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What is the plot of this implicit function: $|sin x|^y+|cos x|^y = 1$
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I'm trying to manually plot the following function:
$$
|sin x|^y+|cos x|^y = 1
$$
My basic approach for implicit functions is to try to express $y$ in terms of $x$ and plot it, or $x$ in terms of $y$ and then plot the inverse. Sometimes it's clear from the first glance if the equations is in some special form (for example a circumference).
For the above I couldn't find an explicit expression. I've tried to manipulate the expression in different ways in order to take logarithms and get rid of the $y$ power. It's even harder to get an insight since neither W|A nor desmos is able to plot it.
Below is the output from Mathematica which I don't really understand:
I'm interested in the ways I could transform the equation above so that it's easier to see what the graph looks like.
upd:
As pointed in the comments the above graph shows various contours. Below is the one which reflects the initial function:
Here is a Mathematica snippet for copy and paste:
ContourPlot[Abs[Cos[x]]^y + Abs[Sin[x]]^y == 1, x, -1, 1, y, -1, 10]
Just to be complete I'm adding the final plot from Mathematica (with some discrepancies which I assume are caused by the way Mathematica calculates the values) which reflects the answer by Michael Seifert.
algebra-precalculus graphing-functions
asked Aug 1 at 13:52
roman
3941312
add a comment |Â
up vote
1
down vote
favorite
I'm trying to manually plot the following function:
$$
|sin x|^y+|cos x|^y = 1
$$
My basic approach for implicit functions is to try to express $y$ in terms of $x$ and plot it, or $x$ in terms of $y$ and then plot the inverse. Sometimes it's clear from the first glance if the equations is in some special form (for example a circumference).
For the above I couldn't find an explicit expression. I've tried to manipulate the expression in different ways in order to take logarithms and get rid of the $y$ power. It's even harder to get an insight since neither W|A nor desmos is able to plot it.
Below is the output from Mathematica which I don't really understand:
I'm interested in the ways I could transform the equation above so that it's easier to see what the graph looks like.
upd:
As pointed in the comments the above graph shows various contours. Below is the one which reflects the initial function:
Here is a Mathematica snippet for copy and paste:
ContourPlot[Abs[Cos[x]]^y + Abs[Sin[x]]^y == 1, x, -1, 1, y, -1, 10]
Just to be complete I'm adding the final plot from Mathematica (with some discrepancies which I assume are caused by the way Mathematica calculates the values) which reflects the answer by Michael Seifert.
algebra-precalculus graphing-functions
asked Aug 1 at 13:52
roman
3941312
Off-topic here, but to get the contour you're asking for in Mathematica, you need to haveAbs[Cos[x]]^y + Abs[Sin[x]]^y - 1 == 0(note the double equals sign.) As it is, Mathematica is plotting several contours of the function $f(x,y) = |sin x|^y+|cos x|^y - 1$, for various values of the contour.
– Michael Seifert
Aug 1 at 14:10
@MichaelSeifert Good point, didn't know that. I will update the OP with a new image
– roman
Aug 1 at 14:14
Also, you may need to zoom out a bit. I believe that no point of the contour lies in the range $-1<x<1$ and $-1<y<1$.
– Michael Seifert
Aug 1 at 14:15
@MichaelSeifert You are right no contour is indeed present in the range you pointed
– roman
Aug 1 at 14:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to manually plot the following function:
$$
|sin x|^y+|cos x|^y = 1
$$
My basic approach for implicit functions is to try to express $y$ in terms of $x$ and plot it, or $x$ in terms of $y$ and then plot the inverse. Sometimes it's clear from the first glance if the equations is in some special form (for example a circumference).
For the above I couldn't find an explicit expression. I've tried to manipulate the expression in different ways in order to take logarithms and get rid of the $y$ power. It's even harder to get an insight since neither W|A nor desmos is able to plot it.
Below is the output from Mathematica which I don't really understand:
I'm interested in the ways I could transform the equation above so that it's easier to see what the graph looks like.
upd:
As pointed in the comments the above graph shows various contours. Below is the one which reflects the initial function:
Here is a Mathematica snippet for copy and paste:
ContourPlot[Abs[Cos[x]]^y + Abs[Sin[x]]^y == 1, x, -1, 1, y, -1, 10]
Just to be complete I'm adding the final plot from Mathematica (with some discrepancies which I assume are caused by the way Mathematica calculates the values) which reflects the answer by Michael Seifert.
algebra-precalculus graphing-functions
asked Aug 1 at 13:52
roman
3941312
I'm trying to manually plot the following function:
$$
|sin x|^y+|cos x|^y = 1
$$
My basic approach for implicit functions is to try to express $y$ in terms of $x$ and plot it, or $x$ in terms of $y$ and then plot the inverse. Sometimes it's clear from the first glance if the equations is in some special form (for example a circumference).
For the above I couldn't find an explicit expression. I've tried to manipulate the expression in different ways in order to take logarithms and get rid of the $y$ power. It's even harder to get an insight since neither W|A nor desmos is able to plot it.
Below is the output from Mathematica which I don't really understand:
I'm interested in the ways I could transform the equation above so that it's easier to see what the graph looks like.
upd:
As pointed in the comments the above graph shows various contours. Below is the one which reflects the initial function:
Here is a Mathematica snippet for copy and paste:
ContourPlot[Abs[Cos[x]]^y + Abs[Sin[x]]^y == 1, x, -1, 1, y, -1, 10]
Just to be complete I'm adding the final plot from Mathematica (with some discrepancies which I assume are caused by the way Mathematica calculates the values) which reflects the answer by Michael Seifert.
algebra-precalculus graphing-functions
asked Aug 1 at 13:52
roman
3941312
edited Aug 1 at 14:49
asked Aug 1 at 13:52
roman
3941312
asked Aug 1 at 13:52
roman
3941312
asked Aug 1 at 13:52
roman
3941312
3941312
Off-topic here, but to get the contour you're asking for in Mathematica, you need to haveAbs[Cos[x]]^y + Abs[Sin[x]]^y - 1 == 0(note the double equals sign.) As it is, Mathematica is plotting several contours of the function $f(x,y) = |sin x|^y+|cos x|^y - 1$, for various values of the contour.
– Michael Seifert
Aug 1 at 14:10
@MichaelSeifert Good point, didn't know that. I will update the OP with a new image
– roman
Aug 1 at 14:14
Also, you may need to zoom out a bit. I believe that no point of the contour lies in the range $-1<x<1$ and $-1<y<1$.
– Michael Seifert
Aug 1 at 14:15
@MichaelSeifert You are right no contour is indeed present in the range you pointed
– roman
Aug 1 at 14:24
add a comment |Â
Off-topic here, but to get the contour you're asking for in Mathematica, you need to haveAbs[Cos[x]]^y + Abs[Sin[x]]^y - 1 == 0(note the double equals sign.) As it is, Mathematica is plotting several contours of the function $f(x,y) = |sin x|^y+|cos x|^y - 1$, for various values of the contour.
– Michael Seifert
Aug 1 at 14:10
@MichaelSeifert Good point, didn't know that. I will update the OP with a new image
– roman
Aug 1 at 14:14
Also, you may need to zoom out a bit. I believe that no point of the contour lies in the range $-1<x<1$ and $-1<y<1$.
– Michael Seifert
Aug 1 at 14:15
@MichaelSeifert You are right no contour is indeed present in the range you pointed
– roman
Aug 1 at 14:24
Off-topic here, but to get the contour you're asking for in Mathematica, you need to have
Abs[Cos[x]]^y + Abs[Sin[x]]^y - 1 == 0 (note the double equals sign.) As it is, Mathematica is plotting several contours of the function $f(x,y) = |sin x|^y+|cos x|^y - 1$, for various values of the contour.– Michael Seifert
Aug 1 at 14:10
Off-topic here, but to get the contour you're asking for in Mathematica, you need to have
Abs[Cos[x]]^y + Abs[Sin[x]]^y - 1 == 0 (note the double equals sign.) As it is, Mathematica is plotting several contours of the function $f(x,y) = |sin x|^y+|cos x|^y - 1$, for various values of the contour.– Michael Seifert
Aug 1 at 14:10
@MichaelSeifert Good point, didn't know that. I will update the OP with a new image
– roman
Aug 1 at 14:14
@MichaelSeifert Good point, didn't know that. I will update the OP with a new image
– roman
Aug 1 at 14:14
Also, you may need to zoom out a bit. I believe that no point of the contour lies in the range $-1<x<1$ and $-1<y<1$.
– Michael Seifert
Aug 1 at 14:15
Also, you may need to zoom out a bit. I believe that no point of the contour lies in the range $-1<x<1$ and $-1<y<1$.
– Michael Seifert
Aug 1 at 14:15
@MichaelSeifert You are right no contour is indeed present in the range you pointed
– roman
Aug 1 at 14:24
@MichaelSeifert You are right no contour is indeed present in the range you pointed
– roman
Aug 1 at 14:24
add a comment |Â
1 Answer
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up vote
2
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We can see an obvious solution for the contour: if $y = 2$, we have $|cos x|^2 + |sin x|^2 = 1$, which is satisfied for all values of $x$. So the line $y = 2$ is part of the solution set.
If $y > 2$, then since $0leq |cos x| leq 1$, we have $|cos x|^y leq |cos x|^2$, with equality iff $|cos x| = 0$ or $|cos x| = 1$. A similar relation holds for $|sin x|$. Thus,
$$
|cos x|^y + |sin x|^y leq |cos x|^2 + |sin x|^2 = 1.
$$
Since equality only holds if both $|cos x|$ and $|sin x|$ are either 0 or 1, we cannot have $|cos x|^y + |sin x|^y = 1$ unless this is so. This occurs when $x = n pi/2$ for some integer $n$.
A similar argument can be made for when $y < 2$; in this case, we have $|cos x|^y geq |cos x|^2$ and similarly for $|sin x|$. Thus, $x = n pi/2$ is a solution when $y < 2$ as well. The only exception is that $0^0$ is indeterminate, so we cannot say that the points $x = n pi/2$, $y = 0$ are part of the contour.
Thus, the solution to the problem is the union of the sets $y = 2 $ and $x = npi/2, y neq 0$ for $n in mathbbZ$.
answered Aug 1 at 14:36
Michael Seifert
4,449623
You'll notice that the results from Mathematica aren't entirely accurate, since they miss the curves $x = n pi/2, y neq 0$ for $n neq 0$. If you're curious as to why this is, feel free to post a question over at Mathematica.SE.
– Michael Seifert
Aug 1 at 14:38
Now I see. I've just tried to play with $x$ and $y$ ranges after which your solutions absolutely makes sense, added the image to the OP. Thank you!
– roman
Aug 1 at 14:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We can see an obvious solution for the contour: if $y = 2$, we have $|cos x|^2 + |sin x|^2 = 1$, which is satisfied for all values of $x$. So the line $y = 2$ is part of the solution set.
If $y > 2$, then since $0leq |cos x| leq 1$, we have $|cos x|^y leq |cos x|^2$, with equality iff $|cos x| = 0$ or $|cos x| = 1$. A similar relation holds for $|sin x|$. Thus,
$$
|cos x|^y + |sin x|^y leq |cos x|^2 + |sin x|^2 = 1.
$$
Since equality only holds if both $|cos x|$ and $|sin x|$ are either 0 or 1, we cannot have $|cos x|^y + |sin x|^y = 1$ unless this is so. This occurs when $x = n pi/2$ for some integer $n$.
A similar argument can be made for when $y < 2$; in this case, we have $|cos x|^y geq |cos x|^2$ and similarly for $|sin x|$. Thus, $x = n pi/2$ is a solution when $y < 2$ as well. The only exception is that $0^0$ is indeterminate, so we cannot say that the points $x = n pi/2$, $y = 0$ are part of the contour.
Thus, the solution to the problem is the union of the sets $y = 2 $ and $x = npi/2, y neq 0$ for $n in mathbbZ$.
answered Aug 1 at 14:36
Michael Seifert
4,449623
You'll notice that the results from Mathematica aren't entirely accurate, since they miss the curves $x = n pi/2, y neq 0$ for $n neq 0$. If you're curious as to why this is, feel free to post a question over at Mathematica.SE.
– Michael Seifert
Aug 1 at 14:38
Now I see. I've just tried to play with $x$ and $y$ ranges after which your solutions absolutely makes sense, added the image to the OP. Thank you!
– roman
Aug 1 at 14:49
add a comment |Â
up vote
2
down vote
accepted
We can see an obvious solution for the contour: if $y = 2$, we have $|cos x|^2 + |sin x|^2 = 1$, which is satisfied for all values of $x$. So the line $y = 2$ is part of the solution set.
If $y > 2$, then since $0leq |cos x| leq 1$, we have $|cos x|^y leq |cos x|^2$, with equality iff $|cos x| = 0$ or $|cos x| = 1$. A similar relation holds for $|sin x|$. Thus,
$$
|cos x|^y + |sin x|^y leq |cos x|^2 + |sin x|^2 = 1.
$$
Since equality only holds if both $|cos x|$ and $|sin x|$ are either 0 or 1, we cannot have $|cos x|^y + |sin x|^y = 1$ unless this is so. This occurs when $x = n pi/2$ for some integer $n$.
A similar argument can be made for when $y < 2$; in this case, we have $|cos x|^y geq |cos x|^2$ and similarly for $|sin x|$. Thus, $x = n pi/2$ is a solution when $y < 2$ as well. The only exception is that $0^0$ is indeterminate, so we cannot say that the points $x = n pi/2$, $y = 0$ are part of the contour.
Thus, the solution to the problem is the union of the sets $y = 2 $ and $x = npi/2, y neq 0$ for $n in mathbbZ$.
answered Aug 1 at 14:36
Michael Seifert
4,449623
You'll notice that the results from Mathematica aren't entirely accurate, since they miss the curves $x = n pi/2, y neq 0$ for $n neq 0$. If you're curious as to why this is, feel free to post a question over at Mathematica.SE.
– Michael Seifert
Aug 1 at 14:38
Now I see. I've just tried to play with $x$ and $y$ ranges after which your solutions absolutely makes sense, added the image to the OP. Thank you!
– roman
Aug 1 at 14:49
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We can see an obvious solution for the contour: if $y = 2$, we have $|cos x|^2 + |sin x|^2 = 1$, which is satisfied for all values of $x$. So the line $y = 2$ is part of the solution set.
If $y > 2$, then since $0leq |cos x| leq 1$, we have $|cos x|^y leq |cos x|^2$, with equality iff $|cos x| = 0$ or $|cos x| = 1$. A similar relation holds for $|sin x|$. Thus,
$$
|cos x|^y + |sin x|^y leq |cos x|^2 + |sin x|^2 = 1.
$$
Since equality only holds if both $|cos x|$ and $|sin x|$ are either 0 or 1, we cannot have $|cos x|^y + |sin x|^y = 1$ unless this is so. This occurs when $x = n pi/2$ for some integer $n$.
A similar argument can be made for when $y < 2$; in this case, we have $|cos x|^y geq |cos x|^2$ and similarly for $|sin x|$. Thus, $x = n pi/2$ is a solution when $y < 2$ as well. The only exception is that $0^0$ is indeterminate, so we cannot say that the points $x = n pi/2$, $y = 0$ are part of the contour.
Thus, the solution to the problem is the union of the sets $y = 2 $ and $x = npi/2, y neq 0$ for $n in mathbbZ$.
answered Aug 1 at 14:36
Michael Seifert
4,449623
We can see an obvious solution for the contour: if $y = 2$, we have $|cos x|^2 + |sin x|^2 = 1$, which is satisfied for all values of $x$. So the line $y = 2$ is part of the solution set.
If $y > 2$, then since $0leq |cos x| leq 1$, we have $|cos x|^y leq |cos x|^2$, with equality iff $|cos x| = 0$ or $|cos x| = 1$. A similar relation holds for $|sin x|$. Thus,
$$
|cos x|^y + |sin x|^y leq |cos x|^2 + |sin x|^2 = 1.
$$
Since equality only holds if both $|cos x|$ and $|sin x|$ are either 0 or 1, we cannot have $|cos x|^y + |sin x|^y = 1$ unless this is so. This occurs when $x = n pi/2$ for some integer $n$.
A similar argument can be made for when $y < 2$; in this case, we have $|cos x|^y geq |cos x|^2$ and similarly for $|sin x|$. Thus, $x = n pi/2$ is a solution when $y < 2$ as well. The only exception is that $0^0$ is indeterminate, so we cannot say that the points $x = n pi/2$, $y = 0$ are part of the contour.
Thus, the solution to the problem is the union of the sets $y = 2 $ and $x = npi/2, y neq 0$ for $n in mathbbZ$.
answered Aug 1 at 14:36
Michael Seifert
4,449623
edited Aug 1 at 14:56
answered Aug 1 at 14:36
Michael Seifert
4,449623
answered Aug 1 at 14:36
Michael Seifert
4,449623
answered Aug 1 at 14:36
Michael Seifert
4,449623
4,449623
You'll notice that the results from Mathematica aren't entirely accurate, since they miss the curves $x = n pi/2, y neq 0$ for $n neq 0$. If you're curious as to why this is, feel free to post a question over at Mathematica.SE.
– Michael Seifert
Aug 1 at 14:38
Now I see. I've just tried to play with $x$ and $y$ ranges after which your solutions absolutely makes sense, added the image to the OP. Thank you!
– roman
Aug 1 at 14:49
add a comment |Â
You'll notice that the results from Mathematica aren't entirely accurate, since they miss the curves $x = n pi/2, y neq 0$ for $n neq 0$. If you're curious as to why this is, feel free to post a question over at Mathematica.SE.
– Michael Seifert
Aug 1 at 14:38
Now I see. I've just tried to play with $x$ and $y$ ranges after which your solutions absolutely makes sense, added the image to the OP. Thank you!
– roman
Aug 1 at 14:49
You'll notice that the results from Mathematica aren't entirely accurate, since they miss the curves $x = n pi/2, y neq 0$ for $n neq 0$. If you're curious as to why this is, feel free to post a question over at Mathematica.SE.
– Michael Seifert
Aug 1 at 14:38
You'll notice that the results from Mathematica aren't entirely accurate, since they miss the curves $x = n pi/2, y neq 0$ for $n neq 0$. If you're curious as to why this is, feel free to post a question over at Mathematica.SE.
– Michael Seifert
Aug 1 at 14:38
Now I see. I've just tried to play with $x$ and $y$ ranges after which your solutions absolutely makes sense, added the image to the OP. Thank you!
– roman
Aug 1 at 14:49
Now I see. I've just tried to play with $x$ and $y$ ranges after which your solutions absolutely makes sense, added the image to the OP. Thank you!
– roman
Aug 1 at 14:49
add a comment |Â
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Off-topic here, but to get the contour you're asking for in Mathematica, you need to have
Abs[Cos[x]]^y + Abs[Sin[x]]^y - 1 == 0(note the double equals sign.) As it is, Mathematica is plotting several contours of the function $f(x,y) = |sin x|^y+|cos x|^y - 1$, for various values of the contour.– Michael Seifert
Aug 1 at 14:10
@MichaelSeifert Good point, didn't know that. I will update the OP with a new image
– roman
Aug 1 at 14:14
Also, you may need to zoom out a bit. I believe that no point of the contour lies in the range $-1<x<1$ and $-1<y<1$.
– Michael Seifert
Aug 1 at 14:15
@MichaelSeifert You are right no contour is indeed present in the range you pointed
– roman
Aug 1 at 14:24