Prove that $T: mathbb{R}^2 rightarrow mathbb{R}^2$ is a bijection
$begingroup$
I have to prove that the following is a bijection: $ T: mathbb{R}^2 rightarrow mathbb{R}^2 $, with $T(x,y)= left(
begin{array}{c}
5x + sin(y)\
5y + arctan(x)
end{array}
right)$
Now, to prove it is surjective I composed it with $x = tan(z)$ obtaining:
$ T^1colon left]-pi /2, pi /2 right[times mathbb{R} rightarrow mathbb{R}^2 $, with $T^1(z,y)= left(
begin{array}{c}
5tan(z) + sin(y)\
5y + z
end{array}
right)
$
Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 Rightarrow z = y_0 - 5y$ and $5 tan(y_0 - 5y) + sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ infty$ and $-infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.
As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') Rightarrow |x-x'|< frac{2}{5}$ and $|y-y'|< frac{2 pi}{5}$ but I can't do any better.
My questions are:
1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)
2) Can somebody give me hints to prove the injectivity?
Thank you very much in advance!
functions
edited Dec 20 '18 at 16:37
Arthur Jr.
476
asked Aug 14 '13 at 11:10
GivAlzGivAlz
30317
$endgroup$
add a comment |
$begingroup$
I have to prove that the following is a bijection: $ T: mathbb{R}^2 rightarrow mathbb{R}^2 $, with $T(x,y)= left(
begin{array}{c}
5x + sin(y)\
5y + arctan(x)
end{array}
right)$
Now, to prove it is surjective I composed it with $x = tan(z)$ obtaining:
$ T^1colon left]-pi /2, pi /2 right[times mathbb{R} rightarrow mathbb{R}^2 $, with $T^1(z,y)= left(
begin{array}{c}
5tan(z) + sin(y)\
5y + z
end{array}
right)
$
Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 Rightarrow z = y_0 - 5y$ and $5 tan(y_0 - 5y) + sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ infty$ and $-infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.
As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') Rightarrow |x-x'|< frac{2}{5}$ and $|y-y'|< frac{2 pi}{5}$ but I can't do any better.
My questions are:
1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)
2) Can somebody give me hints to prove the injectivity?
Thank you very much in advance!
functions
edited Dec 20 '18 at 16:37
Arthur Jr.
476
asked Aug 14 '13 at 11:10
GivAlzGivAlz
30317
$endgroup$
add a comment |
$begingroup$
I have to prove that the following is a bijection: $ T: mathbb{R}^2 rightarrow mathbb{R}^2 $, with $T(x,y)= left(
begin{array}{c}
5x + sin(y)\
5y + arctan(x)
end{array}
right)$
Now, to prove it is surjective I composed it with $x = tan(z)$ obtaining:
$ T^1colon left]-pi /2, pi /2 right[times mathbb{R} rightarrow mathbb{R}^2 $, with $T^1(z,y)= left(
begin{array}{c}
5tan(z) + sin(y)\
5y + z
end{array}
right)
$
Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 Rightarrow z = y_0 - 5y$ and $5 tan(y_0 - 5y) + sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ infty$ and $-infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.
As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') Rightarrow |x-x'|< frac{2}{5}$ and $|y-y'|< frac{2 pi}{5}$ but I can't do any better.
My questions are:
1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)
2) Can somebody give me hints to prove the injectivity?
Thank you very much in advance!
functions
edited Dec 20 '18 at 16:37
Arthur Jr.
476
asked Aug 14 '13 at 11:10
GivAlzGivAlz
30317
$endgroup$
I have to prove that the following is a bijection: $ T: mathbb{R}^2 rightarrow mathbb{R}^2 $, with $T(x,y)= left(
begin{array}{c}
5x + sin(y)\
5y + arctan(x)
end{array}
right)$
Now, to prove it is surjective I composed it with $x = tan(z)$ obtaining:
$ T^1colon left]-pi /2, pi /2 right[times mathbb{R} rightarrow mathbb{R}^2 $, with $T^1(z,y)= left(
begin{array}{c}
5tan(z) + sin(y)\
5y + z
end{array}
right)
$
Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 Rightarrow z = y_0 - 5y$ and $5 tan(y_0 - 5y) + sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ infty$ and $-infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.
As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') Rightarrow |x-x'|< frac{2}{5}$ and $|y-y'|< frac{2 pi}{5}$ but I can't do any better.
My questions are:
1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)
2) Can somebody give me hints to prove the injectivity?
Thank you very much in advance!
functions
functions
edited Dec 20 '18 at 16:37
Arthur Jr.
476
asked Aug 14 '13 at 11:10
GivAlzGivAlz
30317
edited Dec 20 '18 at 16:37
Arthur Jr.
476
asked Aug 14 '13 at 11:10
GivAlzGivAlz
30317
edited Dec 20 '18 at 16:37
Arthur Jr.
476
edited Dec 20 '18 at 16:37
Arthur Jr.
476
edited Dec 20 '18 at 16:37
Arthur Jr.
476
476
asked Aug 14 '13 at 11:10
GivAlzGivAlz
30317
asked Aug 14 '13 at 11:10
GivAlzGivAlz
30317
asked Aug 14 '13 at 11:10
GivAlzGivAlz
30317
30317
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.
However, there may be a better answer to the question.
answered Aug 14 '13 at 12:05
jibounetjibounet
5,6581132
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add a comment |
$begingroup$
Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.
For the Jacobian part, compute it at $(x, y)$
$$
J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
$$
and check if the determinant of $J$ is zero for some pair $(x, y)$.
$$
text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
$$
since $|cos(y)|leq 1$.
$endgroup$
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
$begingroup$
One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.
However, there may be a better answer to the question.
answered Aug 14 '13 at 12:05
jibounetjibounet
5,6581132
$endgroup$
add a comment |
$begingroup$
One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.
However, there may be a better answer to the question.
answered Aug 14 '13 at 12:05
jibounetjibounet
5,6581132
$endgroup$
add a comment |
$begingroup$
One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.
However, there may be a better answer to the question.
answered Aug 14 '13 at 12:05
jibounetjibounet
5,6581132
$endgroup$
One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.
However, there may be a better answer to the question.
answered Aug 14 '13 at 12:05
jibounetjibounet
5,6581132
answered Aug 14 '13 at 12:05
jibounetjibounet
5,6581132
answered Aug 14 '13 at 12:05
jibounetjibounet
5,6581132
answered Aug 14 '13 at 12:05
jibounetjibounet
5,6581132
5,6581132
add a comment |
add a comment |
$begingroup$
Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.
For the Jacobian part, compute it at $(x, y)$
$$
J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
$$
and check if the determinant of $J$ is zero for some pair $(x, y)$.
$$
text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
$$
since $|cos(y)|leq 1$.
$endgroup$
add a comment |
$begingroup$
Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.
For the Jacobian part, compute it at $(x, y)$
$$
J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
$$
and check if the determinant of $J$ is zero for some pair $(x, y)$.
$$
text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
$$
since $|cos(y)|leq 1$.
$endgroup$
add a comment |
$begingroup$
Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.
For the Jacobian part, compute it at $(x, y)$
$$
J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
$$
and check if the determinant of $J$ is zero for some pair $(x, y)$.
$$
text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
$$
since $|cos(y)|leq 1$.
$endgroup$
Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.
For the Jacobian part, compute it at $(x, y)$
$$
J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
$$
and check if the determinant of $J$ is zero for some pair $(x, y)$.
$$
text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
$$
since $|cos(y)|leq 1$.
edited Aug 14 '13 at 12:18
answered Aug 14 '13 at 12:11
user51196
add a comment |
add a comment |
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