If $g(x) ne f(x)$ almost everywhere, then polynomial transformation of $P(g(x)) ne P(f(x))$?
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Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).
Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?
calculus functional-analysis measure-theory
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up vote
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down vote
favorite
Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).
Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?
calculus functional-analysis measure-theory
can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
â 6005
Aug 9 at 22:53
What about $P(f(x)) = P(g(x)) = 0$?
â Adam Francey
Aug 9 at 22:59
1
Are $,f,g,$ assumed to be strictly increasing, maybe?
â dxiv
Aug 9 at 22:59
1
@6005 everywhere on $mathbbR$.
â user1292919
Aug 9 at 23:02
2
I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
â Alex R.
Aug 9 at 23:15
 |Â
show 4 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).
Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?
calculus functional-analysis measure-theory
Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).
Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?
calculus functional-analysis measure-theory
edited Aug 10 at 0:22
6005
34.9k750123
34.9k750123
asked Aug 9 at 22:32
user1292919
650512
650512
can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
â 6005
Aug 9 at 22:53
What about $P(f(x)) = P(g(x)) = 0$?
â Adam Francey
Aug 9 at 22:59
1
Are $,f,g,$ assumed to be strictly increasing, maybe?
â dxiv
Aug 9 at 22:59
1
@6005 everywhere on $mathbbR$.
â user1292919
Aug 9 at 23:02
2
I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
â Alex R.
Aug 9 at 23:15
 |Â
show 4 more comments
can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
â 6005
Aug 9 at 22:53
What about $P(f(x)) = P(g(x)) = 0$?
â Adam Francey
Aug 9 at 22:59
1
Are $,f,g,$ assumed to be strictly increasing, maybe?
â dxiv
Aug 9 at 22:59
1
@6005 everywhere on $mathbbR$.
â user1292919
Aug 9 at 23:02
2
I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
â Alex R.
Aug 9 at 23:15
can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
â 6005
Aug 9 at 22:53
can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
â 6005
Aug 9 at 22:53
What about $P(f(x)) = P(g(x)) = 0$?
â Adam Francey
Aug 9 at 22:59
What about $P(f(x)) = P(g(x)) = 0$?
â Adam Francey
Aug 9 at 22:59
1
1
Are $,f,g,$ assumed to be strictly increasing, maybe?
â dxiv
Aug 9 at 22:59
Are $,f,g,$ assumed to be strictly increasing, maybe?
â dxiv
Aug 9 at 22:59
1
1
@6005 everywhere on $mathbbR$.
â user1292919
Aug 9 at 23:02
@6005 everywhere on $mathbbR$.
â user1292919
Aug 9 at 23:02
2
2
I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
â Alex R.
Aug 9 at 23:15
I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
â Alex R.
Aug 9 at 23:15
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
I think that the following is a counterexample:
Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.
Then
$$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$
I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
â 6005
Aug 9 at 23:40
add a comment |Â
up vote
1
down vote
It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.
To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.
Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
beginalign*
P(X) &= (x-2)^3 - (x-2) \
f(x) &= 2 + frac1sqrt3 cos x + sin x \
g(x) &= 2 + frac1sqrt3 cos x - sin x
endalign*
These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.
In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.
Of course, there are also examples where no such polynomial exists. For instance, take
beginalign*
f(x) &= x^4 \
g(x) &= x^2
endalign*
on the interval $[1,2]$.
Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
I think that the following is a counterexample:
Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.
Then
$$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$
I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
â 6005
Aug 9 at 23:40
add a comment |Â
up vote
6
down vote
accepted
I think that the following is a counterexample:
Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.
Then
$$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$
I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
â 6005
Aug 9 at 23:40
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
I think that the following is a counterexample:
Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.
Then
$$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$
I think that the following is a counterexample:
Let $f(x)=sin(x), g(x)=-cos(x)$ on $[0, fracpi2]$ and $P(X)=X^2(1-X^2)$.
Then
$$P(f(x))=P(g(x))=sin^2(x) cos^2(x)$$
answered Aug 9 at 23:37
N. S.
98k5105197
98k5105197
I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
â 6005
Aug 9 at 23:40
add a comment |Â
I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
â 6005
Aug 9 at 23:40
I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
â 6005
Aug 9 at 23:40
I knew there was a counterexample with sin and cos and $P(X) = X^3 - X$ (or a translation of it), but my algebra keeps not working out. Yours is conceptually easier. Nice work.
â 6005
Aug 9 at 23:40
add a comment |Â
up vote
1
down vote
It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.
To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.
Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
beginalign*
P(X) &= (x-2)^3 - (x-2) \
f(x) &= 2 + frac1sqrt3 cos x + sin x \
g(x) &= 2 + frac1sqrt3 cos x - sin x
endalign*
These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.
In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.
Of course, there are also examples where no such polynomial exists. For instance, take
beginalign*
f(x) &= x^4 \
g(x) &= x^2
endalign*
on the interval $[1,2]$.
Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.
add a comment |Â
up vote
1
down vote
It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.
To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.
Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
beginalign*
P(X) &= (x-2)^3 - (x-2) \
f(x) &= 2 + frac1sqrt3 cos x + sin x \
g(x) &= 2 + frac1sqrt3 cos x - sin x
endalign*
These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.
In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.
Of course, there are also examples where no such polynomial exists. For instance, take
beginalign*
f(x) &= x^4 \
g(x) &= x^2
endalign*
on the interval $[1,2]$.
Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.
To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.
Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
beginalign*
P(X) &= (x-2)^3 - (x-2) \
f(x) &= 2 + frac1sqrt3 cos x + sin x \
g(x) &= 2 + frac1sqrt3 cos x - sin x
endalign*
These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.
In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.
Of course, there are also examples where no such polynomial exists. For instance, take
beginalign*
f(x) &= x^4 \
g(x) &= x^2
endalign*
on the interval $[1,2]$.
Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.
It is possible to have positive, strictly increasing functions $f, g: [a,b] to mathbbR$ such that $f(x) ne g(x)$ for all $x in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x in [a,b]$.
To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:
The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.
Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
beginalign*
P(X) &= (x-2)^3 - (x-2) \
f(x) &= 2 + frac1sqrt3 cos x + sin x \
g(x) &= 2 + frac1sqrt3 cos x - sin x
endalign*
These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2pi/3,-pi/3]$.
The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.
In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.
Of course, there are also examples where no such polynomial exists. For instance, take
beginalign*
f(x) &= x^4 \
g(x) &= x^2
endalign*
on the interval $[1,2]$.
Note that for either example, you can make the function increasing on all of $mathbbR$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$.
edited Aug 10 at 0:10
answered Aug 9 at 23:47
6005
34.9k750123
34.9k750123
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can you clarify the edited requirement? Does "postive everywhere" mean $[a,b]$ or on all of $mathbbR$?
â 6005
Aug 9 at 22:53
What about $P(f(x)) = P(g(x)) = 0$?
â Adam Francey
Aug 9 at 22:59
1
Are $,f,g,$ assumed to be strictly increasing, maybe?
â dxiv
Aug 9 at 22:59
1
@6005 everywhere on $mathbbR$.
â user1292919
Aug 9 at 23:02
2
I'm a bit confused. Let the interval be $[0,pi/2]$, and $f(x)=x$, $g(x)=sin(x)$. Are you suggesting that there's a non-constant polynomial $P$ such that $P(x)=P(sin(x))$? Or are you looking for some kind of additional necessary conditions on $f,g$ to guarantee such a polynomial exists?
â Alex R.
Aug 9 at 23:15