Fulfilling conditions of Inverse function(?)
up vote
1
down vote
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Let's assume that
$$h(g(t))=t$$
What conditions ae needed to say that
$$g(h(t))=t$$
Is satisfied too?
(Provided that $h$ and $g$ are continuous and derivatable, but not knowing whether they have inverse though..)
I am not sure about the title of this question. If you know better title, feel free to let me know or edit this question.
inverse-function constraints inverse-function-theorem
asked Nov 13 at 9:44
KYHSGeekCode
302110
add a comment |
up vote
1
down vote
favorite
Let's assume that
$$h(g(t))=t$$
What conditions ae needed to say that
$$g(h(t))=t$$
Is satisfied too?
(Provided that $h$ and $g$ are continuous and derivatable, but not knowing whether they have inverse though..)
I am not sure about the title of this question. If you know better title, feel free to let me know or edit this question.
inverse-function constraints inverse-function-theorem
asked Nov 13 at 9:44
KYHSGeekCode
302110
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let's assume that
$$h(g(t))=t$$
What conditions ae needed to say that
$$g(h(t))=t$$
Is satisfied too?
(Provided that $h$ and $g$ are continuous and derivatable, but not knowing whether they have inverse though..)
I am not sure about the title of this question. If you know better title, feel free to let me know or edit this question.
inverse-function constraints inverse-function-theorem
asked Nov 13 at 9:44
KYHSGeekCode
302110
Let's assume that
$$h(g(t))=t$$
What conditions ae needed to say that
$$g(h(t))=t$$
Is satisfied too?
(Provided that $h$ and $g$ are continuous and derivatable, but not knowing whether they have inverse though..)
I am not sure about the title of this question. If you know better title, feel free to let me know or edit this question.
inverse-function constraints inverse-function-theorem
inverse-function constraints inverse-function-theorem
asked Nov 13 at 9:44
KYHSGeekCode
302110
asked Nov 13 at 9:44
KYHSGeekCode
302110
asked Nov 13 at 9:44
KYHSGeekCode
302110
asked Nov 13 at 9:44
KYHSGeekCode
302110
asked Nov 13 at 9:44
KYHSGeekCode
302110
302110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If the domain is $mathbb R$ then a necessary and sufficient condition is that $h$ is one-to-one. If $h$ is one-to-one change $t$ to $h(t) $ in the hypothesis to get $h(g(h(t))=h(t)$. This implies $g(h(t))=t$ because $h$ is one-to-one. Conversely the equation $g(h(t))=t$ clearly implies that $h$ is one-to-one.
answered Nov 13 at 9:49
Kavi Rama Murthy
39.3k31748
So easist but loose condition to check is checking if $h'(t)$ does not change its sign, right?
– KYHSGeekCode
Nov 13 at 10:03
Why use derivatives at all? You can ask this question about functions that are not even continuous and the answer is always the same.
– Kavi Rama Murthy
Nov 13 at 10:06
You don't need the assumption that the domain is $mathbb{R}$, either: it's true whenever the domain of $g$ is the codomian of $h$, and the domain of $h$ is the codomain of $g$ (otherwise one of the compositions is undefined), and your proof works unchanged.
– user3482749
Nov 13 at 10:09
@KaviRamaMurthy Thank you for your clarification.(I agree with you(and your proof). The comment was specific only for a function that is analytic but not being able to think its graph at the first glance(in my current curriculum) :) Anyway thanks for your help!
– KYHSGeekCode
Nov 13 at 10:12
@user3482749 Oh thank you.
– KYHSGeekCode
Nov 13 at 10:14
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If the domain is $mathbb R$ then a necessary and sufficient condition is that $h$ is one-to-one. If $h$ is one-to-one change $t$ to $h(t) $ in the hypothesis to get $h(g(h(t))=h(t)$. This implies $g(h(t))=t$ because $h$ is one-to-one. Conversely the equation $g(h(t))=t$ clearly implies that $h$ is one-to-one.
answered Nov 13 at 9:49
Kavi Rama Murthy
39.3k31748
So easist but loose condition to check is checking if $h'(t)$ does not change its sign, right?
– KYHSGeekCode
Nov 13 at 10:03
Why use derivatives at all? You can ask this question about functions that are not even continuous and the answer is always the same.
– Kavi Rama Murthy
Nov 13 at 10:06
You don't need the assumption that the domain is $mathbb{R}$, either: it's true whenever the domain of $g$ is the codomian of $h$, and the domain of $h$ is the codomain of $g$ (otherwise one of the compositions is undefined), and your proof works unchanged.
– user3482749
Nov 13 at 10:09
@KaviRamaMurthy Thank you for your clarification.(I agree with you(and your proof). The comment was specific only for a function that is analytic but not being able to think its graph at the first glance(in my current curriculum) :) Anyway thanks for your help!
– KYHSGeekCode
Nov 13 at 10:12
@user3482749 Oh thank you.
– KYHSGeekCode
Nov 13 at 10:14
|
show 1 more comment
up vote
2
down vote
accepted
If the domain is $mathbb R$ then a necessary and sufficient condition is that $h$ is one-to-one. If $h$ is one-to-one change $t$ to $h(t) $ in the hypothesis to get $h(g(h(t))=h(t)$. This implies $g(h(t))=t$ because $h$ is one-to-one. Conversely the equation $g(h(t))=t$ clearly implies that $h$ is one-to-one.
answered Nov 13 at 9:49
Kavi Rama Murthy
39.3k31748
So easist but loose condition to check is checking if $h'(t)$ does not change its sign, right?
– KYHSGeekCode
Nov 13 at 10:03
Why use derivatives at all? You can ask this question about functions that are not even continuous and the answer is always the same.
– Kavi Rama Murthy
Nov 13 at 10:06
You don't need the assumption that the domain is $mathbb{R}$, either: it's true whenever the domain of $g$ is the codomian of $h$, and the domain of $h$ is the codomain of $g$ (otherwise one of the compositions is undefined), and your proof works unchanged.
– user3482749
Nov 13 at 10:09
@KaviRamaMurthy Thank you for your clarification.(I agree with you(and your proof). The comment was specific only for a function that is analytic but not being able to think its graph at the first glance(in my current curriculum) :) Anyway thanks for your help!
– KYHSGeekCode
Nov 13 at 10:12
@user3482749 Oh thank you.
– KYHSGeekCode
Nov 13 at 10:14
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If the domain is $mathbb R$ then a necessary and sufficient condition is that $h$ is one-to-one. If $h$ is one-to-one change $t$ to $h(t) $ in the hypothesis to get $h(g(h(t))=h(t)$. This implies $g(h(t))=t$ because $h$ is one-to-one. Conversely the equation $g(h(t))=t$ clearly implies that $h$ is one-to-one.
answered Nov 13 at 9:49
Kavi Rama Murthy
39.3k31748
If the domain is $mathbb R$ then a necessary and sufficient condition is that $h$ is one-to-one. If $h$ is one-to-one change $t$ to $h(t) $ in the hypothesis to get $h(g(h(t))=h(t)$. This implies $g(h(t))=t$ because $h$ is one-to-one. Conversely the equation $g(h(t))=t$ clearly implies that $h$ is one-to-one.
answered Nov 13 at 9:49
Kavi Rama Murthy
39.3k31748
answered Nov 13 at 9:49
Kavi Rama Murthy
39.3k31748
answered Nov 13 at 9:49
Kavi Rama Murthy
39.3k31748
answered Nov 13 at 9:49
Kavi Rama Murthy
39.3k31748
39.3k31748
So easist but loose condition to check is checking if $h'(t)$ does not change its sign, right?
– KYHSGeekCode
Nov 13 at 10:03
Why use derivatives at all? You can ask this question about functions that are not even continuous and the answer is always the same.
– Kavi Rama Murthy
Nov 13 at 10:06
You don't need the assumption that the domain is $mathbb{R}$, either: it's true whenever the domain of $g$ is the codomian of $h$, and the domain of $h$ is the codomain of $g$ (otherwise one of the compositions is undefined), and your proof works unchanged.
– user3482749
Nov 13 at 10:09
@KaviRamaMurthy Thank you for your clarification.(I agree with you(and your proof). The comment was specific only for a function that is analytic but not being able to think its graph at the first glance(in my current curriculum) :) Anyway thanks for your help!
– KYHSGeekCode
Nov 13 at 10:12
@user3482749 Oh thank you.
– KYHSGeekCode
Nov 13 at 10:14
|
show 1 more comment
So easist but loose condition to check is checking if $h'(t)$ does not change its sign, right?
– KYHSGeekCode
Nov 13 at 10:03
Why use derivatives at all? You can ask this question about functions that are not even continuous and the answer is always the same.
– Kavi Rama Murthy
Nov 13 at 10:06
You don't need the assumption that the domain is $mathbb{R}$, either: it's true whenever the domain of $g$ is the codomian of $h$, and the domain of $h$ is the codomain of $g$ (otherwise one of the compositions is undefined), and your proof works unchanged.
– user3482749
Nov 13 at 10:09
@KaviRamaMurthy Thank you for your clarification.(I agree with you(and your proof). The comment was specific only for a function that is analytic but not being able to think its graph at the first glance(in my current curriculum) :) Anyway thanks for your help!
– KYHSGeekCode
Nov 13 at 10:12
@user3482749 Oh thank you.
– KYHSGeekCode
Nov 13 at 10:14
So easist but loose condition to check is checking if $h'(t)$ does not change its sign, right?
– KYHSGeekCode
Nov 13 at 10:03
So easist but loose condition to check is checking if $h'(t)$ does not change its sign, right?
– KYHSGeekCode
Nov 13 at 10:03
Why use derivatives at all? You can ask this question about functions that are not even continuous and the answer is always the same.
– Kavi Rama Murthy
Nov 13 at 10:06
Why use derivatives at all? You can ask this question about functions that are not even continuous and the answer is always the same.
– Kavi Rama Murthy
Nov 13 at 10:06
You don't need the assumption that the domain is $mathbb{R}$, either: it's true whenever the domain of $g$ is the codomian of $h$, and the domain of $h$ is the codomain of $g$ (otherwise one of the compositions is undefined), and your proof works unchanged.
– user3482749
Nov 13 at 10:09
You don't need the assumption that the domain is $mathbb{R}$, either: it's true whenever the domain of $g$ is the codomian of $h$, and the domain of $h$ is the codomain of $g$ (otherwise one of the compositions is undefined), and your proof works unchanged.
– user3482749
Nov 13 at 10:09
@KaviRamaMurthy Thank you for your clarification.(I agree with you(and your proof). The comment was specific only for a function that is analytic but not being able to think its graph at the first glance(in my current curriculum) :) Anyway thanks for your help!
– KYHSGeekCode
Nov 13 at 10:12
@KaviRamaMurthy Thank you for your clarification.(I agree with you(and your proof). The comment was specific only for a function that is analytic but not being able to think its graph at the first glance(in my current curriculum) :) Anyway thanks for your help!
– KYHSGeekCode
Nov 13 at 10:12
@user3482749 Oh thank you.
– KYHSGeekCode
Nov 13 at 10:14
@user3482749 Oh thank you.
– KYHSGeekCode
Nov 13 at 10:14
|
show 1 more comment
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