df = [cos(x+y), cos(x+y)]$. So, $Df$ will be a zero vector if $cos(x+y)=0$. But, $x+y = arcsin(c)$. So, $Df$ is a zero vector at $c=pm1$.
– Suhan Shetty
Jul 26 at 4:16
Derivative at where and of what kind?
– Parcly Taxel
Jul 25 at 4:36
Derivative at where and of what kind?
– Parcly Taxel
Jul 25 at 4:36
1
1
A curve, which is a vector function $mathbfr(t)$ of a parameter $t$, is considered not smooth if $dmathbfr/dt$ is ever the zero vector $bf 0$. However, that doesn't mean the gradient $nabla f$ of a scalar function $f$ being the zero vector implies the level curve of $f$ is not smooth.
– anon
Jul 25 at 5:07
A curve, which is a vector function $mathbfr(t)$ of a parameter $t$, is considered not smooth if $dmathbfr/dt$ is ever the zero vector $bf 0$. However, that doesn't mean the gradient $nabla f$ of a scalar function $f$ being the zero vector implies the level curve of $f$ is not smooth.
– anon
Jul 25 at 5:07
Yeah, as anon pointed out in a comment to my failed attempt at this (need more coffee, sorry), the problem seems to be that the usual two families of parallel lines coincide for these values of $c$. This makes IFT fail to diagnose the smoothness because all the solution points are, in a sense, double zeros.
– Jyrki Lahtonen
Jul 25 at 5:10
Yeah, as anon pointed out in a comment to my failed attempt at this (need more coffee, sorry), the problem seems to be that the usual two families of parallel lines coincide for these values of $c$. This makes IFT fail to diagnose the smoothness because all the solution points are, in a sense, double zeros.
– Jyrki Lahtonen
Jul 25 at 5:10
I do not see why $c = pm1$ plays a special role. The derivative $Df$ of $f$ does not depend on $c$. Moreover, the set $N_c = u in mathbbR mid sin(u) = c $ always consists of countably many isolated points $u_n$ so that the solution set decomposes into the parallel lines $y = -x + u_n$.
– Paul Frost
Jul 25 at 7:57
I do not see why $c = pm1$ plays a special role. The derivative $Df$ of $f$ does not depend on $c$. Moreover, the set $N_c = u in mathbbR mid sin(u) = c $ always consists of countably many isolated points $u_n$ so that the solution set decomposes into the parallel lines $y = -x + u_n$.
– Paul Frost
Jul 25 at 7:57
@PaulFrost Let $ f = sin(x+y) - c$ . Then, $Df = [cos(x+y), cos(x+y)]$. So, $Df$ will be a zero vector if $cos(x+y)=0$. But, $x+y = arcsin(c)$. So, $Df$ is a zero vector at $c=pm1$.
– Suhan Shetty
Jul 26 at 4:16
@PaulFrost Let $ f = sin(x+y) - c$ . Then, $Df = [cos(x+y), cos(x+y)]$. So, $Df$ will be a zero vector if $cos(x+y)=0$. But, $x+y = arcsin(c)$. So, $Df$ is a zero vector at $c=pm1$.
– Suhan Shetty
Jul 26 at 4:16
 |Â
show 2 more comments
2 Answers
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up vote
2
down vote
accepted
Presumably you started out by using the implicit function theorem to the zero locus of the function.
$$
F(x,y)=sin(x+y)-c=0.qquad(*)
$$
Assume that $P=(x_0,y_0)$ satisfies $(*)$. IFT promises $y$ as a smooth function of $x$ in a neighborhood of $P$, if $F_y(x_0,y_0)neq0$.
When $c=1$ and, for example $(x_0,y_0)=(pi/2,0)$, we see that the partial
derivative
$$
F_y(x,y)=cos(x+y)
$$
vanishes at $(x_0,y_0)$ because$cos(pi/2)=0$.
What goes wrong even though you have those lines?
The solutions of $sin (x+y)=c$ are $$x+y=arcsin c+ncdot2pi$$ and
$$x+y=pi-arcsin c+ncdot 2pi.$$ All because $sin(pi-alpha)=sinalpha$. Supplementary angles and all that.
[Comment: originally I fumbled this and misguidedly thought that the other family of lines share the slope +1, thanks to @anon for setting this straight.]
So we have two families of parallel lines. Within a family the vertical separation of any two lines is an integer multiple of $2pi$. Furthermore, you get one family from the other with a vertical (or horizontal) translation by $pi-2arcsin c$. So we see that the two families actually coincide when $arcsin c=pmpi/2$, i.e. when $c=pm1$.
IFT diagnoses it when the lines describing the set of solutions "double on themselves".
In other words, the flow of the logic in IFT is one-way-traffic only:
- A non-vanishing Jacobian promises that locally within the set of
solutions the prescribed subset of coordinates is a smooth function
of the remaining coordinates. - But, the set of solutions can be the graph of a smooth function even when the Jacobian vanishes.
My favorite example of this is the smooth curve of intersection formed by a donut lying flat on the table. The points of contact of the two surfaces form a perfectly smooth circle. But, because the surface of the table is the shared plane of tangency of the donut and the table, the IFT breaks down there as the Jacobian determinant vanishes at all points of intersection.
In general the solutions to $sin(x+y)=c$ are $$ beginarrayll x+y & =arcsin(c)+2pi n \ x+y & =[pi-arcsin(c)]+2pi n endarray$$ which are all descending lines. When $c=1$, the values of $arcsin(1)$ and $pi-arcsin(1)$ are the same though. And there's no $x$ inside of an $arcsin$.
– anon
Jul 25 at 5:03
Oops. Thanks, @anon. Need to rethink...
– Jyrki Lahtonen
Jul 25 at 5:05
Thanks, @JyrkiLahtonen. I did not think of the other family of solutions to $sin(x+y) =c$, i.e.,$x+y =pi-arcsin(c)+2pi n$.
– Suhan Shetty
Jul 26 at 4:59
Is there any stronger theorem than IFT which would detect this locus (i.e. $sin(x+y) =pm 1$) as smooth? This is essentially similar to the donut on the table example that you have mentioned; have a look at the plot of the function $sin(x+y)$ here : wolframalpha.com/input/?i=plot+sin(x%2By)
– Suhan Shetty
Jul 26 at 5:47
1
Yes, that makes sense since the intersection of $x+y=pi/2$ and $z=sin(x+y)$ considers a single line which can be detected as smooth by IFT. But, if we consider the intersection of $z=pm 1$ and $z=sin(x+y)$, the issue arises. Thanks a lot, @JyrkiLahtonen - I learned some of the limitations of IFT today.
– Suhan Shetty
Jul 26 at 6:20
 |Â
show 2 more comments
up vote
0
down vote
It seems that Alpha is stuck because it uses a numerical solver to get the curves. As all the roots are double, it has a hard time finding changes of sign.
You can try $$sin(x+y)=1-epsilon$$ for smaller and smaller values of $epsilon$. You will first see pairs of parallel lines, then they fuse with each other, then become lacunar and in the end vanish.
Anyway, the analytical expressions are quite correct.
answered Jul 26 at 16:38
Yves Daoust
111k665203
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Presumably you started out by using the implicit function theorem to the zero locus of the function.
$$
F(x,y)=sin(x+y)-c=0.qquad(*)
$$
Assume that $P=(x_0,y_0)$ satisfies $(*)$. IFT promises $y$ as a smooth function of $x$ in a neighborhood of $P$, if $F_y(x_0,y_0)neq0$.
When $c=1$ and, for example $(x_0,y_0)=(pi/2,0)$, we see that the partial
derivative
$$
F_y(x,y)=cos(x+y)
$$
vanishes at $(x_0,y_0)$ because$cos(pi/2)=0$.
What goes wrong even though you have those lines?
The solutions of $sin (x+y)=c$ are $$x+y=arcsin c+ncdot2pi$$ and
$$x+y=pi-arcsin c+ncdot 2pi.$$ All because $sin(pi-alpha)=sinalpha$. Supplementary angles and all that.
[Comment: originally I fumbled this and misguidedly thought that the other family of lines share the slope +1, thanks to @anon for setting this straight.]
So we have two families of parallel lines. Within a family the vertical separation of any two lines is an integer multiple of $2pi$. Furthermore, you get one family from the other with a vertical (or horizontal) translation by $pi-2arcsin c$. So we see that the two families actually coincide when $arcsin c=pmpi/2$, i.e. when $c=pm1$.
IFT diagnoses it when the lines describing the set of solutions "double on themselves".
In other words, the flow of the logic in IFT is one-way-traffic only:
- A non-vanishing Jacobian promises that locally within the set of
solutions the prescribed subset of coordinates is a smooth function
of the remaining coordinates. - But, the set of solutions can be the graph of a smooth function even when the Jacobian vanishes.
My favorite example of this is the smooth curve of intersection formed by a donut lying flat on the table. The points of contact of the two surfaces form a perfectly smooth circle. But, because the surface of the table is the shared plane of tangency of the donut and the table, the IFT breaks down there as the Jacobian determinant vanishes at all points of intersection.
In general the solutions to $sin(x+y)=c$ are $$ beginarrayll x+y & =arcsin(c)+2pi n \ x+y & =[pi-arcsin(c)]+2pi n endarray$$ which are all descending lines. When $c=1$, the values of $arcsin(1)$ and $pi-arcsin(1)$ are the same though. And there's no $x$ inside of an $arcsin$.
– anon
Jul 25 at 5:03
Oops. Thanks, @anon. Need to rethink...
– Jyrki Lahtonen
Jul 25 at 5:05
Thanks, @JyrkiLahtonen. I did not think of the other family of solutions to $sin(x+y) =c$, i.e.,$x+y =pi-arcsin(c)+2pi n$.
– Suhan Shetty
Jul 26 at 4:59
Is there any stronger theorem than IFT which would detect this locus (i.e. $sin(x+y) =pm 1$) as smooth? This is essentially similar to the donut on the table example that you have mentioned; have a look at the plot of the function $sin(x+y)$ here : wolframalpha.com/input/?i=plot+sin(x%2By)
– Suhan Shetty
Jul 26 at 5:47
1
Yes, that makes sense since the intersection of $x+y=pi/2$ and $z=sin(x+y)$ considers a single line which can be detected as smooth by IFT. But, if we consider the intersection of $z=pm 1$ and $z=sin(x+y)$, the issue arises. Thanks a lot, @JyrkiLahtonen - I learned some of the limitations of IFT today.
– Suhan Shetty
Jul 26 at 6:20
 |Â
show 2 more comments
up vote
2
down vote
accepted
Presumably you started out by using the implicit function theorem to the zero locus of the function.
$$
F(x,y)=sin(x+y)-c=0.qquad(*)
$$
Assume that $P=(x_0,y_0)$ satisfies $(*)$. IFT promises $y$ as a smooth function of $x$ in a neighborhood of $P$, if $F_y(x_0,y_0)neq0$.
When $c=1$ and, for example $(x_0,y_0)=(pi/2,0)$, we see that the partial
derivative
$$
F_y(x,y)=cos(x+y)
$$
vanishes at $(x_0,y_0)$ because$cos(pi/2)=0$.
What goes wrong even though you have those lines?
The solutions of $sin (x+y)=c$ are $$x+y=arcsin c+ncdot2pi$$ and
$$x+y=pi-arcsin c+ncdot 2pi.$$ All because $sin(pi-alpha)=sinalpha$. Supplementary angles and all that.
[Comment: originally I fumbled this and misguidedly thought that the other family of lines share the slope +1, thanks to @anon for setting this straight.]
So we have two families of parallel lines. Within a family the vertical separation of any two lines is an integer multiple of $2pi$. Furthermore, you get one family from the other with a vertical (or horizontal) translation by $pi-2arcsin c$. So we see that the two families actually coincide when $arcsin c=pmpi/2$, i.e. when $c=pm1$.
IFT diagnoses it when the lines describing the set of solutions "double on themselves".
In other words, the flow of the logic in IFT is one-way-traffic only:
- A non-vanishing Jacobian promises that locally within the set of
solutions the prescribed subset of coordinates is a smooth function
of the remaining coordinates. - But, the set of solutions can be the graph of a smooth function even when the Jacobian vanishes.
My favorite example of this is the smooth curve of intersection formed by a donut lying flat on the table. The points of contact of the two surfaces form a perfectly smooth circle. But, because the surface of the table is the shared plane of tangency of the donut and the table, the IFT breaks down there as the Jacobian determinant vanishes at all points of intersection.
In general the solutions to $sin(x+y)=c$ are $$ beginarrayll x+y & =arcsin(c)+2pi n \ x+y & =[pi-arcsin(c)]+2pi n endarray$$ which are all descending lines. When $c=1$, the values of $arcsin(1)$ and $pi-arcsin(1)$ are the same though. And there's no $x$ inside of an $arcsin$.
– anon
Jul 25 at 5:03
Oops. Thanks, @anon. Need to rethink...
– Jyrki Lahtonen
Jul 25 at 5:05
Thanks, @JyrkiLahtonen. I did not think of the other family of solutions to $sin(x+y) =c$, i.e.,$x+y =pi-arcsin(c)+2pi n$.
– Suhan Shetty
Jul 26 at 4:59
Is there any stronger theorem than IFT which would detect this locus (i.e. $sin(x+y) =pm 1$) as smooth? This is essentially similar to the donut on the table example that you have mentioned; have a look at the plot of the function $sin(x+y)$ here : wolframalpha.com/input/?i=plot+sin(x%2By)
– Suhan Shetty
Jul 26 at 5:47
1
Yes, that makes sense since the intersection of $x+y=pi/2$ and $z=sin(x+y)$ considers a single line which can be detected as smooth by IFT. But, if we consider the intersection of $z=pm 1$ and $z=sin(x+y)$, the issue arises. Thanks a lot, @JyrkiLahtonen - I learned some of the limitations of IFT today.
– Suhan Shetty
Jul 26 at 6:20
 |Â
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Presumably you started out by using the implicit function theorem to the zero locus of the function.
$$
F(x,y)=sin(x+y)-c=0.qquad(*)
$$
Assume that $P=(x_0,y_0)$ satisfies $(*)$. IFT promises $y$ as a smooth function of $x$ in a neighborhood of $P$, if $F_y(x_0,y_0)neq0$.
When $c=1$ and, for example $(x_0,y_0)=(pi/2,0)$, we see that the partial
derivative
$$
F_y(x,y)=cos(x+y)
$$
vanishes at $(x_0,y_0)$ because$cos(pi/2)=0$.
What goes wrong even though you have those lines?
The solutions of $sin (x+y)=c$ are $$x+y=arcsin c+ncdot2pi$$ and
$$x+y=pi-arcsin c+ncdot 2pi.$$ All because $sin(pi-alpha)=sinalpha$. Supplementary angles and all that.
[Comment: originally I fumbled this and misguidedly thought that the other family of lines share the slope +1, thanks to @anon for setting this straight.]
So we have two families of parallel lines. Within a family the vertical separation of any two lines is an integer multiple of $2pi$. Furthermore, you get one family from the other with a vertical (or horizontal) translation by $pi-2arcsin c$. So we see that the two families actually coincide when $arcsin c=pmpi/2$, i.e. when $c=pm1$.
IFT diagnoses it when the lines describing the set of solutions "double on themselves".
In other words, the flow of the logic in IFT is one-way-traffic only:
- A non-vanishing Jacobian promises that locally within the set of
solutions the prescribed subset of coordinates is a smooth function
of the remaining coordinates. - But, the set of solutions can be the graph of a smooth function even when the Jacobian vanishes.
My favorite example of this is the smooth curve of intersection formed by a donut lying flat on the table. The points of contact of the two surfaces form a perfectly smooth circle. But, because the surface of the table is the shared plane of tangency of the donut and the table, the IFT breaks down there as the Jacobian determinant vanishes at all points of intersection.
Presumably you started out by using the implicit function theorem to the zero locus of the function.
$$
F(x,y)=sin(x+y)-c=0.qquad(*)
$$
Assume that $P=(x_0,y_0)$ satisfies $(*)$. IFT promises $y$ as a smooth function of $x$ in a neighborhood of $P$, if $F_y(x_0,y_0)neq0$.
When $c=1$ and, for example $(x_0,y_0)=(pi/2,0)$, we see that the partial
derivative
$$
F_y(x,y)=cos(x+y)
$$
vanishes at $(x_0,y_0)$ because$cos(pi/2)=0$.
What goes wrong even though you have those lines?
The solutions of $sin (x+y)=c$ are $$x+y=arcsin c+ncdot2pi$$ and
$$x+y=pi-arcsin c+ncdot 2pi.$$ All because $sin(pi-alpha)=sinalpha$. Supplementary angles and all that.
[Comment: originally I fumbled this and misguidedly thought that the other family of lines share the slope +1, thanks to @anon for setting this straight.]
So we have two families of parallel lines. Within a family the vertical separation of any two lines is an integer multiple of $2pi$. Furthermore, you get one family from the other with a vertical (or horizontal) translation by $pi-2arcsin c$. So we see that the two families actually coincide when $arcsin c=pmpi/2$, i.e. when $c=pm1$.
IFT diagnoses it when the lines describing the set of solutions "double on themselves".
In other words, the flow of the logic in IFT is one-way-traffic only:
- A non-vanishing Jacobian promises that locally within the set of
solutions the prescribed subset of coordinates is a smooth function
of the remaining coordinates. - But, the set of solutions can be the graph of a smooth function even when the Jacobian vanishes.
My favorite example of this is the smooth curve of intersection formed by a donut lying flat on the table. The points of contact of the two surfaces form a perfectly smooth circle. But, because the surface of the table is the shared plane of tangency of the donut and the table, the IFT breaks down there as the Jacobian determinant vanishes at all points of intersection.
edited Jul 28 at 7:39
community wiki
5 revs
Jyrki Lahtonen
In general the solutions to $sin(x+y)=c$ are $$ beginarrayll x+y & =arcsin(c)+2pi n \ x+y & =[pi-arcsin(c)]+2pi n endarray$$ which are all descending lines. When $c=1$, the values of $arcsin(1)$ and $pi-arcsin(1)$ are the same though. And there's no $x$ inside of an $arcsin$.
– anon
Jul 25 at 5:03
Oops. Thanks, @anon. Need to rethink...
– Jyrki Lahtonen
Jul 25 at 5:05
Thanks, @JyrkiLahtonen. I did not think of the other family of solutions to $sin(x+y) =c$, i.e.,$x+y =pi-arcsin(c)+2pi n$.
– Suhan Shetty
Jul 26 at 4:59
Is there any stronger theorem than IFT which would detect this locus (i.e. $sin(x+y) =pm 1$) as smooth? This is essentially similar to the donut on the table example that you have mentioned; have a look at the plot of the function $sin(x+y)$ here : wolframalpha.com/input/?i=plot+sin(x%2By)
– Suhan Shetty
Jul 26 at 5:47
1
Yes, that makes sense since the intersection of $x+y=pi/2$ and $z=sin(x+y)$ considers a single line which can be detected as smooth by IFT. But, if we consider the intersection of $z=pm 1$ and $z=sin(x+y)$, the issue arises. Thanks a lot, @JyrkiLahtonen - I learned some of the limitations of IFT today.
– Suhan Shetty
Jul 26 at 6:20
 |Â
show 2 more comments
In general the solutions to $sin(x+y)=c$ are $$ beginarrayll x+y & =arcsin(c)+2pi n \ x+y & =[pi-arcsin(c)]+2pi n endarray$$ which are all descending lines. When $c=1$, the values of $arcsin(1)$ and $pi-arcsin(1)$ are the same though. And there's no $x$ inside of an $arcsin$.
– anon
Jul 25 at 5:03
Oops. Thanks, @anon. Need to rethink...
– Jyrki Lahtonen
Jul 25 at 5:05
Thanks, @JyrkiLahtonen. I did not think of the other family of solutions to $sin(x+y) =c$, i.e.,$x+y =pi-arcsin(c)+2pi n$.
– Suhan Shetty
Jul 26 at 4:59
Is there any stronger theorem than IFT which would detect this locus (i.e. $sin(x+y) =pm 1$) as smooth? This is essentially similar to the donut on the table example that you have mentioned; have a look at the plot of the function $sin(x+y)$ here : wolframalpha.com/input/?i=plot+sin(x%2By)
– Suhan Shetty
Jul 26 at 5:47
1
Yes, that makes sense since the intersection of $x+y=pi/2$ and $z=sin(x+y)$ considers a single line which can be detected as smooth by IFT. But, if we consider the intersection of $z=pm 1$ and $z=sin(x+y)$, the issue arises. Thanks a lot, @JyrkiLahtonen - I learned some of the limitations of IFT today.
– Suhan Shetty
Jul 26 at 6:20
In general the solutions to $sin(x+y)=c$ are $$ beginarrayll x+y & =arcsin(c)+2pi n \ x+y & =[pi-arcsin(c)]+2pi n endarray$$ which are all descending lines. When $c=1$, the values of $arcsin(1)$ and $pi-arcsin(1)$ are the same though. And there's no $x$ inside of an $arcsin$.
– anon
Jul 25 at 5:03
In general the solutions to $sin(x+y)=c$ are $$ beginarrayll x+y & =arcsin(c)+2pi n \ x+y & =[pi-arcsin(c)]+2pi n endarray$$ which are all descending lines. When $c=1$, the values of $arcsin(1)$ and $pi-arcsin(1)$ are the same though. And there's no $x$ inside of an $arcsin$.
– anon
Jul 25 at 5:03
Oops. Thanks, @anon. Need to rethink...
– Jyrki Lahtonen
Jul 25 at 5:05
Oops. Thanks, @anon. Need to rethink...
– Jyrki Lahtonen
Jul 25 at 5:05
Thanks, @JyrkiLahtonen. I did not think of the other family of solutions to $sin(x+y) =c$, i.e.,$x+y =pi-arcsin(c)+2pi n$.
– Suhan Shetty
Jul 26 at 4:59
Thanks, @JyrkiLahtonen. I did not think of the other family of solutions to $sin(x+y) =c$, i.e.,$x+y =pi-arcsin(c)+2pi n$.
– Suhan Shetty
Jul 26 at 4:59
Is there any stronger theorem than IFT which would detect this locus (i.e. $sin(x+y) =pm 1$) as smooth? This is essentially similar to the donut on the table example that you have mentioned; have a look at the plot of the function $sin(x+y)$ here : wolframalpha.com/input/?i=plot+sin(x%2By)
– Suhan Shetty
Jul 26 at 5:47
Is there any stronger theorem than IFT which would detect this locus (i.e. $sin(x+y) =pm 1$) as smooth? This is essentially similar to the donut on the table example that you have mentioned; have a look at the plot of the function $sin(x+y)$ here : wolframalpha.com/input/?i=plot+sin(x%2By)
– Suhan Shetty
Jul 26 at 5:47
1
1
Yes, that makes sense since the intersection of $x+y=pi/2$ and $z=sin(x+y)$ considers a single line which can be detected as smooth by IFT. But, if we consider the intersection of $z=pm 1$ and $z=sin(x+y)$, the issue arises. Thanks a lot, @JyrkiLahtonen - I learned some of the limitations of IFT today.
– Suhan Shetty
Jul 26 at 6:20
Yes, that makes sense since the intersection of $x+y=pi/2$ and $z=sin(x+y)$ considers a single line which can be detected as smooth by IFT. But, if we consider the intersection of $z=pm 1$ and $z=sin(x+y)$, the issue arises. Thanks a lot, @JyrkiLahtonen - I learned some of the limitations of IFT today.
– Suhan Shetty
Jul 26 at 6:20
 |Â
show 2 more comments
up vote
0
down vote
It seems that Alpha is stuck because it uses a numerical solver to get the curves. As all the roots are double, it has a hard time finding changes of sign.
You can try $$sin(x+y)=1-epsilon$$ for smaller and smaller values of $epsilon$. You will first see pairs of parallel lines, then they fuse with each other, then become lacunar and in the end vanish.
Anyway, the analytical expressions are quite correct.
answered Jul 26 at 16:38
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
It seems that Alpha is stuck because it uses a numerical solver to get the curves. As all the roots are double, it has a hard time finding changes of sign.
You can try $$sin(x+y)=1-epsilon$$ for smaller and smaller values of $epsilon$. You will first see pairs of parallel lines, then they fuse with each other, then become lacunar and in the end vanish.
Anyway, the analytical expressions are quite correct.
answered Jul 26 at 16:38
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It seems that Alpha is stuck because it uses a numerical solver to get the curves. As all the roots are double, it has a hard time finding changes of sign.
You can try $$sin(x+y)=1-epsilon$$ for smaller and smaller values of $epsilon$. You will first see pairs of parallel lines, then they fuse with each other, then become lacunar and in the end vanish.
Anyway, the analytical expressions are quite correct.
answered Jul 26 at 16:38
Yves Daoust
111k665203
It seems that Alpha is stuck because it uses a numerical solver to get the curves. As all the roots are double, it has a hard time finding changes of sign.
You can try $$sin(x+y)=1-epsilon$$ for smaller and smaller values of $epsilon$. You will first see pairs of parallel lines, then they fuse with each other, then become lacunar and in the end vanish.
Anyway, the analytical expressions are quite correct.
answered Jul 26 at 16:38
Yves Daoust
111k665203
answered Jul 26 at 16:38
Yves Daoust
111k665203
answered Jul 26 at 16:38
Yves Daoust
111k665203
answered Jul 26 at 16:38
Yves Daoust
111k665203
111k665203
add a comment |Â
add a comment |Â
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Derivative at where and of what kind?
– Parcly Taxel
Jul 25 at 4:36
1
A curve, which is a vector function $mathbfr(t)$ of a parameter $t$, is considered not smooth if $dmathbfr/dt$ is ever the zero vector $bf 0$. However, that doesn't mean the gradient $nabla f$ of a scalar function $f$ being the zero vector implies the level curve of $f$ is not smooth.
– anon
Jul 25 at 5:07
Yeah, as anon pointed out in a comment to my failed attempt at this (need more coffee, sorry), the problem seems to be that the usual two families of parallel lines coincide for these values of $c$. This makes IFT fail to diagnose the smoothness because all the solution points are, in a sense, double zeros.
– Jyrki Lahtonen
Jul 25 at 5:10
I do not see why $c = pm1$ plays a special role. The derivative $Df$ of $f$ does not depend on $c$. Moreover, the set $N_c = u in mathbbR mid sin(u) = c $ always consists of countably many isolated points $u_n$ so that the solution set decomposes into the parallel lines $y = -x + u_n$.
– Paul Frost
Jul 25 at 7:57
@PaulFrost Let $ f = sin(x+y) - c$ . Then, $Df = [cos(x+y), cos(x+y)]$. So, $Df$ will be a zero vector if $cos(x+y)=0$. But, $x+y = arcsin(c)$. So, $Df$ is a zero vector at $c=pm1$.
– Suhan Shetty
Jul 26 at 4:16