Mixing
Bijection maps from $(0,1)$ to the real numbers
$begingroup$
I'm wondering why $ f(x) = e^{2pi i x + pi/2}$ is a bijection from $(0,1)$ to the real numbers when the domain is $(0,1)$. Is $ f(x) = e^{2pi i x}$ also a bijection? Detailed explanation is appreciated.
functions elementary-set-theory
edited Jan 24 at 21:14
Andrés E. Caicedo
65.7k8160250
asked Jan 24 at 20:55
bt203bt203
114
$endgroup$
add a comment |
$begingroup$
I'm wondering why $ f(x) = e^{2pi i x + pi/2}$ is a bijection from $(0,1)$ to the real numbers when the domain is $(0,1)$. Is $ f(x) = e^{2pi i x}$ also a bijection? Detailed explanation is appreciated.
functions elementary-set-theory
edited Jan 24 at 21:14
Andrés E. Caicedo
65.7k8160250
asked Jan 24 at 20:55
bt203bt203
114
$endgroup$
1
$begingroup$
Wait!!! $f(x) = e^{2pi i x + pi/2} in mathbb{C} neq mathbb{R}$... To have a bijection, I think you should define a function $f : (0,1) to mathbb{R}$, but in this case you defined $f : (0,1) to mathbb{C}$.
$endgroup$
– the_candyman
Jan 24 at 20:58
$begingroup$
This is not a bijection from $(0,1)to mathbb{R}$ as user @the_candyman points out, your function is from $(0,1)to mathbb{C}$.
$endgroup$
– LoveTooNap29
Jan 24 at 21:14
add a comment |
$begingroup$
I'm wondering why $ f(x) = e^{2pi i x + pi/2}$ is a bijection from $(0,1)$ to the real numbers when the domain is $(0,1)$. Is $ f(x) = e^{2pi i x}$ also a bijection? Detailed explanation is appreciated.
functions elementary-set-theory
edited Jan 24 at 21:14
Andrés E. Caicedo
65.7k8160250
asked Jan 24 at 20:55
bt203bt203
114
$endgroup$
I'm wondering why $ f(x) = e^{2pi i x + pi/2}$ is a bijection from $(0,1)$ to the real numbers when the domain is $(0,1)$. Is $ f(x) = e^{2pi i x}$ also a bijection? Detailed explanation is appreciated.
functions elementary-set-theory
functions elementary-set-theory
edited Jan 24 at 21:14
Andrés E. Caicedo
65.7k8160250
asked Jan 24 at 20:55
bt203bt203
114
edited Jan 24 at 21:14
Andrés E. Caicedo
65.7k8160250
asked Jan 24 at 20:55
bt203bt203
114
edited Jan 24 at 21:14
Andrés E. Caicedo
65.7k8160250
edited Jan 24 at 21:14
Andrés E. Caicedo
65.7k8160250
edited Jan 24 at 21:14
Andrés E. Caicedo
65.7k8160250
65.7k8160250
asked Jan 24 at 20:55
bt203bt203
114
asked Jan 24 at 20:55
bt203bt203
114
asked Jan 24 at 20:55
bt203bt203
114
114
1
$begingroup$
Wait!!! $f(x) = e^{2pi i x + pi/2} in mathbb{C} neq mathbb{R}$... To have a bijection, I think you should define a function $f : (0,1) to mathbb{R}$, but in this case you defined $f : (0,1) to mathbb{C}$.
$endgroup$
– the_candyman
Jan 24 at 20:58
$begingroup$
This is not a bijection from $(0,1)to mathbb{R}$ as user @the_candyman points out, your function is from $(0,1)to mathbb{C}$.
$endgroup$
– LoveTooNap29
Jan 24 at 21:14
add a comment |
1
$begingroup$
Wait!!! $f(x) = e^{2pi i x + pi/2} in mathbb{C} neq mathbb{R}$... To have a bijection, I think you should define a function $f : (0,1) to mathbb{R}$, but in this case you defined $f : (0,1) to mathbb{C}$.
$endgroup$
– the_candyman
Jan 24 at 20:58
$begingroup$
This is not a bijection from $(0,1)to mathbb{R}$ as user @the_candyman points out, your function is from $(0,1)to mathbb{C}$.
$endgroup$
– LoveTooNap29
Jan 24 at 21:14
1
1
$begingroup$
Wait!!! $f(x) = e^{2pi i x + pi/2} in mathbb{C} neq mathbb{R}$... To have a bijection, I think you should define a function $f : (0,1) to mathbb{R}$, but in this case you defined $f : (0,1) to mathbb{C}$.
$endgroup$
– the_candyman
Jan 24 at 20:58
$begingroup$
Wait!!! $f(x) = e^{2pi i x + pi/2} in mathbb{C} neq mathbb{R}$... To have a bijection, I think you should define a function $f : (0,1) to mathbb{R}$, but in this case you defined $f : (0,1) to mathbb{C}$.
$endgroup$
– the_candyman
Jan 24 at 20:58
$begingroup$
This is not a bijection from $(0,1)to mathbb{R}$ as user @the_candyman points out, your function is from $(0,1)to mathbb{C}$.
$endgroup$
– LoveTooNap29
Jan 24 at 21:14
$begingroup$
This is not a bijection from $(0,1)to mathbb{R}$ as user @the_candyman points out, your function is from $(0,1)to mathbb{C}$.
$endgroup$
– LoveTooNap29
Jan 24 at 21:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you apply Euler's formula you get
$$e^{2pi i x}=cos {2 pi x}+isin{2 pi x}$$
so the result is not real unless $x$ is an integer or half integer. It cannot be a bijection. Adding $frac pi 2$ to the exponent multiplies the right side by $e^{pi/2}$ but that does not change the argument.
answered Jan 24 at 21:02
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
You have
$$
f(x)
= e^{pi/2} left[ e^{2pi i x} right]
= e^{pi/2} left[ cos(2pi x) + isin(2pi x) right]
$$
which is mapping $(0,1)$ to a circle in the complex plane with radius $e^{pi/2}$. Take note that since neither $0$ nor $1$ are in the domain, the point $1+0i$ will not be included in the image.
This is indeed a bijection between the real interval and the indicated part of the complex circle, but not sure this is what you expect.
Your suggested curve is a similar circle with unit radius...
answered Jan 24 at 21:00
gt6989bgt6989b
34.9k22557
$endgroup$
$begingroup$
Yes. That's what my question is supposed to ask.
$endgroup$
– bt203
Jan 24 at 21:05
$begingroup$
But is $ f(x) = e^{2pi i x}$ also a bijection?
$endgroup$
– bt203
Jan 24 at 21:08
$begingroup$
@bt203 I am not doing your work for you. There is enough information in this answer to figure it our yourself. If there is a particular part of this problem you do not understand, please ask again about which part.
$endgroup$
– gt6989b
Jan 24 at 21:11
$begingroup$
I think it is. My reasoning is that $ f(x) = e^{2pi i x}$ is a bijection from $(0,1)$ to the unit circle. Then I can define a second bijection from the points on the unit circle except $1+0i$ to the real numbers. Is that correct?
$endgroup$
– bt203
Jan 24 at 21:17
$begingroup$
@bt203 correct indeed
$endgroup$
– gt6989b
Jan 24 at 22:21
add a comment |
$begingroup$
A second step is needed to establish bijection. Hint: think about how you can establish a bijection geometrically between the unit circle(minus one point) and the real numbers
answered Jan 24 at 21:08
Tian HeTian He
277
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you apply Euler's formula you get
$$e^{2pi i x}=cos {2 pi x}+isin{2 pi x}$$
so the result is not real unless $x$ is an integer or half integer. It cannot be a bijection. Adding $frac pi 2$ to the exponent multiplies the right side by $e^{pi/2}$ but that does not change the argument.
answered Jan 24 at 21:02
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
If you apply Euler's formula you get
$$e^{2pi i x}=cos {2 pi x}+isin{2 pi x}$$
so the result is not real unless $x$ is an integer or half integer. It cannot be a bijection. Adding $frac pi 2$ to the exponent multiplies the right side by $e^{pi/2}$ but that does not change the argument.
answered Jan 24 at 21:02
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
If you apply Euler's formula you get
$$e^{2pi i x}=cos {2 pi x}+isin{2 pi x}$$
so the result is not real unless $x$ is an integer or half integer. It cannot be a bijection. Adding $frac pi 2$ to the exponent multiplies the right side by $e^{pi/2}$ but that does not change the argument.
answered Jan 24 at 21:02
Ross MillikanRoss Millikan
299k24200374
$endgroup$
If you apply Euler's formula you get
$$e^{2pi i x}=cos {2 pi x}+isin{2 pi x}$$
so the result is not real unless $x$ is an integer or half integer. It cannot be a bijection. Adding $frac pi 2$ to the exponent multiplies the right side by $e^{pi/2}$ but that does not change the argument.
answered Jan 24 at 21:02
Ross MillikanRoss Millikan
299k24200374
answered Jan 24 at 21:02
Ross MillikanRoss Millikan
299k24200374
answered Jan 24 at 21:02
Ross MillikanRoss Millikan
299k24200374
answered Jan 24 at 21:02
Ross MillikanRoss Millikan
299k24200374
299k24200374
add a comment |
add a comment |
$begingroup$
You have
$$
f(x)
= e^{pi/2} left[ e^{2pi i x} right]
= e^{pi/2} left[ cos(2pi x) + isin(2pi x) right]
$$
which is mapping $(0,1)$ to a circle in the complex plane with radius $e^{pi/2}$. Take note that since neither $0$ nor $1$ are in the domain, the point $1+0i$ will not be included in the image.
This is indeed a bijection between the real interval and the indicated part of the complex circle, but not sure this is what you expect.
Your suggested curve is a similar circle with unit radius...
answered Jan 24 at 21:00
gt6989bgt6989b
34.9k22557
$endgroup$
$begingroup$
Yes. That's what my question is supposed to ask.
$endgroup$
– bt203
Jan 24 at 21:05
$begingroup$
But is $ f(x) = e^{2pi i x}$ also a bijection?
$endgroup$
– bt203
Jan 24 at 21:08
$begingroup$
@bt203 I am not doing your work for you. There is enough information in this answer to figure it our yourself. If there is a particular part of this problem you do not understand, please ask again about which part.
$endgroup$
– gt6989b
Jan 24 at 21:11
$begingroup$
I think it is. My reasoning is that $ f(x) = e^{2pi i x}$ is a bijection from $(0,1)$ to the unit circle. Then I can define a second bijection from the points on the unit circle except $1+0i$ to the real numbers. Is that correct?
$endgroup$
– bt203
Jan 24 at 21:17
$begingroup$
@bt203 correct indeed
$endgroup$
– gt6989b
Jan 24 at 22:21
add a comment |
$begingroup$
You have
$$
f(x)
= e^{pi/2} left[ e^{2pi i x} right]
= e^{pi/2} left[ cos(2pi x) + isin(2pi x) right]
$$
which is mapping $(0,1)$ to a circle in the complex plane with radius $e^{pi/2}$. Take note that since neither $0$ nor $1$ are in the domain, the point $1+0i$ will not be included in the image.
This is indeed a bijection between the real interval and the indicated part of the complex circle, but not sure this is what you expect.
Your suggested curve is a similar circle with unit radius...
answered Jan 24 at 21:00
gt6989bgt6989b
34.9k22557
$endgroup$
$begingroup$
Yes. That's what my question is supposed to ask.
$endgroup$
– bt203
Jan 24 at 21:05
$begingroup$
But is $ f(x) = e^{2pi i x}$ also a bijection?
$endgroup$
– bt203
Jan 24 at 21:08
$begingroup$
@bt203 I am not doing your work for you. There is enough information in this answer to figure it our yourself. If there is a particular part of this problem you do not understand, please ask again about which part.
$endgroup$
– gt6989b
Jan 24 at 21:11
$begingroup$
I think it is. My reasoning is that $ f(x) = e^{2pi i x}$ is a bijection from $(0,1)$ to the unit circle. Then I can define a second bijection from the points on the unit circle except $1+0i$ to the real numbers. Is that correct?
$endgroup$
– bt203
Jan 24 at 21:17
$begingroup$
@bt203 correct indeed
$endgroup$
– gt6989b
Jan 24 at 22:21
add a comment |
$begingroup$
You have
$$
f(x)
= e^{pi/2} left[ e^{2pi i x} right]
= e^{pi/2} left[ cos(2pi x) + isin(2pi x) right]
$$
which is mapping $(0,1)$ to a circle in the complex plane with radius $e^{pi/2}$. Take note that since neither $0$ nor $1$ are in the domain, the point $1+0i$ will not be included in the image.
This is indeed a bijection between the real interval and the indicated part of the complex circle, but not sure this is what you expect.
Your suggested curve is a similar circle with unit radius...
answered Jan 24 at 21:00
gt6989bgt6989b
34.9k22557
$endgroup$
You have
$$
f(x)
= e^{pi/2} left[ e^{2pi i x} right]
= e^{pi/2} left[ cos(2pi x) + isin(2pi x) right]
$$
which is mapping $(0,1)$ to a circle in the complex plane with radius $e^{pi/2}$. Take note that since neither $0$ nor $1$ are in the domain, the point $1+0i$ will not be included in the image.
This is indeed a bijection between the real interval and the indicated part of the complex circle, but not sure this is what you expect.
Your suggested curve is a similar circle with unit radius...
answered Jan 24 at 21:00
gt6989bgt6989b
34.9k22557
edited Jan 24 at 21:07
answered Jan 24 at 21:00
gt6989bgt6989b
34.9k22557
answered Jan 24 at 21:00
gt6989bgt6989b
34.9k22557
answered Jan 24 at 21:00
gt6989bgt6989b
34.9k22557
34.9k22557
$begingroup$
Yes. That's what my question is supposed to ask.
$endgroup$
– bt203
Jan 24 at 21:05
$begingroup$
But is $ f(x) = e^{2pi i x}$ also a bijection?
$endgroup$
– bt203
Jan 24 at 21:08
$begingroup$
@bt203 I am not doing your work for you. There is enough information in this answer to figure it our yourself. If there is a particular part of this problem you do not understand, please ask again about which part.
$endgroup$
– gt6989b
Jan 24 at 21:11
$begingroup$
I think it is. My reasoning is that $ f(x) = e^{2pi i x}$ is a bijection from $(0,1)$ to the unit circle. Then I can define a second bijection from the points on the unit circle except $1+0i$ to the real numbers. Is that correct?
$endgroup$
– bt203
Jan 24 at 21:17
$begingroup$
@bt203 correct indeed
$endgroup$
– gt6989b
Jan 24 at 22:21
add a comment |
$begingroup$
Yes. That's what my question is supposed to ask.
$endgroup$
– bt203
Jan 24 at 21:05
$begingroup$
But is $ f(x) = e^{2pi i x}$ also a bijection?
$endgroup$
– bt203
Jan 24 at 21:08
$begingroup$
@bt203 I am not doing your work for you. There is enough information in this answer to figure it our yourself. If there is a particular part of this problem you do not understand, please ask again about which part.
$endgroup$
– gt6989b
Jan 24 at 21:11
$begingroup$
I think it is. My reasoning is that $ f(x) = e^{2pi i x}$ is a bijection from $(0,1)$ to the unit circle. Then I can define a second bijection from the points on the unit circle except $1+0i$ to the real numbers. Is that correct?
$endgroup$
– bt203
Jan 24 at 21:17
$begingroup$
@bt203 correct indeed
$endgroup$
– gt6989b
Jan 24 at 22:21
$begingroup$
Yes. That's what my question is supposed to ask.
$endgroup$
– bt203
Jan 24 at 21:05
$begingroup$
Yes. That's what my question is supposed to ask.
$endgroup$
– bt203
Jan 24 at 21:05
$begingroup$
But is $ f(x) = e^{2pi i x}$ also a bijection?
$endgroup$
– bt203
Jan 24 at 21:08
$begingroup$
But is $ f(x) = e^{2pi i x}$ also a bijection?
$endgroup$
– bt203
Jan 24 at 21:08
$begingroup$
@bt203 I am not doing your work for you. There is enough information in this answer to figure it our yourself. If there is a particular part of this problem you do not understand, please ask again about which part.
$endgroup$
– gt6989b
Jan 24 at 21:11
$begingroup$
@bt203 I am not doing your work for you. There is enough information in this answer to figure it our yourself. If there is a particular part of this problem you do not understand, please ask again about which part.
$endgroup$
– gt6989b
Jan 24 at 21:11
$begingroup$
I think it is. My reasoning is that $ f(x) = e^{2pi i x}$ is a bijection from $(0,1)$ to the unit circle. Then I can define a second bijection from the points on the unit circle except $1+0i$ to the real numbers. Is that correct?
$endgroup$
– bt203
Jan 24 at 21:17
$begingroup$
I think it is. My reasoning is that $ f(x) = e^{2pi i x}$ is a bijection from $(0,1)$ to the unit circle. Then I can define a second bijection from the points on the unit circle except $1+0i$ to the real numbers. Is that correct?
$endgroup$
– bt203
Jan 24 at 21:17
$begingroup$
@bt203 correct indeed
$endgroup$
– gt6989b
Jan 24 at 22:21
$begingroup$
@bt203 correct indeed
$endgroup$
– gt6989b
Jan 24 at 22:21
add a comment |
$begingroup$
A second step is needed to establish bijection. Hint: think about how you can establish a bijection geometrically between the unit circle(minus one point) and the real numbers
answered Jan 24 at 21:08
Tian HeTian He
277
$endgroup$
add a comment |
$begingroup$
A second step is needed to establish bijection. Hint: think about how you can establish a bijection geometrically between the unit circle(minus one point) and the real numbers
answered Jan 24 at 21:08
Tian HeTian He
277
$endgroup$
add a comment |
$begingroup$
A second step is needed to establish bijection. Hint: think about how you can establish a bijection geometrically between the unit circle(minus one point) and the real numbers
answered Jan 24 at 21:08
Tian HeTian He
277
$endgroup$
A second step is needed to establish bijection. Hint: think about how you can establish a bijection geometrically between the unit circle(minus one point) and the real numbers
answered Jan 24 at 21:08
Tian HeTian He
277
answered Jan 24 at 21:08
Tian HeTian He
277
answered Jan 24 at 21:08
Tian HeTian He
277
answered Jan 24 at 21:08
Tian HeTian He
277
277
add a comment |
add a comment |
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$begingroup$
Wait!!! $f(x) = e^{2pi i x + pi/2} in mathbb{C} neq mathbb{R}$... To have a bijection, I think you should define a function $f : (0,1) to mathbb{R}$, but in this case you defined $f : (0,1) to mathbb{C}$.
$endgroup$
– the_candyman
Jan 24 at 20:58
$begingroup$
This is not a bijection from $(0,1)to mathbb{R}$ as user @the_candyman points out, your function is from $(0,1)to mathbb{C}$.
$endgroup$
– LoveTooNap29
Jan 24 at 21:14