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How to use the Implicit Function Theorem for show that a set is open?
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Consider $a_i,; i=1,...,n$, real distinct numbers and the polynomial of degree odd $n$,
$$p(x) = prod_1^n(x-a_i).$$
Define
$$A = binmathbbR mid p(x) = btext has $n$ distinct roots.$$
(i) Show that $0 in A$.
(ii) Show that $A$ is bounded.
(iii) Use the Implicit Function Theorem to show that $A$ is open.
(i) is trivial
(ii) I take a sequence $(b_i)$ such that $b_i to 0$ to get a contradiction, but I couldn't. I just wanted a hint, some better way.
(iii) I never solved any questions involving IFT in order to classify a set. I have no idea how to use it. I wish someone could help me with that. I imagine that I should define $f: mathbbR^p to mathbbR^q$ ($p leq q$) such that $A subset mathbbR^p$ and take $x_0 in A$ and to use the IFT to get $varphi: B_delta(x_0) subset A to mathbbR^q in C^1$. It's the only way I can imagine using IFT. Anyway, I appreciate any help!
real-analysis implicit-function-theorem
asked Aug 1 at 20:47
Lucas Corrêa
1,071319
add a comment |Â
up vote
1
down vote
favorite
Consider $a_i,; i=1,...,n$, real distinct numbers and the polynomial of degree odd $n$,
$$p(x) = prod_1^n(x-a_i).$$
Define
$$A = binmathbbR mid p(x) = btext has $n$ distinct roots.$$
(i) Show that $0 in A$.
(ii) Show that $A$ is bounded.
(iii) Use the Implicit Function Theorem to show that $A$ is open.
(i) is trivial
(ii) I take a sequence $(b_i)$ such that $b_i to 0$ to get a contradiction, but I couldn't. I just wanted a hint, some better way.
(iii) I never solved any questions involving IFT in order to classify a set. I have no idea how to use it. I wish someone could help me with that. I imagine that I should define $f: mathbbR^p to mathbbR^q$ ($p leq q$) such that $A subset mathbbR^p$ and take $x_0 in A$ and to use the IFT to get $varphi: B_delta(x_0) subset A to mathbbR^q in C^1$. It's the only way I can imagine using IFT. Anyway, I appreciate any help!
real-analysis implicit-function-theorem
asked Aug 1 at 20:47
Lucas Corrêa
1,071319
1
(ii) is false as stated. I guess it means to assume that $n>1$ and by "roots" it means real roots?
– Eric Wofsey
Aug 1 at 20:54
1
@EricWofsey, good point! I also think it is $n> 1$. As for the roots, it was not really clear that if they are only real. The hypothesis $n$ is odd serves to ensure that it has at least one real root so, perhaps, be considered complex roots.
– Lucas Corrêa
Aug 1 at 21:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider $a_i,; i=1,...,n$, real distinct numbers and the polynomial of degree odd $n$,
$$p(x) = prod_1^n(x-a_i).$$
Define
$$A = binmathbbR mid p(x) = btext has $n$ distinct roots.$$
(i) Show that $0 in A$.
(ii) Show that $A$ is bounded.
(iii) Use the Implicit Function Theorem to show that $A$ is open.
(i) is trivial
(ii) I take a sequence $(b_i)$ such that $b_i to 0$ to get a contradiction, but I couldn't. I just wanted a hint, some better way.
(iii) I never solved any questions involving IFT in order to classify a set. I have no idea how to use it. I wish someone could help me with that. I imagine that I should define $f: mathbbR^p to mathbbR^q$ ($p leq q$) such that $A subset mathbbR^p$ and take $x_0 in A$ and to use the IFT to get $varphi: B_delta(x_0) subset A to mathbbR^q in C^1$. It's the only way I can imagine using IFT. Anyway, I appreciate any help!
real-analysis implicit-function-theorem
asked Aug 1 at 20:47
Lucas Corrêa
1,071319
Consider $a_i,; i=1,...,n$, real distinct numbers and the polynomial of degree odd $n$,
$$p(x) = prod_1^n(x-a_i).$$
Define
$$A = binmathbbR mid p(x) = btext has $n$ distinct roots.$$
(i) Show that $0 in A$.
(ii) Show that $A$ is bounded.
(iii) Use the Implicit Function Theorem to show that $A$ is open.
(i) is trivial
(ii) I take a sequence $(b_i)$ such that $b_i to 0$ to get a contradiction, but I couldn't. I just wanted a hint, some better way.
(iii) I never solved any questions involving IFT in order to classify a set. I have no idea how to use it. I wish someone could help me with that. I imagine that I should define $f: mathbbR^p to mathbbR^q$ ($p leq q$) such that $A subset mathbbR^p$ and take $x_0 in A$ and to use the IFT to get $varphi: B_delta(x_0) subset A to mathbbR^q in C^1$. It's the only way I can imagine using IFT. Anyway, I appreciate any help!
real-analysis implicit-function-theorem
asked Aug 1 at 20:47
Lucas Corrêa
1,071319
asked Aug 1 at 20:47
Lucas Corrêa
1,071319
asked Aug 1 at 20:47
Lucas Corrêa
1,071319
asked Aug 1 at 20:47
Lucas Corrêa
1,071319
1,071319
1
(ii) is false as stated. I guess it means to assume that $n>1$ and by "roots" it means real roots?
– Eric Wofsey
Aug 1 at 20:54
1
@EricWofsey, good point! I also think it is $n> 1$. As for the roots, it was not really clear that if they are only real. The hypothesis $n$ is odd serves to ensure that it has at least one real root so, perhaps, be considered complex roots.
– Lucas Corrêa
Aug 1 at 21:01
add a comment |Â
1
(ii) is false as stated. I guess it means to assume that $n>1$ and by "roots" it means real roots?
– Eric Wofsey
Aug 1 at 20:54
1
@EricWofsey, good point! I also think it is $n> 1$. As for the roots, it was not really clear that if they are only real. The hypothesis $n$ is odd serves to ensure that it has at least one real root so, perhaps, be considered complex roots.
– Lucas Corrêa
Aug 1 at 21:01
1
1
(ii) is false as stated. I guess it means to assume that $n>1$ and by "roots" it means real roots?
– Eric Wofsey
Aug 1 at 20:54
(ii) is false as stated. I guess it means to assume that $n>1$ and by "roots" it means real roots?
– Eric Wofsey
Aug 1 at 20:54
1
1
@EricWofsey, good point! I also think it is $n> 1$. As for the roots, it was not really clear that if they are only real. The hypothesis $n$ is odd serves to ensure that it has at least one real root so, perhaps, be considered complex roots.
– Lucas Corrêa
Aug 1 at 21:01
@EricWofsey, good point! I also think it is $n> 1$. As for the roots, it was not really clear that if they are only real. The hypothesis $n$ is odd serves to ensure that it has at least one real root so, perhaps, be considered complex roots.
– Lucas Corrêa
Aug 1 at 21:01
add a comment |Â
1 Answer
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You can actually also apply the IFT $n$ times on $p:mathbbR^1 rightarrow mathbbR^1$ as follows:
Proof of (iii): Let $b_0 in A$. then $p(x)=b_0$ has $n$ distinct roots $x_0,i$. As $p$ is of degree exactly $n$, the roots must all be of multiplicity 1 and therefore $p'(x_i) neq 0$. Hence, the IFT can be applied $n$ times to $p(x)$ at the positions $x_0,i$. This means we get continuous functions
$$ x_i= x_i(b), $$
defined for $|b-b_0|$ small enough, such that
$$p(x_i(b)) = b$$
and
$$x_i(b_0) = x_0,i.$$
That means that for small variations in $b$ around $b_0$ the roots of $p(x)=b$ vary also small and are still distinct if the variation is small enough. (I'm too lazy to carry out the epsilon delta formulations here ... :) )
Hence $A$ contains a neighborhood around $b_0$ and is open.
answered Aug 1 at 21:14
til
694
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You can actually also apply the IFT $n$ times on $p:mathbbR^1 rightarrow mathbbR^1$ as follows:
Proof of (iii): Let $b_0 in A$. then $p(x)=b_0$ has $n$ distinct roots $x_0,i$. As $p$ is of degree exactly $n$, the roots must all be of multiplicity 1 and therefore $p'(x_i) neq 0$. Hence, the IFT can be applied $n$ times to $p(x)$ at the positions $x_0,i$. This means we get continuous functions
$$ x_i= x_i(b), $$
defined for $|b-b_0|$ small enough, such that
$$p(x_i(b)) = b$$
and
$$x_i(b_0) = x_0,i.$$
That means that for small variations in $b$ around $b_0$ the roots of $p(x)=b$ vary also small and are still distinct if the variation is small enough. (I'm too lazy to carry out the epsilon delta formulations here ... :) )
Hence $A$ contains a neighborhood around $b_0$ and is open.
answered Aug 1 at 21:14
til
694
add a comment |Â
up vote
0
down vote
You can actually also apply the IFT $n$ times on $p:mathbbR^1 rightarrow mathbbR^1$ as follows:
Proof of (iii): Let $b_0 in A$. then $p(x)=b_0$ has $n$ distinct roots $x_0,i$. As $p$ is of degree exactly $n$, the roots must all be of multiplicity 1 and therefore $p'(x_i) neq 0$. Hence, the IFT can be applied $n$ times to $p(x)$ at the positions $x_0,i$. This means we get continuous functions
$$ x_i= x_i(b), $$
defined for $|b-b_0|$ small enough, such that
$$p(x_i(b)) = b$$
and
$$x_i(b_0) = x_0,i.$$
That means that for small variations in $b$ around $b_0$ the roots of $p(x)=b$ vary also small and are still distinct if the variation is small enough. (I'm too lazy to carry out the epsilon delta formulations here ... :) )
Hence $A$ contains a neighborhood around $b_0$ and is open.
answered Aug 1 at 21:14
til
694
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can actually also apply the IFT $n$ times on $p:mathbbR^1 rightarrow mathbbR^1$ as follows:
Proof of (iii): Let $b_0 in A$. then $p(x)=b_0$ has $n$ distinct roots $x_0,i$. As $p$ is of degree exactly $n$, the roots must all be of multiplicity 1 and therefore $p'(x_i) neq 0$. Hence, the IFT can be applied $n$ times to $p(x)$ at the positions $x_0,i$. This means we get continuous functions
$$ x_i= x_i(b), $$
defined for $|b-b_0|$ small enough, such that
$$p(x_i(b)) = b$$
and
$$x_i(b_0) = x_0,i.$$
That means that for small variations in $b$ around $b_0$ the roots of $p(x)=b$ vary also small and are still distinct if the variation is small enough. (I'm too lazy to carry out the epsilon delta formulations here ... :) )
Hence $A$ contains a neighborhood around $b_0$ and is open.
answered Aug 1 at 21:14
til
694
You can actually also apply the IFT $n$ times on $p:mathbbR^1 rightarrow mathbbR^1$ as follows:
Proof of (iii): Let $b_0 in A$. then $p(x)=b_0$ has $n$ distinct roots $x_0,i$. As $p$ is of degree exactly $n$, the roots must all be of multiplicity 1 and therefore $p'(x_i) neq 0$. Hence, the IFT can be applied $n$ times to $p(x)$ at the positions $x_0,i$. This means we get continuous functions
$$ x_i= x_i(b), $$
defined for $|b-b_0|$ small enough, such that
$$p(x_i(b)) = b$$
and
$$x_i(b_0) = x_0,i.$$
That means that for small variations in $b$ around $b_0$ the roots of $p(x)=b$ vary also small and are still distinct if the variation is small enough. (I'm too lazy to carry out the epsilon delta formulations here ... :) )
Hence $A$ contains a neighborhood around $b_0$ and is open.
answered Aug 1 at 21:14
til
694
edited Aug 1 at 21:19
answered Aug 1 at 21:14
til
694
answered Aug 1 at 21:14
til
694
answered Aug 1 at 21:14
til
694
694
add a comment |Â
add a comment |Â
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1
(ii) is false as stated. I guess it means to assume that $n>1$ and by "roots" it means real roots?
– Eric Wofsey
Aug 1 at 20:54
1
@EricWofsey, good point! I also think it is $n> 1$. As for the roots, it was not really clear that if they are only real. The hypothesis $n$ is odd serves to ensure that it has at least one real root so, perhaps, be considered complex roots.
– Lucas Corrêa
Aug 1 at 21:01