Using the Weierstrass M-test, show that the series converges uniformly on the given domain
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
$sum_k geq 0 fracz^kz^k+1$ on the domain $overlineD[0, r]$, where $0 leq r < 1$
I'm honestly not sure how to do this. My text mentions the Weierstrass M-test but the example they gave after stating it uses a completely different method (looks like a repeat of a previous example) and looks nothing like the M-test.
complex-analysis
asked May 27 '16 at 21:43
Wolfram Hicks
362
add a comment |Â
up vote
4
down vote
favorite
$sum_k geq 0 fracz^kz^k+1$ on the domain $overlineD[0, r]$, where $0 leq r < 1$
I'm honestly not sure how to do this. My text mentions the Weierstrass M-test but the example they gave after stating it uses a completely different method (looks like a repeat of a previous example) and looks nothing like the M-test.
complex-analysis
asked May 27 '16 at 21:43
Wolfram Hicks
362
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$sum_k geq 0 fracz^kz^k+1$ on the domain $overlineD[0, r]$, where $0 leq r < 1$
I'm honestly not sure how to do this. My text mentions the Weierstrass M-test but the example they gave after stating it uses a completely different method (looks like a repeat of a previous example) and looks nothing like the M-test.
complex-analysis
asked May 27 '16 at 21:43
Wolfram Hicks
362
$sum_k geq 0 fracz^kz^k+1$ on the domain $overlineD[0, r]$, where $0 leq r < 1$
I'm honestly not sure how to do this. My text mentions the Weierstrass M-test but the example they gave after stating it uses a completely different method (looks like a repeat of a previous example) and looks nothing like the M-test.
complex-analysis
asked May 27 '16 at 21:43
Wolfram Hicks
362
asked May 27 '16 at 21:43
Wolfram Hicks
362
asked May 27 '16 at 21:43
Wolfram Hicks
362
asked May 27 '16 at 21:43
Wolfram Hicks
362
362
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
From the notation and complex analysis tag I would guess we should assume $zin mathbb C,|z|le r.$ For $k>0,$ we can then say $|1+z^k|ge 1-|z^k| = 1-|z|^k ge 1- |z| ge 1-r.$ Thus
$$left |fracz^k1+z^kright | = fracz^k le fracr^k1-r.$$
Since $sum_k=1^infty r^k/(1-r) < 1/(1-r)^2,$ we have uniform convergence in $|z|le r$ by Weierstrass M.
answered May 27 '16 at 22:21
zhw.
66.1k42870
zhw., do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC user1337 has a valid point. Your bound on $$left |fracz^k1+z^kright |$$ was not correct.
– zhw.
Aug 8 at 15:15
Right right, but I meant to ask about the $fracr^k1-r^k$. Anyhoo, Mark Viola answered. Thanks anyhoo!
– BCLC
Aug 8 at 15:17
add a comment |Â
up vote
4
down vote
For $|z|in [0,r]$, with $0le r<1$, we have
$$left|fracz^k1+z^kright|le fracr^k1-r$$
and $sum_k=0^infty r^k=frac11-r<infty$
answered May 27 '16 at 21:48
Mark Viola
126k1172167
Mark Viola, do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC That solution is fine. But note that for $0le r<1$, $r^k<r$ so that $fracr^k1-r^kle fracr^k1-r$.
– Mark Viola
Aug 8 at 14:37
Thanks Mark Viola! ^-^
– BCLC
Aug 8 at 14:38
1
@BCLC You are welcome.
– Mark Viola
Aug 8 at 14:39
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
From the notation and complex analysis tag I would guess we should assume $zin mathbb C,|z|le r.$ For $k>0,$ we can then say $|1+z^k|ge 1-|z^k| = 1-|z|^k ge 1- |z| ge 1-r.$ Thus
$$left |fracz^k1+z^kright | = fracz^k le fracr^k1-r.$$
Since $sum_k=1^infty r^k/(1-r) < 1/(1-r)^2,$ we have uniform convergence in $|z|le r$ by Weierstrass M.
answered May 27 '16 at 22:21
zhw.
66.1k42870
zhw., do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC user1337 has a valid point. Your bound on $$left |fracz^k1+z^kright |$$ was not correct.
– zhw.
Aug 8 at 15:15
Right right, but I meant to ask about the $fracr^k1-r^k$. Anyhoo, Mark Viola answered. Thanks anyhoo!
– BCLC
Aug 8 at 15:17
add a comment |Â
up vote
4
down vote
From the notation and complex analysis tag I would guess we should assume $zin mathbb C,|z|le r.$ For $k>0,$ we can then say $|1+z^k|ge 1-|z^k| = 1-|z|^k ge 1- |z| ge 1-r.$ Thus
$$left |fracz^k1+z^kright | = fracz^k le fracr^k1-r.$$
Since $sum_k=1^infty r^k/(1-r) < 1/(1-r)^2,$ we have uniform convergence in $|z|le r$ by Weierstrass M.
answered May 27 '16 at 22:21
zhw.
66.1k42870
zhw., do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC user1337 has a valid point. Your bound on $$left |fracz^k1+z^kright |$$ was not correct.
– zhw.
Aug 8 at 15:15
Right right, but I meant to ask about the $fracr^k1-r^k$. Anyhoo, Mark Viola answered. Thanks anyhoo!
– BCLC
Aug 8 at 15:17
add a comment |Â
up vote
4
down vote
up vote
4
down vote
From the notation and complex analysis tag I would guess we should assume $zin mathbb C,|z|le r.$ For $k>0,$ we can then say $|1+z^k|ge 1-|z^k| = 1-|z|^k ge 1- |z| ge 1-r.$ Thus
$$left |fracz^k1+z^kright | = fracz^k le fracr^k1-r.$$
Since $sum_k=1^infty r^k/(1-r) < 1/(1-r)^2,$ we have uniform convergence in $|z|le r$ by Weierstrass M.
answered May 27 '16 at 22:21
zhw.
66.1k42870
From the notation and complex analysis tag I would guess we should assume $zin mathbb C,|z|le r.$ For $k>0,$ we can then say $|1+z^k|ge 1-|z^k| = 1-|z|^k ge 1- |z| ge 1-r.$ Thus
$$left |fracz^k1+z^kright | = fracz^k le fracr^k1-r.$$
Since $sum_k=1^infty r^k/(1-r) < 1/(1-r)^2,$ we have uniform convergence in $|z|le r$ by Weierstrass M.
answered May 27 '16 at 22:21
zhw.
66.1k42870
edited May 28 '16 at 16:52
answered May 27 '16 at 22:21
zhw.
66.1k42870
answered May 27 '16 at 22:21
zhw.
66.1k42870
answered May 27 '16 at 22:21
zhw.
66.1k42870
66.1k42870
zhw., do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC user1337 has a valid point. Your bound on $$left |fracz^k1+z^kright |$$ was not correct.
– zhw.
Aug 8 at 15:15
Right right, but I meant to ask about the $fracr^k1-r^k$. Anyhoo, Mark Viola answered. Thanks anyhoo!
– BCLC
Aug 8 at 15:17
add a comment |Â
zhw., do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC user1337 has a valid point. Your bound on $$left |fracz^k1+z^kright |$$ was not correct.
– zhw.
Aug 8 at 15:15
Right right, but I meant to ask about the $fracr^k1-r^k$. Anyhoo, Mark Viola answered. Thanks anyhoo!
– BCLC
Aug 8 at 15:17
zhw., do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
zhw., do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
1
@BCLC user1337 has a valid point. Your bound on $$left |fracz^k1+z^kright |$$ was not correct.
– zhw.
Aug 8 at 15:15
@BCLC user1337 has a valid point. Your bound on $$left |fracz^k1+z^kright |$$ was not correct.
– zhw.
Aug 8 at 15:15
Right right, but I meant to ask about the $fracr^k1-r^k$. Anyhoo, Mark Viola answered. Thanks anyhoo!
– BCLC
Aug 8 at 15:17
Right right, but I meant to ask about the $fracr^k1-r^k$. Anyhoo, Mark Viola answered. Thanks anyhoo!
– BCLC
Aug 8 at 15:17
add a comment |Â
up vote
4
down vote
For $|z|in [0,r]$, with $0le r<1$, we have
$$left|fracz^k1+z^kright|le fracr^k1-r$$
and $sum_k=0^infty r^k=frac11-r<infty$
answered May 27 '16 at 21:48
Mark Viola
126k1172167
Mark Viola, do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC That solution is fine. But note that for $0le r<1$, $r^k<r$ so that $fracr^k1-r^kle fracr^k1-r$.
– Mark Viola
Aug 8 at 14:37
Thanks Mark Viola! ^-^
– BCLC
Aug 8 at 14:38
1
@BCLC You are welcome.
– Mark Viola
Aug 8 at 14:39
add a comment |Â
up vote
4
down vote
For $|z|in [0,r]$, with $0le r<1$, we have
$$left|fracz^k1+z^kright|le fracr^k1-r$$
and $sum_k=0^infty r^k=frac11-r<infty$
answered May 27 '16 at 21:48
Mark Viola
126k1172167
Mark Viola, do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC That solution is fine. But note that for $0le r<1$, $r^k<r$ so that $fracr^k1-r^kle fracr^k1-r$.
– Mark Viola
Aug 8 at 14:37
Thanks Mark Viola! ^-^
– BCLC
Aug 8 at 14:38
1
@BCLC You are welcome.
– Mark Viola
Aug 8 at 14:39
add a comment |Â
up vote
4
down vote
up vote
4
down vote
For $|z|in [0,r]$, with $0le r<1$, we have
$$left|fracz^k1+z^kright|le fracr^k1-r$$
and $sum_k=0^infty r^k=frac11-r<infty$
answered May 27 '16 at 21:48
Mark Viola
126k1172167
For $|z|in [0,r]$, with $0le r<1$, we have
$$left|fracz^k1+z^kright|le fracr^k1-r$$
and $sum_k=0^infty r^k=frac11-r<infty$
answered May 27 '16 at 21:48
Mark Viola
126k1172167
edited Aug 8 at 14:33
answered May 27 '16 at 21:48
Mark Viola
126k1172167
answered May 27 '16 at 21:48
Mark Viola
126k1172167
answered May 27 '16 at 21:48
Mark Viola
126k1172167
126k1172167
Mark Viola, do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC That solution is fine. But note that for $0le r<1$, $r^k<r$ so that $fracr^k1-r^kle fracr^k1-r$.
– Mark Viola
Aug 8 at 14:37
Thanks Mark Viola! ^-^
– BCLC
Aug 8 at 14:38
1
@BCLC You are welcome.
– Mark Viola
Aug 8 at 14:39
add a comment |Â
Mark Viola, do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
@BCLC That solution is fine. But note that for $0le r<1$, $r^k<r$ so that $fracr^k1-r^kle fracr^k1-r$.
– Mark Viola
Aug 8 at 14:37
Thanks Mark Viola! ^-^
– BCLC
Aug 8 at 14:38
1
@BCLC You are welcome.
– Mark Viola
Aug 8 at 14:39
Mark Viola, do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
Mark Viola, do you disagree with user1337? math.stackexchange.com/questions/2870028
– BCLC
Aug 8 at 12:51
1
1
@BCLC That solution is fine. But note that for $0le r<1$, $r^k<r$ so that $fracr^k1-r^kle fracr^k1-r$.
– Mark Viola
Aug 8 at 14:37
@BCLC That solution is fine. But note that for $0le r<1$, $r^k<r$ so that $fracr^k1-r^kle fracr^k1-r$.
– Mark Viola
Aug 8 at 14:37
Thanks Mark Viola! ^-^
– BCLC
Aug 8 at 14:38
Thanks Mark Viola! ^-^
– BCLC
Aug 8 at 14:38
1
1
@BCLC You are welcome.
– Mark Viola
Aug 8 at 14:39
@BCLC You are welcome.
– Mark Viola
Aug 8 at 14:39
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%2f1802710%2fusing-the-weierstrass-m-test-show-that-the-series-converges-uniformly-on-the-gi%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
