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2 Exercises on Weierstrass M-test for power series
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka 7.29(c),7.30
Please point out errors.
Exer 7.29
(c) $M_k = r^k$ because $$frac^kr^k le 1 le |z^k + 1| $$
Elaboration: I believe that $|fracz^kz^k+1| le r^k$.
Pf: For $|z| le r,$ we have $$|z|^k le r^k implies frac^kr^k le 1. tag1$$
But for $|z| in [0,infty),$ $$|z^k+1| ge ||z^k|-|1|| = ||z|^k-1| ge |0-1| = 1 implies 1 le |z^k+1| tag2$$
Thus,
$$(1) wedge (2) implies frac^kr^k le 1 le |z^k+1| implies frac^k le r^k$$
$$therefore, |fracz^kz^k+1| = frac = frac^k le r^k textQED$$
Exer 7.30
Weierstrass M-Test: $|fraczw|^k le frac^kr^k = |fraczr|^k =: M_k$
real-analysis complex-analysis convergence power-series uniform-convergence
asked Aug 2 at 12:38
BCLC
6,98221973
add a comment |Â
up vote
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka 7.29(c),7.30
Please point out errors.
Exer 7.29
(c) $M_k = r^k$ because $$frac^kr^k le 1 le |z^k + 1| $$
Elaboration: I believe that $|fracz^kz^k+1| le r^k$.
Pf: For $|z| le r,$ we have $$|z|^k le r^k implies frac^kr^k le 1. tag1$$
But for $|z| in [0,infty),$ $$|z^k+1| ge ||z^k|-|1|| = ||z|^k-1| ge |0-1| = 1 implies 1 le |z^k+1| tag2$$
Thus,
$$(1) wedge (2) implies frac^kr^k le 1 le |z^k+1| implies frac^k le r^k$$
$$therefore, |fracz^kz^k+1| = frac = frac^k le r^k textQED$$
Exer 7.30
Weierstrass M-Test: $|fraczw|^k le frac^kr^k = |fraczr|^k =: M_k$
real-analysis complex-analysis convergence power-series uniform-convergence
asked Aug 2 at 12:38
BCLC
6,98221973
add a comment |Â
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 7.29(c),7.30
Please point out errors.
Exer 7.29
(c) $M_k = r^k$ because $$frac^kr^k le 1 le |z^k + 1| $$
Elaboration: I believe that $|fracz^kz^k+1| le r^k$.
Pf: For $|z| le r,$ we have $$|z|^k le r^k implies frac^kr^k le 1. tag1$$
But for $|z| in [0,infty),$ $$|z^k+1| ge ||z^k|-|1|| = ||z|^k-1| ge |0-1| = 1 implies 1 le |z^k+1| tag2$$
Thus,
$$(1) wedge (2) implies frac^kr^k le 1 le |z^k+1| implies frac^k le r^k$$
$$therefore, |fracz^kz^k+1| = frac = frac^k le r^k textQED$$
Exer 7.30
Weierstrass M-Test: $|fraczw|^k le frac^kr^k = |fraczr|^k =: M_k$
real-analysis complex-analysis convergence power-series uniform-convergence
asked Aug 2 at 12:38
BCLC
6,98221973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka 7.29(c),7.30
Please point out errors.
Exer 7.29
(c) $M_k = r^k$ because $$frac^kr^k le 1 le |z^k + 1| $$
Elaboration: I believe that $|fracz^kz^k+1| le r^k$.
Pf: For $|z| le r,$ we have $$|z|^k le r^k implies frac^kr^k le 1. tag1$$
But for $|z| in [0,infty),$ $$|z^k+1| ge ||z^k|-|1|| = ||z|^k-1| ge |0-1| = 1 implies 1 le |z^k+1| tag2$$
Thus,
$$(1) wedge (2) implies frac^kr^k le 1 le |z^k+1| implies frac^k le r^k$$
$$therefore, |fracz^kz^k+1| = frac = frac^k le r^k textQED$$
Exer 7.30
Weierstrass M-Test: $|fraczw|^k le frac^kr^k = |fraczr|^k =: M_k$
real-analysis complex-analysis convergence power-series uniform-convergence
asked Aug 2 at 12:38
BCLC
6,98221973
edited 2 days ago
asked Aug 2 at 12:38
BCLC
6,98221973
asked Aug 2 at 12:38
BCLC
6,98221973
asked Aug 2 at 12:38
BCLC
6,98221973
6,98221973
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
In exercise 7.29 (c)
$$leftlvert fracz^kz^k+1 rightrvert leq r^k$$
doesn't hold for all of the $z in overlineD[0,r]$. According to the maximum modulus principle, the maximum occurs on the boundary where $z=r expi theta$. The functions to be maximized then become
$$fracr^k,$$
and the maxima occur when $expi k theta=-1$. Hence $M_k$ really should be $$M_k:=fracr^k1-r^k .$$
Exericse 7.30 looks ok to me.
answered Aug 2 at 13:42
user1337
16.4k42989
Thanks user1337, but what exactly is wrong with my argument please?
– BCLC
Aug 4 at 2:26
@BCLC Your choice of $M_k$.
– user1337
Aug 4 at 17:22
lol I mean why? I believe it's an upper bound whose series is convergent
– BCLC
Aug 4 at 20:11
1
@BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance.
– user1337
Aug 4 at 21:02
1
@BCLC $||z|^k-1|ge|0-1|$ is not necessarily true, consider $z=0.1$
– Erlang Wiratama Surya
2 days ago
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In exercise 7.29 (c)
$$leftlvert fracz^kz^k+1 rightrvert leq r^k$$
doesn't hold for all of the $z in overlineD[0,r]$. According to the maximum modulus principle, the maximum occurs on the boundary where $z=r expi theta$. The functions to be maximized then become
$$fracr^k,$$
and the maxima occur when $expi k theta=-1$. Hence $M_k$ really should be $$M_k:=fracr^k1-r^k .$$
Exericse 7.30 looks ok to me.
answered Aug 2 at 13:42
user1337
16.4k42989
Thanks user1337, but what exactly is wrong with my argument please?
– BCLC
Aug 4 at 2:26
@BCLC Your choice of $M_k$.
– user1337
Aug 4 at 17:22
lol I mean why? I believe it's an upper bound whose series is convergent
– BCLC
Aug 4 at 20:11
1
@BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance.
– user1337
Aug 4 at 21:02
1
@BCLC $||z|^k-1|ge|0-1|$ is not necessarily true, consider $z=0.1$
– Erlang Wiratama Surya
2 days ago
 |Â
show 2 more comments
up vote
1
down vote
accepted
In exercise 7.29 (c)
$$leftlvert fracz^kz^k+1 rightrvert leq r^k$$
doesn't hold for all of the $z in overlineD[0,r]$. According to the maximum modulus principle, the maximum occurs on the boundary where $z=r expi theta$. The functions to be maximized then become
$$fracr^k,$$
and the maxima occur when $expi k theta=-1$. Hence $M_k$ really should be $$M_k:=fracr^k1-r^k .$$
Exericse 7.30 looks ok to me.
answered Aug 2 at 13:42
user1337
16.4k42989
Thanks user1337, but what exactly is wrong with my argument please?
– BCLC
Aug 4 at 2:26
@BCLC Your choice of $M_k$.
– user1337
Aug 4 at 17:22
lol I mean why? I believe it's an upper bound whose series is convergent
– BCLC
Aug 4 at 20:11
1
@BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance.
– user1337
Aug 4 at 21:02
1
@BCLC $||z|^k-1|ge|0-1|$ is not necessarily true, consider $z=0.1$
– Erlang Wiratama Surya
2 days ago
 |Â
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In exercise 7.29 (c)
$$leftlvert fracz^kz^k+1 rightrvert leq r^k$$
doesn't hold for all of the $z in overlineD[0,r]$. According to the maximum modulus principle, the maximum occurs on the boundary where $z=r expi theta$. The functions to be maximized then become
$$fracr^k,$$
and the maxima occur when $expi k theta=-1$. Hence $M_k$ really should be $$M_k:=fracr^k1-r^k .$$
Exericse 7.30 looks ok to me.
answered Aug 2 at 13:42
user1337
16.4k42989
In exercise 7.29 (c)
$$leftlvert fracz^kz^k+1 rightrvert leq r^k$$
doesn't hold for all of the $z in overlineD[0,r]$. According to the maximum modulus principle, the maximum occurs on the boundary where $z=r expi theta$. The functions to be maximized then become
$$fracr^k,$$
and the maxima occur when $expi k theta=-1$. Hence $M_k$ really should be $$M_k:=fracr^k1-r^k .$$
Exericse 7.30 looks ok to me.
answered Aug 2 at 13:42
user1337
16.4k42989
answered Aug 2 at 13:42
user1337
16.4k42989
answered Aug 2 at 13:42
user1337
16.4k42989
answered Aug 2 at 13:42
user1337
16.4k42989
16.4k42989
Thanks user1337, but what exactly is wrong with my argument please?
– BCLC
Aug 4 at 2:26
@BCLC Your choice of $M_k$.
– user1337
Aug 4 at 17:22
lol I mean why? I believe it's an upper bound whose series is convergent
– BCLC
Aug 4 at 20:11
1
@BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance.
– user1337
Aug 4 at 21:02
1
@BCLC $||z|^k-1|ge|0-1|$ is not necessarily true, consider $z=0.1$
– Erlang Wiratama Surya
2 days ago
 |Â
show 2 more comments
Thanks user1337, but what exactly is wrong with my argument please?
– BCLC
Aug 4 at 2:26
@BCLC Your choice of $M_k$.
– user1337
Aug 4 at 17:22
lol I mean why? I believe it's an upper bound whose series is convergent
– BCLC
Aug 4 at 20:11
1
@BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance.
– user1337
Aug 4 at 21:02
1
@BCLC $||z|^k-1|ge|0-1|$ is not necessarily true, consider $z=0.1$
– Erlang Wiratama Surya
2 days ago
Thanks user1337, but what exactly is wrong with my argument please?
– BCLC
Aug 4 at 2:26
Thanks user1337, but what exactly is wrong with my argument please?
– BCLC
Aug 4 at 2:26
@BCLC Your choice of $M_k$.
– user1337
Aug 4 at 17:22
@BCLC Your choice of $M_k$.
– user1337
Aug 4 at 17:22
lol I mean why? I believe it's an upper bound whose series is convergent
– BCLC
Aug 4 at 20:11
lol I mean why? I believe it's an upper bound whose series is convergent
– BCLC
Aug 4 at 20:11
1
1
@BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance.
– user1337
Aug 4 at 21:02
@BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance.
– user1337
Aug 4 at 21:02
1
1
@BCLC $||z|^k-1|ge|0-1|$ is not necessarily true, consider $z=0.1$
– Erlang Wiratama Surya
2 days ago
@BCLC $||z|^k-1|ge|0-1|$ is not necessarily true, consider $z=0.1$
– Erlang Wiratama Surya
2 days ago
 |Â
show 2 more comments
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