Mixing
How can $f(x,y)$ be written as a function of $gleft(frac{y}{x}right)$?
$begingroup$
If $f(tx,ty)=f(x,y)$, how does it imply that:
$$f(x,y)=gleft(dfrac {y}{x}right)$$
Yesterday I asked a question on ordinary differential equations and got an answer. The answerer used the above stated "theorem" but I didn't understand why it is always true. Also why is $t$ (which is a constant) set to $1/x?$ Why can't we set $t=1/(x^2+1)$ or anything else? Is it that we can set $t$ to any function of $x$ that makes my life easy? Why is it so?
calculus ordinary-differential-equations functions substitution
edited Feb 3 at 10:54
Martin Sleziak
45k10123277
asked Feb 3 at 5:50
OliverOliver
225
$endgroup$
add a comment |
$begingroup$
If $f(tx,ty)=f(x,y)$, how does it imply that:
$$f(x,y)=gleft(dfrac {y}{x}right)$$
Yesterday I asked a question on ordinary differential equations and got an answer. The answerer used the above stated "theorem" but I didn't understand why it is always true. Also why is $t$ (which is a constant) set to $1/x?$ Why can't we set $t=1/(x^2+1)$ or anything else? Is it that we can set $t$ to any function of $x$ that makes my life easy? Why is it so?
calculus ordinary-differential-equations functions substitution
edited Feb 3 at 10:54
Martin Sleziak
45k10123277
asked Feb 3 at 5:50
OliverOliver
225
$endgroup$
1
$begingroup$
There is a good question here, but it would be good if you provided some context, outside of the tags (to motivate people to answer you and help others find your question). Perhaps you can write a little bit why this is relevant to the study of ODEs and substitution?
$endgroup$
– Theo Bendit
Feb 3 at 5:52
$begingroup$
This is in general not true because part of what defines a function is its domain. Functions of the form $g(y/x)$ don't allow $x = 0$ points (y-axis points) of the plane while $f$ don't have this restriction
$endgroup$
– DWade64
Feb 3 at 17:51
$begingroup$
GReyes answer is really good. From a function of the form $g(y/x)$ you can conclude that it has the property $g(x,y) = g(tx,ty)$. However I don't think the logic works the other way around. Just given $f$ with the property $f(x,y) = f(tx, ty)$, the domain of this function is whatever it is (some blob, maybe all of the plane. But the domain could include $x = 0$ points). You can imagine such a function like this. $f$ would have some constant value for a vertical line through the origin. But the domain of $g(y/x)$ prohibits the domain from containing a line through the origin.
$endgroup$
– DWade64
Feb 3 at 18:01
$begingroup$
Given $f$ you can definitely write it as $g(y/x)$ for every single line, except the vertical line. But given the question as it stands, technically, you can't come to this conclusion. If the question stated given $f$ with the property $f(x,y) = f(tx,ty)$, can $f$ be written as $g(y/x)$ for all points $xneq 0$ and some constant $c$ for points $x = 0$. Then I think the answer would be yes. We need more details on the domain of $f$
$endgroup$
– DWade64
Feb 3 at 18:04
$begingroup$
Also adding on to what GReyes said, given $f(x,y)=f(tx,ty)$, either $f$ is all one big constant, otherwise the only thing we know about the domain of f from this statement is that f can't be defined at the origin. f can't be multi-valued. But other $x=0$ points could be in the domain of $f$. We don't know. Also $t$ is a parameter independent of $x$ and $y$, so I don't see how you can write it as a function of $x$
$endgroup$
– DWade64
Feb 3 at 18:25
add a comment |
$begingroup$
If $f(tx,ty)=f(x,y)$, how does it imply that:
$$f(x,y)=gleft(dfrac {y}{x}right)$$
Yesterday I asked a question on ordinary differential equations and got an answer. The answerer used the above stated "theorem" but I didn't understand why it is always true. Also why is $t$ (which is a constant) set to $1/x?$ Why can't we set $t=1/(x^2+1)$ or anything else? Is it that we can set $t$ to any function of $x$ that makes my life easy? Why is it so?
calculus ordinary-differential-equations functions substitution
edited Feb 3 at 10:54
Martin Sleziak
45k10123277
asked Feb 3 at 5:50
OliverOliver
225
$endgroup$
If $f(tx,ty)=f(x,y)$, how does it imply that:
$$f(x,y)=gleft(dfrac {y}{x}right)$$
Yesterday I asked a question on ordinary differential equations and got an answer. The answerer used the above stated "theorem" but I didn't understand why it is always true. Also why is $t$ (which is a constant) set to $1/x?$ Why can't we set $t=1/(x^2+1)$ or anything else? Is it that we can set $t$ to any function of $x$ that makes my life easy? Why is it so?
calculus ordinary-differential-equations functions substitution
calculus ordinary-differential-equations functions substitution
edited Feb 3 at 10:54
Martin Sleziak
45k10123277
asked Feb 3 at 5:50
OliverOliver
225
edited Feb 3 at 10:54
Martin Sleziak
45k10123277
asked Feb 3 at 5:50
OliverOliver
225
edited Feb 3 at 10:54
Martin Sleziak
45k10123277
edited Feb 3 at 10:54
Martin Sleziak
45k10123277
edited Feb 3 at 10:54
Martin Sleziak
45k10123277
45k10123277
asked Feb 3 at 5:50
OliverOliver
225
asked Feb 3 at 5:50
OliverOliver
225
asked Feb 3 at 5:50
OliverOliver
225
225
1
$begingroup$
There is a good question here, but it would be good if you provided some context, outside of the tags (to motivate people to answer you and help others find your question). Perhaps you can write a little bit why this is relevant to the study of ODEs and substitution?
$endgroup$
– Theo Bendit
Feb 3 at 5:52
$begingroup$
This is in general not true because part of what defines a function is its domain. Functions of the form $g(y/x)$ don't allow $x = 0$ points (y-axis points) of the plane while $f$ don't have this restriction
$endgroup$
– DWade64
Feb 3 at 17:51
$begingroup$
GReyes answer is really good. From a function of the form $g(y/x)$ you can conclude that it has the property $g(x,y) = g(tx,ty)$. However I don't think the logic works the other way around. Just given $f$ with the property $f(x,y) = f(tx, ty)$, the domain of this function is whatever it is (some blob, maybe all of the plane. But the domain could include $x = 0$ points). You can imagine such a function like this. $f$ would have some constant value for a vertical line through the origin. But the domain of $g(y/x)$ prohibits the domain from containing a line through the origin.
$endgroup$
– DWade64
Feb 3 at 18:01
$begingroup$
Given $f$ you can definitely write it as $g(y/x)$ for every single line, except the vertical line. But given the question as it stands, technically, you can't come to this conclusion. If the question stated given $f$ with the property $f(x,y) = f(tx,ty)$, can $f$ be written as $g(y/x)$ for all points $xneq 0$ and some constant $c$ for points $x = 0$. Then I think the answer would be yes. We need more details on the domain of $f$
$endgroup$
– DWade64
Feb 3 at 18:04
$begingroup$
Also adding on to what GReyes said, given $f(x,y)=f(tx,ty)$, either $f$ is all one big constant, otherwise the only thing we know about the domain of f from this statement is that f can't be defined at the origin. f can't be multi-valued. But other $x=0$ points could be in the domain of $f$. We don't know. Also $t$ is a parameter independent of $x$ and $y$, so I don't see how you can write it as a function of $x$
$endgroup$
– DWade64
Feb 3 at 18:25
add a comment |
1
$begingroup$
There is a good question here, but it would be good if you provided some context, outside of the tags (to motivate people to answer you and help others find your question). Perhaps you can write a little bit why this is relevant to the study of ODEs and substitution?
$endgroup$
– Theo Bendit
Feb 3 at 5:52
$begingroup$
This is in general not true because part of what defines a function is its domain. Functions of the form $g(y/x)$ don't allow $x = 0$ points (y-axis points) of the plane while $f$ don't have this restriction
$endgroup$
– DWade64
Feb 3 at 17:51
$begingroup$
GReyes answer is really good. From a function of the form $g(y/x)$ you can conclude that it has the property $g(x,y) = g(tx,ty)$. However I don't think the logic works the other way around. Just given $f$ with the property $f(x,y) = f(tx, ty)$, the domain of this function is whatever it is (some blob, maybe all of the plane. But the domain could include $x = 0$ points). You can imagine such a function like this. $f$ would have some constant value for a vertical line through the origin. But the domain of $g(y/x)$ prohibits the domain from containing a line through the origin.
$endgroup$
– DWade64
Feb 3 at 18:01
$begingroup$
Given $f$ you can definitely write it as $g(y/x)$ for every single line, except the vertical line. But given the question as it stands, technically, you can't come to this conclusion. If the question stated given $f$ with the property $f(x,y) = f(tx,ty)$, can $f$ be written as $g(y/x)$ for all points $xneq 0$ and some constant $c$ for points $x = 0$. Then I think the answer would be yes. We need more details on the domain of $f$
$endgroup$
– DWade64
Feb 3 at 18:04
$begingroup$
Also adding on to what GReyes said, given $f(x,y)=f(tx,ty)$, either $f$ is all one big constant, otherwise the only thing we know about the domain of f from this statement is that f can't be defined at the origin. f can't be multi-valued. But other $x=0$ points could be in the domain of $f$. We don't know. Also $t$ is a parameter independent of $x$ and $y$, so I don't see how you can write it as a function of $x$
$endgroup$
– DWade64
Feb 3 at 18:25
1
1
$begingroup$
There is a good question here, but it would be good if you provided some context, outside of the tags (to motivate people to answer you and help others find your question). Perhaps you can write a little bit why this is relevant to the study of ODEs and substitution?
$endgroup$
– Theo Bendit
Feb 3 at 5:52
$begingroup$
There is a good question here, but it would be good if you provided some context, outside of the tags (to motivate people to answer you and help others find your question). Perhaps you can write a little bit why this is relevant to the study of ODEs and substitution?
$endgroup$
– Theo Bendit
Feb 3 at 5:52
$begingroup$
This is in general not true because part of what defines a function is its domain. Functions of the form $g(y/x)$ don't allow $x = 0$ points (y-axis points) of the plane while $f$ don't have this restriction
$endgroup$
– DWade64
Feb 3 at 17:51
$begingroup$
This is in general not true because part of what defines a function is its domain. Functions of the form $g(y/x)$ don't allow $x = 0$ points (y-axis points) of the plane while $f$ don't have this restriction
$endgroup$
– DWade64
Feb 3 at 17:51
$begingroup$
GReyes answer is really good. From a function of the form $g(y/x)$ you can conclude that it has the property $g(x,y) = g(tx,ty)$. However I don't think the logic works the other way around. Just given $f$ with the property $f(x,y) = f(tx, ty)$, the domain of this function is whatever it is (some blob, maybe all of the plane. But the domain could include $x = 0$ points). You can imagine such a function like this. $f$ would have some constant value for a vertical line through the origin. But the domain of $g(y/x)$ prohibits the domain from containing a line through the origin.
$endgroup$
– DWade64
Feb 3 at 18:01
$begingroup$
GReyes answer is really good. From a function of the form $g(y/x)$ you can conclude that it has the property $g(x,y) = g(tx,ty)$. However I don't think the logic works the other way around. Just given $f$ with the property $f(x,y) = f(tx, ty)$, the domain of this function is whatever it is (some blob, maybe all of the plane. But the domain could include $x = 0$ points). You can imagine such a function like this. $f$ would have some constant value for a vertical line through the origin. But the domain of $g(y/x)$ prohibits the domain from containing a line through the origin.
$endgroup$
– DWade64
Feb 3 at 18:01
$begingroup$
Given $f$ you can definitely write it as $g(y/x)$ for every single line, except the vertical line. But given the question as it stands, technically, you can't come to this conclusion. If the question stated given $f$ with the property $f(x,y) = f(tx,ty)$, can $f$ be written as $g(y/x)$ for all points $xneq 0$ and some constant $c$ for points $x = 0$. Then I think the answer would be yes. We need more details on the domain of $f$
$endgroup$
– DWade64
Feb 3 at 18:04
$begingroup$
Given $f$ you can definitely write it as $g(y/x)$ for every single line, except the vertical line. But given the question as it stands, technically, you can't come to this conclusion. If the question stated given $f$ with the property $f(x,y) = f(tx,ty)$, can $f$ be written as $g(y/x)$ for all points $xneq 0$ and some constant $c$ for points $x = 0$. Then I think the answer would be yes. We need more details on the domain of $f$
$endgroup$
– DWade64
Feb 3 at 18:04
$begingroup$
Also adding on to what GReyes said, given $f(x,y)=f(tx,ty)$, either $f$ is all one big constant, otherwise the only thing we know about the domain of f from this statement is that f can't be defined at the origin. f can't be multi-valued. But other $x=0$ points could be in the domain of $f$. We don't know. Also $t$ is a parameter independent of $x$ and $y$, so I don't see how you can write it as a function of $x$
$endgroup$
– DWade64
Feb 3 at 18:25
$begingroup$
Also adding on to what GReyes said, given $f(x,y)=f(tx,ty)$, either $f$ is all one big constant, otherwise the only thing we know about the domain of f from this statement is that f can't be defined at the origin. f can't be multi-valued. But other $x=0$ points could be in the domain of $f$. We don't know. Also $t$ is a parameter independent of $x$ and $y$, so I don't see how you can write it as a function of $x$
$endgroup$
– DWade64
Feb 3 at 18:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For any fixed $(x,y)neq (0,0)$, the set ${(tx,ty), tinmathbb{R}}$ represents the whole line through the origin in the direction of $(x,y)$. The condition $f(tx,ty)=f(x,y)$ means that $f$ is constant along such line and thus, it only depends on its slope $y/x$. These functions are called homogeneous (more exactly, homogeneous of order zero). Since they are constant along lines through the origin, they are discontinuous at the origin unless they are constant on the whole space. In the context of differential equations, you introduce a new variable $u=y/x$ which is precisely the slope of the line to reduce to a separable equation.
answered Feb 3 at 6:32
GReyesGReyes
2,54815
$endgroup$
$begingroup$
From your answer, I have got a visual proof of my question. So, if the condition is obeyed, can we always write $f(x,y)$ as a function of $y/x$?
$endgroup$
– Oliver
Feb 3 at 9:46
$begingroup$
Only if it is scale invariant that is if it's value does not change when we multiply x and y by the same t: $f(tx,ty)=f(x,y)$
$endgroup$
– GReyes
Feb 3 at 19:00
add a comment |
$begingroup$
Hint: it is easy. first set $t$ to $frac{1}{x}$. You will get $f(1,frac{y}{x}) = f(x,y)$.
answered Feb 3 at 5:52
OmGOmG
2,5371824
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any fixed $(x,y)neq (0,0)$, the set ${(tx,ty), tinmathbb{R}}$ represents the whole line through the origin in the direction of $(x,y)$. The condition $f(tx,ty)=f(x,y)$ means that $f$ is constant along such line and thus, it only depends on its slope $y/x$. These functions are called homogeneous (more exactly, homogeneous of order zero). Since they are constant along lines through the origin, they are discontinuous at the origin unless they are constant on the whole space. In the context of differential equations, you introduce a new variable $u=y/x$ which is precisely the slope of the line to reduce to a separable equation.
answered Feb 3 at 6:32
GReyesGReyes
2,54815
$endgroup$
$begingroup$
From your answer, I have got a visual proof of my question. So, if the condition is obeyed, can we always write $f(x,y)$ as a function of $y/x$?
$endgroup$
– Oliver
Feb 3 at 9:46
$begingroup$
Only if it is scale invariant that is if it's value does not change when we multiply x and y by the same t: $f(tx,ty)=f(x,y)$
$endgroup$
– GReyes
Feb 3 at 19:00
add a comment |
$begingroup$
For any fixed $(x,y)neq (0,0)$, the set ${(tx,ty), tinmathbb{R}}$ represents the whole line through the origin in the direction of $(x,y)$. The condition $f(tx,ty)=f(x,y)$ means that $f$ is constant along such line and thus, it only depends on its slope $y/x$. These functions are called homogeneous (more exactly, homogeneous of order zero). Since they are constant along lines through the origin, they are discontinuous at the origin unless they are constant on the whole space. In the context of differential equations, you introduce a new variable $u=y/x$ which is precisely the slope of the line to reduce to a separable equation.
answered Feb 3 at 6:32
GReyesGReyes
2,54815
$endgroup$
$begingroup$
From your answer, I have got a visual proof of my question. So, if the condition is obeyed, can we always write $f(x,y)$ as a function of $y/x$?
$endgroup$
– Oliver
Feb 3 at 9:46
$begingroup$
Only if it is scale invariant that is if it's value does not change when we multiply x and y by the same t: $f(tx,ty)=f(x,y)$
$endgroup$
– GReyes
Feb 3 at 19:00
add a comment |
$begingroup$
For any fixed $(x,y)neq (0,0)$, the set ${(tx,ty), tinmathbb{R}}$ represents the whole line through the origin in the direction of $(x,y)$. The condition $f(tx,ty)=f(x,y)$ means that $f$ is constant along such line and thus, it only depends on its slope $y/x$. These functions are called homogeneous (more exactly, homogeneous of order zero). Since they are constant along lines through the origin, they are discontinuous at the origin unless they are constant on the whole space. In the context of differential equations, you introduce a new variable $u=y/x$ which is precisely the slope of the line to reduce to a separable equation.
answered Feb 3 at 6:32
GReyesGReyes
2,54815
$endgroup$
For any fixed $(x,y)neq (0,0)$, the set ${(tx,ty), tinmathbb{R}}$ represents the whole line through the origin in the direction of $(x,y)$. The condition $f(tx,ty)=f(x,y)$ means that $f$ is constant along such line and thus, it only depends on its slope $y/x$. These functions are called homogeneous (more exactly, homogeneous of order zero). Since they are constant along lines through the origin, they are discontinuous at the origin unless they are constant on the whole space. In the context of differential equations, you introduce a new variable $u=y/x$ which is precisely the slope of the line to reduce to a separable equation.
answered Feb 3 at 6:32
GReyesGReyes
2,54815
answered Feb 3 at 6:32
GReyesGReyes
2,54815
answered Feb 3 at 6:32
GReyesGReyes
2,54815
answered Feb 3 at 6:32
GReyesGReyes
2,54815
2,54815
$begingroup$
From your answer, I have got a visual proof of my question. So, if the condition is obeyed, can we always write $f(x,y)$ as a function of $y/x$?
$endgroup$
– Oliver
Feb 3 at 9:46
$begingroup$
Only if it is scale invariant that is if it's value does not change when we multiply x and y by the same t: $f(tx,ty)=f(x,y)$
$endgroup$
– GReyes
Feb 3 at 19:00
add a comment |
$begingroup$
From your answer, I have got a visual proof of my question. So, if the condition is obeyed, can we always write $f(x,y)$ as a function of $y/x$?
$endgroup$
– Oliver
Feb 3 at 9:46
$begingroup$
Only if it is scale invariant that is if it's value does not change when we multiply x and y by the same t: $f(tx,ty)=f(x,y)$
$endgroup$
– GReyes
Feb 3 at 19:00
$begingroup$
From your answer, I have got a visual proof of my question. So, if the condition is obeyed, can we always write $f(x,y)$ as a function of $y/x$?
$endgroup$
– Oliver
Feb 3 at 9:46
$begingroup$
From your answer, I have got a visual proof of my question. So, if the condition is obeyed, can we always write $f(x,y)$ as a function of $y/x$?
$endgroup$
– Oliver
Feb 3 at 9:46
$begingroup$
Only if it is scale invariant that is if it's value does not change when we multiply x and y by the same t: $f(tx,ty)=f(x,y)$
$endgroup$
– GReyes
Feb 3 at 19:00
$begingroup$
Only if it is scale invariant that is if it's value does not change when we multiply x and y by the same t: $f(tx,ty)=f(x,y)$
$endgroup$
– GReyes
Feb 3 at 19:00
add a comment |
$begingroup$
Hint: it is easy. first set $t$ to $frac{1}{x}$. You will get $f(1,frac{y}{x}) = f(x,y)$.
answered Feb 3 at 5:52
OmGOmG
2,5371824
$endgroup$
add a comment |
$begingroup$
Hint: it is easy. first set $t$ to $frac{1}{x}$. You will get $f(1,frac{y}{x}) = f(x,y)$.
answered Feb 3 at 5:52
OmGOmG
2,5371824
$endgroup$
add a comment |
$begingroup$
Hint: it is easy. first set $t$ to $frac{1}{x}$. You will get $f(1,frac{y}{x}) = f(x,y)$.
answered Feb 3 at 5:52
OmGOmG
2,5371824
$endgroup$
Hint: it is easy. first set $t$ to $frac{1}{x}$. You will get $f(1,frac{y}{x}) = f(x,y)$.
answered Feb 3 at 5:52
OmGOmG
2,5371824
answered Feb 3 at 5:52
OmGOmG
2,5371824
answered Feb 3 at 5:52
OmGOmG
2,5371824
answered Feb 3 at 5:52
OmGOmG
2,5371824
2,5371824
add a comment |
add a comment |
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There is a good question here, but it would be good if you provided some context, outside of the tags (to motivate people to answer you and help others find your question). Perhaps you can write a little bit why this is relevant to the study of ODEs and substitution?
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– Theo Bendit
Feb 3 at 5:52
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This is in general not true because part of what defines a function is its domain. Functions of the form $g(y/x)$ don't allow $x = 0$ points (y-axis points) of the plane while $f$ don't have this restriction
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– DWade64
Feb 3 at 17:51
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GReyes answer is really good. From a function of the form $g(y/x)$ you can conclude that it has the property $g(x,y) = g(tx,ty)$. However I don't think the logic works the other way around. Just given $f$ with the property $f(x,y) = f(tx, ty)$, the domain of this function is whatever it is (some blob, maybe all of the plane. But the domain could include $x = 0$ points). You can imagine such a function like this. $f$ would have some constant value for a vertical line through the origin. But the domain of $g(y/x)$ prohibits the domain from containing a line through the origin.
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– DWade64
Feb 3 at 18:01
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Given $f$ you can definitely write it as $g(y/x)$ for every single line, except the vertical line. But given the question as it stands, technically, you can't come to this conclusion. If the question stated given $f$ with the property $f(x,y) = f(tx,ty)$, can $f$ be written as $g(y/x)$ for all points $xneq 0$ and some constant $c$ for points $x = 0$. Then I think the answer would be yes. We need more details on the domain of $f$
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– DWade64
Feb 3 at 18:04
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Also adding on to what GReyes said, given $f(x,y)=f(tx,ty)$, either $f$ is all one big constant, otherwise the only thing we know about the domain of f from this statement is that f can't be defined at the origin. f can't be multi-valued. But other $x=0$ points could be in the domain of $f$. We don't know. Also $t$ is a parameter independent of $x$ and $y$, so I don't see how you can write it as a function of $x$
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– DWade64
Feb 3 at 18:25