Mixing
Global inverse using Inverse Function Theorem
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I'm tasked with solving the following question: let $0 le alpha < 1$ be a constant and consider the function $f: mathbbR to mathbbR$ defined by $f(x) = x - alpha sin x$. Show that $f$ is one-to-one and onto and that its inverse function is smooth.
So far I've tried using the Inverse Function Theorem for a single variable. My version of the Theorem is: suppose $f : mathbbR to mathbbR$ is continuously differentiable. Suppose that for $a in mathbbR$ we have $f^prime (a) ne 0$ and that $f(a) = b$. Then there exist open neighborhoods $U$ of $a$ and $W$ of $b$ such that (i) $f(U) = W$ and $f|_U$ is one-to-one; (ii) if $h$ is the inverse of $f|_U$ then $h$ is continuously differentiable; (iii) for each $xin U$ we have $h^prime (f(x)) = frac1f^prime (x)$.
So to use the IFT we check the condition: it's clear that $f(x) = x - alpha sin x$ is continuously differentiable regardless of the value of $alpha$. For any fixed $a in mathbbR$ we have $f^prime (a) = 1 - alpha cos a$. Now, if $alpha = 0$ then $f^prime (a) = 1 ne 0$ and the condition is automatically satisfied. So suppose that $0 < alpha < 1$. The condition that $0 = f^prime (a) = 1 - alpha cos a$ is equivalent to $cos a = frac1alpha$, but $0 < alpha <1$ implies $1 < frac1alpha < infty$ so in this case there's no solution to $cos a = frac1alpha$. Thus in every possible case $f^prime (a) ne 0$. Therefore the Inverse Function Theorem guarantees a local (smooth) inverse about every point. To me it seems that the problem statement is asking about some global inverse however. Is there some way we can glue together local inverses to obtain a global one?
calculus inverse-function
asked Jul 29 at 20:06
aherring
21917
add a comment |Â
up vote
0
down vote
favorite
I'm tasked with solving the following question: let $0 le alpha < 1$ be a constant and consider the function $f: mathbbR to mathbbR$ defined by $f(x) = x - alpha sin x$. Show that $f$ is one-to-one and onto and that its inverse function is smooth.
So far I've tried using the Inverse Function Theorem for a single variable. My version of the Theorem is: suppose $f : mathbbR to mathbbR$ is continuously differentiable. Suppose that for $a in mathbbR$ we have $f^prime (a) ne 0$ and that $f(a) = b$. Then there exist open neighborhoods $U$ of $a$ and $W$ of $b$ such that (i) $f(U) = W$ and $f|_U$ is one-to-one; (ii) if $h$ is the inverse of $f|_U$ then $h$ is continuously differentiable; (iii) for each $xin U$ we have $h^prime (f(x)) = frac1f^prime (x)$.
So to use the IFT we check the condition: it's clear that $f(x) = x - alpha sin x$ is continuously differentiable regardless of the value of $alpha$. For any fixed $a in mathbbR$ we have $f^prime (a) = 1 - alpha cos a$. Now, if $alpha = 0$ then $f^prime (a) = 1 ne 0$ and the condition is automatically satisfied. So suppose that $0 < alpha < 1$. The condition that $0 = f^prime (a) = 1 - alpha cos a$ is equivalent to $cos a = frac1alpha$, but $0 < alpha <1$ implies $1 < frac1alpha < infty$ so in this case there's no solution to $cos a = frac1alpha$. Thus in every possible case $f^prime (a) ne 0$. Therefore the Inverse Function Theorem guarantees a local (smooth) inverse about every point. To me it seems that the problem statement is asking about some global inverse however. Is there some way we can glue together local inverses to obtain a global one?
calculus inverse-function
asked Jul 29 at 20:06
aherring
21917
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm tasked with solving the following question: let $0 le alpha < 1$ be a constant and consider the function $f: mathbbR to mathbbR$ defined by $f(x) = x - alpha sin x$. Show that $f$ is one-to-one and onto and that its inverse function is smooth.
So far I've tried using the Inverse Function Theorem for a single variable. My version of the Theorem is: suppose $f : mathbbR to mathbbR$ is continuously differentiable. Suppose that for $a in mathbbR$ we have $f^prime (a) ne 0$ and that $f(a) = b$. Then there exist open neighborhoods $U$ of $a$ and $W$ of $b$ such that (i) $f(U) = W$ and $f|_U$ is one-to-one; (ii) if $h$ is the inverse of $f|_U$ then $h$ is continuously differentiable; (iii) for each $xin U$ we have $h^prime (f(x)) = frac1f^prime (x)$.
So to use the IFT we check the condition: it's clear that $f(x) = x - alpha sin x$ is continuously differentiable regardless of the value of $alpha$. For any fixed $a in mathbbR$ we have $f^prime (a) = 1 - alpha cos a$. Now, if $alpha = 0$ then $f^prime (a) = 1 ne 0$ and the condition is automatically satisfied. So suppose that $0 < alpha < 1$. The condition that $0 = f^prime (a) = 1 - alpha cos a$ is equivalent to $cos a = frac1alpha$, but $0 < alpha <1$ implies $1 < frac1alpha < infty$ so in this case there's no solution to $cos a = frac1alpha$. Thus in every possible case $f^prime (a) ne 0$. Therefore the Inverse Function Theorem guarantees a local (smooth) inverse about every point. To me it seems that the problem statement is asking about some global inverse however. Is there some way we can glue together local inverses to obtain a global one?
calculus inverse-function
asked Jul 29 at 20:06
aherring
21917
I'm tasked with solving the following question: let $0 le alpha < 1$ be a constant and consider the function $f: mathbbR to mathbbR$ defined by $f(x) = x - alpha sin x$. Show that $f$ is one-to-one and onto and that its inverse function is smooth.
So far I've tried using the Inverse Function Theorem for a single variable. My version of the Theorem is: suppose $f : mathbbR to mathbbR$ is continuously differentiable. Suppose that for $a in mathbbR$ we have $f^prime (a) ne 0$ and that $f(a) = b$. Then there exist open neighborhoods $U$ of $a$ and $W$ of $b$ such that (i) $f(U) = W$ and $f|_U$ is one-to-one; (ii) if $h$ is the inverse of $f|_U$ then $h$ is continuously differentiable; (iii) for each $xin U$ we have $h^prime (f(x)) = frac1f^prime (x)$.
So to use the IFT we check the condition: it's clear that $f(x) = x - alpha sin x$ is continuously differentiable regardless of the value of $alpha$. For any fixed $a in mathbbR$ we have $f^prime (a) = 1 - alpha cos a$. Now, if $alpha = 0$ then $f^prime (a) = 1 ne 0$ and the condition is automatically satisfied. So suppose that $0 < alpha < 1$. The condition that $0 = f^prime (a) = 1 - alpha cos a$ is equivalent to $cos a = frac1alpha$, but $0 < alpha <1$ implies $1 < frac1alpha < infty$ so in this case there's no solution to $cos a = frac1alpha$. Thus in every possible case $f^prime (a) ne 0$. Therefore the Inverse Function Theorem guarantees a local (smooth) inverse about every point. To me it seems that the problem statement is asking about some global inverse however. Is there some way we can glue together local inverses to obtain a global one?
calculus inverse-function
asked Jul 29 at 20:06
aherring
21917
asked Jul 29 at 20:06
aherring
21917
asked Jul 29 at 20:06
aherring
21917
asked Jul 29 at 20:06
aherring
21917
21917
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Note that:
$$f'(x)=1-alpha cos(x)>0$$
so $f$ is always increasing, hence it is (globally) invertible. Then you can apply inverse function theorem for differentiability of the inverse.
To expand ever so slightly: one-to-one follows immediately from the increasing property. For onto, we can use a very simple Intermediate Value Theorem argument (again using the increasing property).
– aherring
Jul 30 at 15:09
As for differentiability of the inverse, would it be something like the following: apply IFT to a point to find a local smooth inverse and then argue that the local inverse must coincide with the global inverse? Each local inverse is smooth on its domain of definition and thus so too is the global inverse smooth?
– aherring
Jul 30 at 15:16
add a comment |Â
up vote
1
down vote
Thanks to the ordering of the real line, there is a global version of the inverse function theorem in the one dimensional case. In what follows, open intervals are allowed to be infinite in extent, such that the case $mathbbR = (-infty, +infty)$ is also covered.
Global Inverse Function Theorem
Let $I$ be a non-empty open interval in $mathbbR$, and suppose that $f colon I to mathbbR$ is a continuously differentiable function such that $f'$ does not assume the value zero. Then:
The derivative $f'$ has constant sign on $I$, and $f$ is strictly monotone.
The image $f(I) = f(x) : x in I $ of $f$ is an open interval.
Furthermore, if $g colon f(I) to I$ is the inverse function of $f$ (which exists by 1.), then $g$ is continuously differentiable, has the same monotonicity as $f$, and $$g' = frac1f' circ g.$$
If we suppose that the open interval $I$ has lower endpoint $a$ and upper endpoint $b$ (both of which may be infinite), the strict monotonicity of $f$ along with the intermediate value theorem allows us to conclude that the image of $f$ is the open interval $f(I) = (m, M)$, where
$$m = inf_x in I f(x) quad textand quad M = sup_x in I f(x).$$
If $f$ is strictly increasing, one has the additional formulas
$$m = lim_x downarrow a f(x) quad textand quad M = lim_x uparrow b f(x),$$
and, if $f$ is strictly decreasing, the formulas read
$$m = lim_x uparrow b f(x) quad textand quad M = lim_x downarrow a f(x).$$
Combining the above results should enable you to solve the problem.
answered Jul 29 at 22:27
Qeeko
7510
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
Note that:
$$f'(x)=1-alpha cos(x)>0$$
so $f$ is always increasing, hence it is (globally) invertible. Then you can apply inverse function theorem for differentiability of the inverse.
To expand ever so slightly: one-to-one follows immediately from the increasing property. For onto, we can use a very simple Intermediate Value Theorem argument (again using the increasing property).
– aherring
Jul 30 at 15:09
As for differentiability of the inverse, would it be something like the following: apply IFT to a point to find a local smooth inverse and then argue that the local inverse must coincide with the global inverse? Each local inverse is smooth on its domain of definition and thus so too is the global inverse smooth?
– aherring
Jul 30 at 15:16
add a comment |Â
up vote
2
down vote
accepted
Note that:
$$f'(x)=1-alpha cos(x)>0$$
so $f$ is always increasing, hence it is (globally) invertible. Then you can apply inverse function theorem for differentiability of the inverse.
To expand ever so slightly: one-to-one follows immediately from the increasing property. For onto, we can use a very simple Intermediate Value Theorem argument (again using the increasing property).
– aherring
Jul 30 at 15:09
As for differentiability of the inverse, would it be something like the following: apply IFT to a point to find a local smooth inverse and then argue that the local inverse must coincide with the global inverse? Each local inverse is smooth on its domain of definition and thus so too is the global inverse smooth?
– aherring
Jul 30 at 15:16
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Note that:
$$f'(x)=1-alpha cos(x)>0$$
so $f$ is always increasing, hence it is (globally) invertible. Then you can apply inverse function theorem for differentiability of the inverse.
Note that:
$$f'(x)=1-alpha cos(x)>0$$
so $f$ is always increasing, hence it is (globally) invertible. Then you can apply inverse function theorem for differentiability of the inverse.
edited Jul 29 at 21:48
answered Jul 29 at 20:58
user223391
To expand ever so slightly: one-to-one follows immediately from the increasing property. For onto, we can use a very simple Intermediate Value Theorem argument (again using the increasing property).
– aherring
Jul 30 at 15:09
As for differentiability of the inverse, would it be something like the following: apply IFT to a point to find a local smooth inverse and then argue that the local inverse must coincide with the global inverse? Each local inverse is smooth on its domain of definition and thus so too is the global inverse smooth?
– aherring
Jul 30 at 15:16
add a comment |Â
To expand ever so slightly: one-to-one follows immediately from the increasing property. For onto, we can use a very simple Intermediate Value Theorem argument (again using the increasing property).
– aherring
Jul 30 at 15:09
As for differentiability of the inverse, would it be something like the following: apply IFT to a point to find a local smooth inverse and then argue that the local inverse must coincide with the global inverse? Each local inverse is smooth on its domain of definition and thus so too is the global inverse smooth?
– aherring
Jul 30 at 15:16
To expand ever so slightly: one-to-one follows immediately from the increasing property. For onto, we can use a very simple Intermediate Value Theorem argument (again using the increasing property).
– aherring
Jul 30 at 15:09
To expand ever so slightly: one-to-one follows immediately from the increasing property. For onto, we can use a very simple Intermediate Value Theorem argument (again using the increasing property).
– aherring
Jul 30 at 15:09
As for differentiability of the inverse, would it be something like the following: apply IFT to a point to find a local smooth inverse and then argue that the local inverse must coincide with the global inverse? Each local inverse is smooth on its domain of definition and thus so too is the global inverse smooth?
– aherring
Jul 30 at 15:16
As for differentiability of the inverse, would it be something like the following: apply IFT to a point to find a local smooth inverse and then argue that the local inverse must coincide with the global inverse? Each local inverse is smooth on its domain of definition and thus so too is the global inverse smooth?
– aherring
Jul 30 at 15:16
add a comment |Â
up vote
1
down vote
Thanks to the ordering of the real line, there is a global version of the inverse function theorem in the one dimensional case. In what follows, open intervals are allowed to be infinite in extent, such that the case $mathbbR = (-infty, +infty)$ is also covered.
Global Inverse Function Theorem
Let $I$ be a non-empty open interval in $mathbbR$, and suppose that $f colon I to mathbbR$ is a continuously differentiable function such that $f'$ does not assume the value zero. Then:
The derivative $f'$ has constant sign on $I$, and $f$ is strictly monotone.
The image $f(I) = f(x) : x in I $ of $f$ is an open interval.
Furthermore, if $g colon f(I) to I$ is the inverse function of $f$ (which exists by 1.), then $g$ is continuously differentiable, has the same monotonicity as $f$, and $$g' = frac1f' circ g.$$
If we suppose that the open interval $I$ has lower endpoint $a$ and upper endpoint $b$ (both of which may be infinite), the strict monotonicity of $f$ along with the intermediate value theorem allows us to conclude that the image of $f$ is the open interval $f(I) = (m, M)$, where
$$m = inf_x in I f(x) quad textand quad M = sup_x in I f(x).$$
If $f$ is strictly increasing, one has the additional formulas
$$m = lim_x downarrow a f(x) quad textand quad M = lim_x uparrow b f(x),$$
and, if $f$ is strictly decreasing, the formulas read
$$m = lim_x uparrow b f(x) quad textand quad M = lim_x downarrow a f(x).$$
Combining the above results should enable you to solve the problem.
answered Jul 29 at 22:27
Qeeko
7510
add a comment |Â
up vote
1
down vote
Thanks to the ordering of the real line, there is a global version of the inverse function theorem in the one dimensional case. In what follows, open intervals are allowed to be infinite in extent, such that the case $mathbbR = (-infty, +infty)$ is also covered.
Global Inverse Function Theorem
Let $I$ be a non-empty open interval in $mathbbR$, and suppose that $f colon I to mathbbR$ is a continuously differentiable function such that $f'$ does not assume the value zero. Then:
The derivative $f'$ has constant sign on $I$, and $f$ is strictly monotone.
The image $f(I) = f(x) : x in I $ of $f$ is an open interval.
Furthermore, if $g colon f(I) to I$ is the inverse function of $f$ (which exists by 1.), then $g$ is continuously differentiable, has the same monotonicity as $f$, and $$g' = frac1f' circ g.$$
If we suppose that the open interval $I$ has lower endpoint $a$ and upper endpoint $b$ (both of which may be infinite), the strict monotonicity of $f$ along with the intermediate value theorem allows us to conclude that the image of $f$ is the open interval $f(I) = (m, M)$, where
$$m = inf_x in I f(x) quad textand quad M = sup_x in I f(x).$$
If $f$ is strictly increasing, one has the additional formulas
$$m = lim_x downarrow a f(x) quad textand quad M = lim_x uparrow b f(x),$$
and, if $f$ is strictly decreasing, the formulas read
$$m = lim_x uparrow b f(x) quad textand quad M = lim_x downarrow a f(x).$$
Combining the above results should enable you to solve the problem.
answered Jul 29 at 22:27
Qeeko
7510
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Thanks to the ordering of the real line, there is a global version of the inverse function theorem in the one dimensional case. In what follows, open intervals are allowed to be infinite in extent, such that the case $mathbbR = (-infty, +infty)$ is also covered.
Global Inverse Function Theorem
Let $I$ be a non-empty open interval in $mathbbR$, and suppose that $f colon I to mathbbR$ is a continuously differentiable function such that $f'$ does not assume the value zero. Then:
The derivative $f'$ has constant sign on $I$, and $f$ is strictly monotone.
The image $f(I) = f(x) : x in I $ of $f$ is an open interval.
Furthermore, if $g colon f(I) to I$ is the inverse function of $f$ (which exists by 1.), then $g$ is continuously differentiable, has the same monotonicity as $f$, and $$g' = frac1f' circ g.$$
If we suppose that the open interval $I$ has lower endpoint $a$ and upper endpoint $b$ (both of which may be infinite), the strict monotonicity of $f$ along with the intermediate value theorem allows us to conclude that the image of $f$ is the open interval $f(I) = (m, M)$, where
$$m = inf_x in I f(x) quad textand quad M = sup_x in I f(x).$$
If $f$ is strictly increasing, one has the additional formulas
$$m = lim_x downarrow a f(x) quad textand quad M = lim_x uparrow b f(x),$$
and, if $f$ is strictly decreasing, the formulas read
$$m = lim_x uparrow b f(x) quad textand quad M = lim_x downarrow a f(x).$$
Combining the above results should enable you to solve the problem.
answered Jul 29 at 22:27
Qeeko
7510
Thanks to the ordering of the real line, there is a global version of the inverse function theorem in the one dimensional case. In what follows, open intervals are allowed to be infinite in extent, such that the case $mathbbR = (-infty, +infty)$ is also covered.
Global Inverse Function Theorem
Let $I$ be a non-empty open interval in $mathbbR$, and suppose that $f colon I to mathbbR$ is a continuously differentiable function such that $f'$ does not assume the value zero. Then:
The derivative $f'$ has constant sign on $I$, and $f$ is strictly monotone.
The image $f(I) = f(x) : x in I $ of $f$ is an open interval.
Furthermore, if $g colon f(I) to I$ is the inverse function of $f$ (which exists by 1.), then $g$ is continuously differentiable, has the same monotonicity as $f$, and $$g' = frac1f' circ g.$$
If we suppose that the open interval $I$ has lower endpoint $a$ and upper endpoint $b$ (both of which may be infinite), the strict monotonicity of $f$ along with the intermediate value theorem allows us to conclude that the image of $f$ is the open interval $f(I) = (m, M)$, where
$$m = inf_x in I f(x) quad textand quad M = sup_x in I f(x).$$
If $f$ is strictly increasing, one has the additional formulas
$$m = lim_x downarrow a f(x) quad textand quad M = lim_x uparrow b f(x),$$
and, if $f$ is strictly decreasing, the formulas read
$$m = lim_x uparrow b f(x) quad textand quad M = lim_x downarrow a f(x).$$
Combining the above results should enable you to solve the problem.
answered Jul 29 at 22:27
Qeeko
7510
edited Jul 29 at 22:49
answered Jul 29 at 22:27
Qeeko
7510
answered Jul 29 at 22:27
Qeeko
7510
answered Jul 29 at 22:27
Qeeko
7510
7510
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866383%2fglobal-inverse-using-inverse-function-theorem%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password