How to argue that a series that isn't a power series is analytic?
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Let $f(z)=sum_n=0^inftyfracz^n1-z^n$. Then $f$ converges in the unit disc. I want to show that it is analytic in this region as well, but since it is not a power series I don't have any theorems to apply. My only thought was to try and use similar reasoning as with power series: show that converges is uniform so that we can switch integral and sum to show that the integral of the series is 0, then apply Morera's Theorem. But the problem is, according to Wolfram Alpha, $int_gammafracz^n1-z^n=int_0^2pifrace^intie^it1-e^intdt$ doesn't converge.
EDIT: To apply Morera's Theorem $gamma$ should be closed path contained in the unit circle. (It probably should have been obvious that the integral doesn't converge since the unit circle contains 1...). So the above strategy does works (yay!) since the function is holomorphic in any closed disk with radius less than 1.
sequences-and-series complex-analysis analytic-functions
asked Aug 11 at 15:08
SihOASHoihd
18112
 |Â
show 4 more comments
up vote
3
down vote
favorite
Let $f(z)=sum_n=0^inftyfracz^n1-z^n$. Then $f$ converges in the unit disc. I want to show that it is analytic in this region as well, but since it is not a power series I don't have any theorems to apply. My only thought was to try and use similar reasoning as with power series: show that converges is uniform so that we can switch integral and sum to show that the integral of the series is 0, then apply Morera's Theorem. But the problem is, according to Wolfram Alpha, $int_gammafracz^n1-z^n=int_0^2pifrace^intie^it1-e^intdt$ doesn't converge.
EDIT: To apply Morera's Theorem $gamma$ should be closed path contained in the unit circle. (It probably should have been obvious that the integral doesn't converge since the unit circle contains 1...). So the above strategy does works (yay!) since the function is holomorphic in any closed disk with radius less than 1.
sequences-and-series complex-analysis analytic-functions
asked Aug 11 at 15:08
SihOASHoihd
18112
7
Of course the sum should start at $n=1$, not $n=0$. To show this is a power series, note that, if $|z|<1$, $$f(z)=sum_n=1^infty z^nsum_k=0^infty z^kn=sum_ngeqslant1,kgeqslant0z^n(k+1)=sum_n=1^inftysigma_1(n)z^n$$ where $sigma_1(n)$ denotes the number of divisors of $n$.
– Did
Aug 11 at 15:15
How are you getting the first equality?
– SihOASHoihd
Aug 11 at 15:23
Any one of the five people who upvoted should feel free to answer
– SihOASHoihd
Aug 11 at 15:33
@Did How did you get the last equality?
– Szeto
Aug 11 at 15:35
@SihOASHoihd That’s just geometric series.
– Szeto
Aug 11 at 15:35
 |Â
show 4 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $f(z)=sum_n=0^inftyfracz^n1-z^n$. Then $f$ converges in the unit disc. I want to show that it is analytic in this region as well, but since it is not a power series I don't have any theorems to apply. My only thought was to try and use similar reasoning as with power series: show that converges is uniform so that we can switch integral and sum to show that the integral of the series is 0, then apply Morera's Theorem. But the problem is, according to Wolfram Alpha, $int_gammafracz^n1-z^n=int_0^2pifrace^intie^it1-e^intdt$ doesn't converge.
EDIT: To apply Morera's Theorem $gamma$ should be closed path contained in the unit circle. (It probably should have been obvious that the integral doesn't converge since the unit circle contains 1...). So the above strategy does works (yay!) since the function is holomorphic in any closed disk with radius less than 1.
sequences-and-series complex-analysis analytic-functions
asked Aug 11 at 15:08
SihOASHoihd
18112
Let $f(z)=sum_n=0^inftyfracz^n1-z^n$. Then $f$ converges in the unit disc. I want to show that it is analytic in this region as well, but since it is not a power series I don't have any theorems to apply. My only thought was to try and use similar reasoning as with power series: show that converges is uniform so that we can switch integral and sum to show that the integral of the series is 0, then apply Morera's Theorem. But the problem is, according to Wolfram Alpha, $int_gammafracz^n1-z^n=int_0^2pifrace^intie^it1-e^intdt$ doesn't converge.
EDIT: To apply Morera's Theorem $gamma$ should be closed path contained in the unit circle. (It probably should have been obvious that the integral doesn't converge since the unit circle contains 1...). So the above strategy does works (yay!) since the function is holomorphic in any closed disk with radius less than 1.
sequences-and-series complex-analysis analytic-functions
sequences-and-series complex-analysis analytic-functions
asked Aug 11 at 15:08
SihOASHoihd
18112
asked Aug 11 at 15:08
SihOASHoihd
18112
edited Aug 11 at 15:57
asked Aug 11 at 15:08
SihOASHoihd
18112
asked Aug 11 at 15:08
SihOASHoihd
18112
asked Aug 11 at 15:08
SihOASHoihd
18112
18112
7
Of course the sum should start at $n=1$, not $n=0$. To show this is a power series, note that, if $|z|<1$, $$f(z)=sum_n=1^infty z^nsum_k=0^infty z^kn=sum_ngeqslant1,kgeqslant0z^n(k+1)=sum_n=1^inftysigma_1(n)z^n$$ where $sigma_1(n)$ denotes the number of divisors of $n$.
– Did
Aug 11 at 15:15
How are you getting the first equality?
– SihOASHoihd
Aug 11 at 15:23
Any one of the five people who upvoted should feel free to answer
– SihOASHoihd
Aug 11 at 15:33
@Did How did you get the last equality?
– Szeto
Aug 11 at 15:35
@SihOASHoihd That’s just geometric series.
– Szeto
Aug 11 at 15:35
 |Â
show 4 more comments
7
Of course the sum should start at $n=1$, not $n=0$. To show this is a power series, note that, if $|z|<1$, $$f(z)=sum_n=1^infty z^nsum_k=0^infty z^kn=sum_ngeqslant1,kgeqslant0z^n(k+1)=sum_n=1^inftysigma_1(n)z^n$$ where $sigma_1(n)$ denotes the number of divisors of $n$.
– Did
Aug 11 at 15:15
How are you getting the first equality?
– SihOASHoihd
Aug 11 at 15:23
Any one of the five people who upvoted should feel free to answer
– SihOASHoihd
Aug 11 at 15:33
@Did How did you get the last equality?
– Szeto
Aug 11 at 15:35
@SihOASHoihd That’s just geometric series.
– Szeto
Aug 11 at 15:35
7
7
Of course the sum should start at $n=1$, not $n=0$. To show this is a power series, note that, if $|z|<1$, $$f(z)=sum_n=1^infty z^nsum_k=0^infty z^kn=sum_ngeqslant1,kgeqslant0z^n(k+1)=sum_n=1^inftysigma_1(n)z^n$$ where $sigma_1(n)$ denotes the number of divisors of $n$.
– Did
Aug 11 at 15:15
Of course the sum should start at $n=1$, not $n=0$. To show this is a power series, note that, if $|z|<1$, $$f(z)=sum_n=1^infty z^nsum_k=0^infty z^kn=sum_ngeqslant1,kgeqslant0z^n(k+1)=sum_n=1^inftysigma_1(n)z^n$$ where $sigma_1(n)$ denotes the number of divisors of $n$.
– Did
Aug 11 at 15:15
How are you getting the first equality?
– SihOASHoihd
Aug 11 at 15:23
How are you getting the first equality?
– SihOASHoihd
Aug 11 at 15:23
Any one of the five people who upvoted should feel free to answer
– SihOASHoihd
Aug 11 at 15:33
Any one of the five people who upvoted should feel free to answer
– SihOASHoihd
Aug 11 at 15:33
@Did How did you get the last equality?
– Szeto
Aug 11 at 15:35
@Did How did you get the last equality?
– Szeto
Aug 11 at 15:35
@SihOASHoihd That’s just geometric series.
– Szeto
Aug 11 at 15:35
@SihOASHoihd That’s just geometric series.
– Szeto
Aug 11 at 15:35
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
Let $C$ be a compact subset of the open unit disk and let $M=sup_zin C|z|$. Then $M<1$ and, for each $ninmathbb N$,$$left|fracz^n1-z^nright|=frac1-z^nleqslantfracM^n1-leqslantfracM^n1-M^n.$$Since the series $sum_n=1^inftyfracM^n1-M^n$ converges, your series converges uniformly on $C$. Since this holds for every compact subset of $D_1(0)$, the sum of the series is analytic.
answered Aug 11 at 15:16
José Carlos Santos
122k16101186
I was wondering what other theorems there are. This is not one I think I knew.
– SihOASHoihd
Aug 11 at 15:22
2
Just to be clear: the theorem that I used here is due to Weierstrass and says that if a sequence $(f_n)_ninmathbb N$ of analytic functions converges uniformly to a function $f$ on each compact, then $f$ is analytic too.
– José Carlos Santos
Aug 11 at 15:27
After posting I realized that to apply Morera's Theorem I only need uniform convergence in some closed disk contained in the unit circle. So the integral above probably does equal zero on such a closed disk.
– SihOASHoihd
Aug 11 at 15:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Let $C$ be a compact subset of the open unit disk and let $M=sup_zin C|z|$. Then $M<1$ and, for each $ninmathbb N$,$$left|fracz^n1-z^nright|=frac1-z^nleqslantfracM^n1-leqslantfracM^n1-M^n.$$Since the series $sum_n=1^inftyfracM^n1-M^n$ converges, your series converges uniformly on $C$. Since this holds for every compact subset of $D_1(0)$, the sum of the series is analytic.
answered Aug 11 at 15:16
José Carlos Santos
122k16101186
I was wondering what other theorems there are. This is not one I think I knew.
– SihOASHoihd
Aug 11 at 15:22
2
Just to be clear: the theorem that I used here is due to Weierstrass and says that if a sequence $(f_n)_ninmathbb N$ of analytic functions converges uniformly to a function $f$ on each compact, then $f$ is analytic too.
– José Carlos Santos
Aug 11 at 15:27
After posting I realized that to apply Morera's Theorem I only need uniform convergence in some closed disk contained in the unit circle. So the integral above probably does equal zero on such a closed disk.
– SihOASHoihd
Aug 11 at 15:31
add a comment |Â
up vote
8
down vote
accepted
Let $C$ be a compact subset of the open unit disk and let $M=sup_zin C|z|$. Then $M<1$ and, for each $ninmathbb N$,$$left|fracz^n1-z^nright|=frac1-z^nleqslantfracM^n1-leqslantfracM^n1-M^n.$$Since the series $sum_n=1^inftyfracM^n1-M^n$ converges, your series converges uniformly on $C$. Since this holds for every compact subset of $D_1(0)$, the sum of the series is analytic.
answered Aug 11 at 15:16
José Carlos Santos
122k16101186
I was wondering what other theorems there are. This is not one I think I knew.
– SihOASHoihd
Aug 11 at 15:22
2
Just to be clear: the theorem that I used here is due to Weierstrass and says that if a sequence $(f_n)_ninmathbb N$ of analytic functions converges uniformly to a function $f$ on each compact, then $f$ is analytic too.
– José Carlos Santos
Aug 11 at 15:27
After posting I realized that to apply Morera's Theorem I only need uniform convergence in some closed disk contained in the unit circle. So the integral above probably does equal zero on such a closed disk.
– SihOASHoihd
Aug 11 at 15:31
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Let $C$ be a compact subset of the open unit disk and let $M=sup_zin C|z|$. Then $M<1$ and, for each $ninmathbb N$,$$left|fracz^n1-z^nright|=frac1-z^nleqslantfracM^n1-leqslantfracM^n1-M^n.$$Since the series $sum_n=1^inftyfracM^n1-M^n$ converges, your series converges uniformly on $C$. Since this holds for every compact subset of $D_1(0)$, the sum of the series is analytic.
answered Aug 11 at 15:16
José Carlos Santos
122k16101186
Let $C$ be a compact subset of the open unit disk and let $M=sup_zin C|z|$. Then $M<1$ and, for each $ninmathbb N$,$$left|fracz^n1-z^nright|=frac1-z^nleqslantfracM^n1-leqslantfracM^n1-M^n.$$Since the series $sum_n=1^inftyfracM^n1-M^n$ converges, your series converges uniformly on $C$. Since this holds for every compact subset of $D_1(0)$, the sum of the series is analytic.
answered Aug 11 at 15:16
José Carlos Santos
122k16101186
answered Aug 11 at 15:16
José Carlos Santos
122k16101186
answered Aug 11 at 15:16
José Carlos Santos
122k16101186
answered Aug 11 at 15:16
José Carlos Santos
122k16101186
122k16101186
I was wondering what other theorems there are. This is not one I think I knew.
– SihOASHoihd
Aug 11 at 15:22
2
Just to be clear: the theorem that I used here is due to Weierstrass and says that if a sequence $(f_n)_ninmathbb N$ of analytic functions converges uniformly to a function $f$ on each compact, then $f$ is analytic too.
– José Carlos Santos
Aug 11 at 15:27
After posting I realized that to apply Morera's Theorem I only need uniform convergence in some closed disk contained in the unit circle. So the integral above probably does equal zero on such a closed disk.
– SihOASHoihd
Aug 11 at 15:31
add a comment |Â
I was wondering what other theorems there are. This is not one I think I knew.
– SihOASHoihd
Aug 11 at 15:22
2
Just to be clear: the theorem that I used here is due to Weierstrass and says that if a sequence $(f_n)_ninmathbb N$ of analytic functions converges uniformly to a function $f$ on each compact, then $f$ is analytic too.
– José Carlos Santos
Aug 11 at 15:27
After posting I realized that to apply Morera's Theorem I only need uniform convergence in some closed disk contained in the unit circle. So the integral above probably does equal zero on such a closed disk.
– SihOASHoihd
Aug 11 at 15:31
I was wondering what other theorems there are. This is not one I think I knew.
– SihOASHoihd
Aug 11 at 15:22
I was wondering what other theorems there are. This is not one I think I knew.
– SihOASHoihd
Aug 11 at 15:22
2
2
Just to be clear: the theorem that I used here is due to Weierstrass and says that if a sequence $(f_n)_ninmathbb N$ of analytic functions converges uniformly to a function $f$ on each compact, then $f$ is analytic too.
– José Carlos Santos
Aug 11 at 15:27
Just to be clear: the theorem that I used here is due to Weierstrass and says that if a sequence $(f_n)_ninmathbb N$ of analytic functions converges uniformly to a function $f$ on each compact, then $f$ is analytic too.
– José Carlos Santos
Aug 11 at 15:27
After posting I realized that to apply Morera's Theorem I only need uniform convergence in some closed disk contained in the unit circle. So the integral above probably does equal zero on such a closed disk.
– SihOASHoihd
Aug 11 at 15:31
After posting I realized that to apply Morera's Theorem I only need uniform convergence in some closed disk contained in the unit circle. So the integral above probably does equal zero on such a closed disk.
– SihOASHoihd
Aug 11 at 15:31
add a comment |Â
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7
Of course the sum should start at $n=1$, not $n=0$. To show this is a power series, note that, if $|z|<1$, $$f(z)=sum_n=1^infty z^nsum_k=0^infty z^kn=sum_ngeqslant1,kgeqslant0z^n(k+1)=sum_n=1^inftysigma_1(n)z^n$$ where $sigma_1(n)$ denotes the number of divisors of $n$.
– Did
Aug 11 at 15:15
How are you getting the first equality?
– SihOASHoihd
Aug 11 at 15:23
Any one of the five people who upvoted should feel free to answer
– SihOASHoihd
Aug 11 at 15:33
@Did How did you get the last equality?
– Szeto
Aug 11 at 15:35
@SihOASHoihd That’s just geometric series.
– Szeto
Aug 11 at 15:35