Mixing
characterizing the boundary of the convergent region of $f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$
$begingroup$
Let
$$f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$$
A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:
The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.
The code that was used to generate this is (using mpmath and matplotlib):
from mpmath import *
import pylab
def f(z):
return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])
fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)
complex-analysis
edited Jan 4 at 8:24
Glorfindel
3,41981830
asked Mar 6 '12 at 20:28
graveolensagraveolensa
3,07711736
$endgroup$
add a comment |
$begingroup$
Let
$$f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$$
A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:
The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.
The code that was used to generate this is (using mpmath and matplotlib):
from mpmath import *
import pylab
def f(z):
return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])
fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)
complex-analysis
edited Jan 4 at 8:24
Glorfindel
3,41981830
asked Mar 6 '12 at 20:28
graveolensagraveolensa
3,07711736
$endgroup$
$begingroup$
The series converges for $$left| {{z^{1/z}}} right| < 1$$
$endgroup$
– Pedro Tamaroff♦
Mar 6 '12 at 21:03
1
$begingroup$
No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
$endgroup$
– Robert Israel
Mar 6 '12 at 21:22
add a comment |
$begingroup$
Let
$$f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$$
A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:
The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.
The code that was used to generate this is (using mpmath and matplotlib):
from mpmath import *
import pylab
def f(z):
return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])
fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)
complex-analysis
edited Jan 4 at 8:24
Glorfindel
3,41981830
asked Mar 6 '12 at 20:28
graveolensagraveolensa
3,07711736
$endgroup$
Let
$$f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$$
A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:
The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.
The code that was used to generate this is (using mpmath and matplotlib):
from mpmath import *
import pylab
def f(z):
return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])
fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)
complex-analysis
complex-analysis
edited Jan 4 at 8:24
Glorfindel
3,41981830
asked Mar 6 '12 at 20:28
graveolensagraveolensa
3,07711736
edited Jan 4 at 8:24
Glorfindel
3,41981830
asked Mar 6 '12 at 20:28
graveolensagraveolensa
3,07711736
edited Jan 4 at 8:24
Glorfindel
3,41981830
edited Jan 4 at 8:24
Glorfindel
3,41981830
edited Jan 4 at 8:24
Glorfindel
3,41981830
3,41981830
asked Mar 6 '12 at 20:28
graveolensagraveolensa
3,07711736
asked Mar 6 '12 at 20:28
graveolensagraveolensa
3,07711736
asked Mar 6 '12 at 20:28
graveolensagraveolensa
3,07711736
3,07711736
$begingroup$
The series converges for $$left| {{z^{1/z}}} right| < 1$$
$endgroup$
– Pedro Tamaroff♦
Mar 6 '12 at 21:03
1
$begingroup$
No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
$endgroup$
– Robert Israel
Mar 6 '12 at 21:22
add a comment |
$begingroup$
The series converges for $$left| {{z^{1/z}}} right| < 1$$
$endgroup$
– Pedro Tamaroff♦
Mar 6 '12 at 21:03
1
$begingroup$
No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
$endgroup$
– Robert Israel
Mar 6 '12 at 21:22
$begingroup$
The series converges for $$left| {{z^{1/z}}} right| < 1$$
$endgroup$
– Pedro Tamaroff♦
Mar 6 '12 at 21:03
$begingroup$
The series converges for $$left| {{z^{1/z}}} right| < 1$$
$endgroup$
– Pedro Tamaroff♦
Mar 6 '12 at 21:03
1
1
$begingroup$
No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
$endgroup$
– Robert Israel
Mar 6 '12 at 21:22
$begingroup$
No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
$endgroup$
– Robert Israel
Mar 6 '12 at 21:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
$log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.
So the thickness of your red region is really just an artifact of the numerical methods being used.
answered Mar 6 '12 at 21:20
Robert IsraelRobert Israel
320k23209460
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
$log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.
So the thickness of your red region is really just an artifact of the numerical methods being used.
answered Mar 6 '12 at 21:20
Robert IsraelRobert Israel
320k23209460
$endgroup$
add a comment |
$begingroup$
$|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
$log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.
So the thickness of your red region is really just an artifact of the numerical methods being used.
answered Mar 6 '12 at 21:20
Robert IsraelRobert Israel
320k23209460
$endgroup$
add a comment |
$begingroup$
$|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
$log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.
So the thickness of your red region is really just an artifact of the numerical methods being used.
answered Mar 6 '12 at 21:20
Robert IsraelRobert Israel
320k23209460
$endgroup$
$|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
$log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.
So the thickness of your red region is really just an artifact of the numerical methods being used.
answered Mar 6 '12 at 21:20
Robert IsraelRobert Israel
320k23209460
answered Mar 6 '12 at 21:20
Robert IsraelRobert Israel
320k23209460
answered Mar 6 '12 at 21:20
Robert IsraelRobert Israel
320k23209460
answered Mar 6 '12 at 21:20
Robert IsraelRobert Israel
320k23209460
320k23209460
add a comment |
add a comment |
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$begingroup$
The series converges for $$left| {{z^{1/z}}} right| < 1$$
$endgroup$
– Pedro Tamaroff♦
Mar 6 '12 at 21:03
1
$begingroup$
No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
$endgroup$
– Robert Israel
Mar 6 '12 at 21:22