Mixing
$supsin(2pi t)_t in mathbb Q setminus mathbb Z = supsin(2pi t)_t in mathbb Q$?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 7.12,7.15
$$Re(e^2 pi i t) = sin(2 pi t)$$
For $t in mathbb Q$, $sup=max=1$, achieved only $forall t equiv 1/4 mod 2$
For $t in mathbb Z$, $sup=max...$ still $=1$ and still only $forall t equiv 1/4 mod 2$?
For $t in mathbb Q setminus mathbb Z$, $sup=max...$ still $=1$ and still only for $forall t equiv 1/4 mod 2$?
Should this perhaps be $Re(e^pi i t/2) = sin(pi t/2)$? I think that's where we say things like:
For $t in mathbb Q setminus mathbb Z$, $max$ dne, but $sup=...$ still $1$? I guess it would be 1 only $forall t equiv 1/4 mod 2, exists q_n_n in mathbb N: q_n in mathbb Q, q_n to t$ and then we discuss if such a sequence exists, possibly related to Exer 7.13?
(7.12) Where are the errors, if any?
Pf:
$leftarrow$
$$|c_n| le |Re(c_n)| + |Im(c_n)| < frac varepsilon 2 + frac varepsilon 2 < varepsilon$$
for $N = maxN_Re,N_Im$ where $N_Re$ keeps $|Re(c_n)| < frac varepsilon 2$ and $N_Im$ keeps $|Im(c_n)| < frac varepsilon 2$
$rightarrow$
$|Re(c_n)|, |Im(c_n)| le sqrtRe^2(c_n) + Im^2(c_n) = |Re(c_n) + iIm(c_n)| = |c_n| < varepsilon$
for the same $n ge N$ that keeps $|c_n| < varepsilon$
QED
(7.13) Not really complex analysis: How do you prove that Z (the set of integers) equipped with the Euclidean metric (induced from real numbers) is complete?
sequences-and-series complex-analysis trigonometry convergence supremum-and-infimum
asked Aug 2 at 9:58
BCLC
6,98221973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-12 13:09:40Z">in 4 days.
This question has not received enough attention.
 |Â
show 1 more comment
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 7.12,7.15
$$Re(e^2 pi i t) = sin(2 pi t)$$
For $t in mathbb Q$, $sup=max=1$, achieved only $forall t equiv 1/4 mod 2$
For $t in mathbb Z$, $sup=max...$ still $=1$ and still only $forall t equiv 1/4 mod 2$?
For $t in mathbb Q setminus mathbb Z$, $sup=max...$ still $=1$ and still only for $forall t equiv 1/4 mod 2$?
Should this perhaps be $Re(e^pi i t/2) = sin(pi t/2)$? I think that's where we say things like:
For $t in mathbb Q setminus mathbb Z$, $max$ dne, but $sup=...$ still $1$? I guess it would be 1 only $forall t equiv 1/4 mod 2, exists q_n_n in mathbb N: q_n in mathbb Q, q_n to t$ and then we discuss if such a sequence exists, possibly related to Exer 7.13?
(7.12) Where are the errors, if any?
Pf:
$leftarrow$
$$|c_n| le |Re(c_n)| + |Im(c_n)| < frac varepsilon 2 + frac varepsilon 2 < varepsilon$$
for $N = maxN_Re,N_Im$ where $N_Re$ keeps $|Re(c_n)| < frac varepsilon 2$ and $N_Im$ keeps $|Im(c_n)| < frac varepsilon 2$
$rightarrow$
$|Re(c_n)|, |Im(c_n)| le sqrtRe^2(c_n) + Im^2(c_n) = |Re(c_n) + iIm(c_n)| = |c_n| < varepsilon$
for the same $n ge N$ that keeps $|c_n| < varepsilon$
QED
(7.13) Not really complex analysis: How do you prove that Z (the set of integers) equipped with the Euclidean metric (induced from real numbers) is complete?
sequences-and-series complex-analysis trigonometry convergence supremum-and-infimum
asked Aug 2 at 9:58
BCLC
6,98221973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-12 13:09:40Z">in 4 days.
This question has not received enough attention.
1
What do Cauchy sequences of integers look like?!
– Lord Shark the Unknown
Aug 2 at 10:25
@LordSharktheUnknown Edited. I didn't mean to convey that I didn't have any idea. (My assumption is that that was your interpretation). Thanks
– BCLC
Aug 2 at 10:34
1
Completeness is about the convergence of Cauchy sequences; you need to prove that each Cauchy sequence of integers converges to an integer.
– Lord Shark the Unknown
Aug 2 at 10:43
@LordSharktheUnknown I'm a dumbass. I could've just looked it up. Thanks. How about the others please?
– BCLC
Aug 2 at 10:47
"$Re(e^2 pi i t) = sin(2 pi t)$"?? Should be $cos(2pi t)$
– zhw.
2 days ago
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 7.12,7.15
$$Re(e^2 pi i t) = sin(2 pi t)$$
For $t in mathbb Q$, $sup=max=1$, achieved only $forall t equiv 1/4 mod 2$
For $t in mathbb Z$, $sup=max...$ still $=1$ and still only $forall t equiv 1/4 mod 2$?
For $t in mathbb Q setminus mathbb Z$, $sup=max...$ still $=1$ and still only for $forall t equiv 1/4 mod 2$?
Should this perhaps be $Re(e^pi i t/2) = sin(pi t/2)$? I think that's where we say things like:
For $t in mathbb Q setminus mathbb Z$, $max$ dne, but $sup=...$ still $1$? I guess it would be 1 only $forall t equiv 1/4 mod 2, exists q_n_n in mathbb N: q_n in mathbb Q, q_n to t$ and then we discuss if such a sequence exists, possibly related to Exer 7.13?
(7.12) Where are the errors, if any?
Pf:
$leftarrow$
$$|c_n| le |Re(c_n)| + |Im(c_n)| < frac varepsilon 2 + frac varepsilon 2 < varepsilon$$
for $N = maxN_Re,N_Im$ where $N_Re$ keeps $|Re(c_n)| < frac varepsilon 2$ and $N_Im$ keeps $|Im(c_n)| < frac varepsilon 2$
$rightarrow$
$|Re(c_n)|, |Im(c_n)| le sqrtRe^2(c_n) + Im^2(c_n) = |Re(c_n) + iIm(c_n)| = |c_n| < varepsilon$
for the same $n ge N$ that keeps $|c_n| < varepsilon$
QED
(7.13) Not really complex analysis: How do you prove that Z (the set of integers) equipped with the Euclidean metric (induced from real numbers) is complete?
sequences-and-series complex-analysis trigonometry convergence supremum-and-infimum
asked Aug 2 at 9:58
BCLC
6,98221973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 7.12,7.15
$$Re(e^2 pi i t) = sin(2 pi t)$$
For $t in mathbb Q$, $sup=max=1$, achieved only $forall t equiv 1/4 mod 2$
For $t in mathbb Z$, $sup=max...$ still $=1$ and still only $forall t equiv 1/4 mod 2$?
For $t in mathbb Q setminus mathbb Z$, $sup=max...$ still $=1$ and still only for $forall t equiv 1/4 mod 2$?
Should this perhaps be $Re(e^pi i t/2) = sin(pi t/2)$? I think that's where we say things like:
For $t in mathbb Q setminus mathbb Z$, $max$ dne, but $sup=...$ still $1$? I guess it would be 1 only $forall t equiv 1/4 mod 2, exists q_n_n in mathbb N: q_n in mathbb Q, q_n to t$ and then we discuss if such a sequence exists, possibly related to Exer 7.13?
(7.12) Where are the errors, if any?
Pf:
$leftarrow$
$$|c_n| le |Re(c_n)| + |Im(c_n)| < frac varepsilon 2 + frac varepsilon 2 < varepsilon$$
for $N = maxN_Re,N_Im$ where $N_Re$ keeps $|Re(c_n)| < frac varepsilon 2$ and $N_Im$ keeps $|Im(c_n)| < frac varepsilon 2$
$rightarrow$
$|Re(c_n)|, |Im(c_n)| le sqrtRe^2(c_n) + Im^2(c_n) = |Re(c_n) + iIm(c_n)| = |c_n| < varepsilon$
for the same $n ge N$ that keeps $|c_n| < varepsilon$
QED
(7.13) Not really complex analysis: How do you prove that Z (the set of integers) equipped with the Euclidean metric (induced from real numbers) is complete?
sequences-and-series complex-analysis trigonometry convergence supremum-and-infimum
asked Aug 2 at 9:58
BCLC
6,98221973
edited 2 days ago
asked Aug 2 at 9:58
BCLC
6,98221973
asked Aug 2 at 9:58
BCLC
6,98221973
asked Aug 2 at 9:58
BCLC
6,98221973
6,98221973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-12 13:09:40Z">in 4 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-12 13:09:40Z">in 4 days.
This question has not received enough attention.
1
What do Cauchy sequences of integers look like?!
– Lord Shark the Unknown
Aug 2 at 10:25
@LordSharktheUnknown Edited. I didn't mean to convey that I didn't have any idea. (My assumption is that that was your interpretation). Thanks
– BCLC
Aug 2 at 10:34
1
Completeness is about the convergence of Cauchy sequences; you need to prove that each Cauchy sequence of integers converges to an integer.
– Lord Shark the Unknown
Aug 2 at 10:43
@LordSharktheUnknown I'm a dumbass. I could've just looked it up. Thanks. How about the others please?
– BCLC
Aug 2 at 10:47
"$Re(e^2 pi i t) = sin(2 pi t)$"?? Should be $cos(2pi t)$
– zhw.
2 days ago
 |Â
show 1 more comment
1
What do Cauchy sequences of integers look like?!
– Lord Shark the Unknown
Aug 2 at 10:25
@LordSharktheUnknown Edited. I didn't mean to convey that I didn't have any idea. (My assumption is that that was your interpretation). Thanks
– BCLC
Aug 2 at 10:34
1
Completeness is about the convergence of Cauchy sequences; you need to prove that each Cauchy sequence of integers converges to an integer.
– Lord Shark the Unknown
Aug 2 at 10:43
@LordSharktheUnknown I'm a dumbass. I could've just looked it up. Thanks. How about the others please?
– BCLC
Aug 2 at 10:47
"$Re(e^2 pi i t) = sin(2 pi t)$"?? Should be $cos(2pi t)$
– zhw.
2 days ago
1
1
What do Cauchy sequences of integers look like?!
– Lord Shark the Unknown
Aug 2 at 10:25
What do Cauchy sequences of integers look like?!
– Lord Shark the Unknown
Aug 2 at 10:25
@LordSharktheUnknown Edited. I didn't mean to convey that I didn't have any idea. (My assumption is that that was your interpretation). Thanks
– BCLC
Aug 2 at 10:34
@LordSharktheUnknown Edited. I didn't mean to convey that I didn't have any idea. (My assumption is that that was your interpretation). Thanks
– BCLC
Aug 2 at 10:34
1
1
Completeness is about the convergence of Cauchy sequences; you need to prove that each Cauchy sequence of integers converges to an integer.
– Lord Shark the Unknown
Aug 2 at 10:43
Completeness is about the convergence of Cauchy sequences; you need to prove that each Cauchy sequence of integers converges to an integer.
– Lord Shark the Unknown
Aug 2 at 10:43
@LordSharktheUnknown I'm a dumbass. I could've just looked it up. Thanks. How about the others please?
– BCLC
Aug 2 at 10:47
@LordSharktheUnknown I'm a dumbass. I could've just looked it up. Thanks. How about the others please?
– BCLC
Aug 2 at 10:47
"$Re(e^2 pi i t) = sin(2 pi t)$"?? Should be $cos(2pi t)$
– zhw.
2 days ago
"$Re(e^2 pi i t) = sin(2 pi t)$"?? Should be $cos(2pi t)$
– zhw.
2 days ago
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
Assuming you meant $text Re e^2pi i t= cos(2pi t),$ note first $cos(2pi t)le 1$ for any real $t,$ so $1$ is an upper bound for the set $E=cos(2pi t):tin mathbb Qsetminus Z.$ Now let $s < 1.$ Because $cos(2pi(1/n)) to 1$ as $nto infty,$ we have $cos(2pi(1/n)) >s$ for large $n.$ Thus $s$ is not an upper bound of $E.$ By definition then, $1$ is the least upper bound of $E.$
Your proof of 7.12 is not correct. In fact I can't see what you are trying to do. Let $c_n=x_n+iy_n.$ You are trying to show $c_n$ converges in $mathbb C$ to $c=x+iy$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ This is exactly the same as showing $(x_n,y_n)to (x,y)$ in $mathbb R^2$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ You must have seen a proof of the latter sometime, somewhere ...
answered 2 days ago
zhw.
65.1k42769
Thanks for pointing out cos. I'll try on my own then analyse later
– BCLC
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assuming you meant $text Re e^2pi i t= cos(2pi t),$ note first $cos(2pi t)le 1$ for any real $t,$ so $1$ is an upper bound for the set $E=cos(2pi t):tin mathbb Qsetminus Z.$ Now let $s < 1.$ Because $cos(2pi(1/n)) to 1$ as $nto infty,$ we have $cos(2pi(1/n)) >s$ for large $n.$ Thus $s$ is not an upper bound of $E.$ By definition then, $1$ is the least upper bound of $E.$
Your proof of 7.12 is not correct. In fact I can't see what you are trying to do. Let $c_n=x_n+iy_n.$ You are trying to show $c_n$ converges in $mathbb C$ to $c=x+iy$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ This is exactly the same as showing $(x_n,y_n)to (x,y)$ in $mathbb R^2$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ You must have seen a proof of the latter sometime, somewhere ...
answered 2 days ago
zhw.
65.1k42769
Thanks for pointing out cos. I'll try on my own then analyse later
– BCLC
2 days ago
add a comment |Â
up vote
0
down vote
Assuming you meant $text Re e^2pi i t= cos(2pi t),$ note first $cos(2pi t)le 1$ for any real $t,$ so $1$ is an upper bound for the set $E=cos(2pi t):tin mathbb Qsetminus Z.$ Now let $s < 1.$ Because $cos(2pi(1/n)) to 1$ as $nto infty,$ we have $cos(2pi(1/n)) >s$ for large $n.$ Thus $s$ is not an upper bound of $E.$ By definition then, $1$ is the least upper bound of $E.$
Your proof of 7.12 is not correct. In fact I can't see what you are trying to do. Let $c_n=x_n+iy_n.$ You are trying to show $c_n$ converges in $mathbb C$ to $c=x+iy$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ This is exactly the same as showing $(x_n,y_n)to (x,y)$ in $mathbb R^2$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ You must have seen a proof of the latter sometime, somewhere ...
answered 2 days ago
zhw.
65.1k42769
Thanks for pointing out cos. I'll try on my own then analyse later
– BCLC
2 days ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assuming you meant $text Re e^2pi i t= cos(2pi t),$ note first $cos(2pi t)le 1$ for any real $t,$ so $1$ is an upper bound for the set $E=cos(2pi t):tin mathbb Qsetminus Z.$ Now let $s < 1.$ Because $cos(2pi(1/n)) to 1$ as $nto infty,$ we have $cos(2pi(1/n)) >s$ for large $n.$ Thus $s$ is not an upper bound of $E.$ By definition then, $1$ is the least upper bound of $E.$
Your proof of 7.12 is not correct. In fact I can't see what you are trying to do. Let $c_n=x_n+iy_n.$ You are trying to show $c_n$ converges in $mathbb C$ to $c=x+iy$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ This is exactly the same as showing $(x_n,y_n)to (x,y)$ in $mathbb R^2$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ You must have seen a proof of the latter sometime, somewhere ...
answered 2 days ago
zhw.
65.1k42769
Assuming you meant $text Re e^2pi i t= cos(2pi t),$ note first $cos(2pi t)le 1$ for any real $t,$ so $1$ is an upper bound for the set $E=cos(2pi t):tin mathbb Qsetminus Z.$ Now let $s < 1.$ Because $cos(2pi(1/n)) to 1$ as $nto infty,$ we have $cos(2pi(1/n)) >s$ for large $n.$ Thus $s$ is not an upper bound of $E.$ By definition then, $1$ is the least upper bound of $E.$
Your proof of 7.12 is not correct. In fact I can't see what you are trying to do. Let $c_n=x_n+iy_n.$ You are trying to show $c_n$ converges in $mathbb C$ to $c=x+iy$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ This is exactly the same as showing $(x_n,y_n)to (x,y)$ in $mathbb R^2$ iff $x_nto x$ in $mathbb R $ and $y_nto y$ in $mathbb R .$ You must have seen a proof of the latter sometime, somewhere ...
answered 2 days ago
zhw.
65.1k42769
answered 2 days ago
zhw.
65.1k42769
answered 2 days ago
zhw.
65.1k42769
answered 2 days ago
zhw.
65.1k42769
65.1k42769
Thanks for pointing out cos. I'll try on my own then analyse later
– BCLC
2 days ago
add a comment |Â
Thanks for pointing out cos. I'll try on my own then analyse later
– BCLC
2 days ago
Thanks for pointing out cos. I'll try on my own then analyse later
– BCLC
2 days ago
Thanks for pointing out cos. I'll try on my own then analyse later
– BCLC
2 days ago
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%2f2869902%2fsup-sin2-pi-t-t-in-mathbb-q-setminus-mathbb-z-sup-sin2-pi-t%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
1
What do Cauchy sequences of integers look like?!
– Lord Shark the Unknown
Aug 2 at 10:25
@LordSharktheUnknown Edited. I didn't mean to convey that I didn't have any idea. (My assumption is that that was your interpretation). Thanks
– BCLC
Aug 2 at 10:34
1
Completeness is about the convergence of Cauchy sequences; you need to prove that each Cauchy sequence of integers converges to an integer.
– Lord Shark the Unknown
Aug 2 at 10:43
@LordSharktheUnknown I'm a dumbass. I could've just looked it up. Thanks. How about the others please?
– BCLC
Aug 2 at 10:47
"$Re(e^2 pi i t) = sin(2 pi t)$"?? Should be $cos(2pi t)$
– zhw.
2 days ago