Mixing
$e^x+x=0$ has countable infinite many solutions
I have shown that the following transcendental equation has only one real root. But I am looking for the argument to show that it has infinitely many numbers of roots in the complex plane. Moreover, these roots are countable.
$$e^x+x=0$$
And real root dominate all other roots in the sense of absolute value. Or real part of all complex roots is less than the real root.
Would be grateful for the hint and remarks.
real-analysis complex-analysis transcendental-equations
asked Nov 18 '18 at 15:45
JohnJohn
7810
add a comment |
I have shown that the following transcendental equation has only one real root. But I am looking for the argument to show that it has infinitely many numbers of roots in the complex plane. Moreover, these roots are countable.
$$e^x+x=0$$
And real root dominate all other roots in the sense of absolute value. Or real part of all complex roots is less than the real root.
Would be grateful for the hint and remarks.
real-analysis complex-analysis transcendental-equations
asked Nov 18 '18 at 15:45
JohnJohn
7810
2
An non-constant entire function has at most countably many zeros, since if it had uncountably many then you could find a sequence tending to some point such that $f$ was zero on that sequence. By the identity theorem, that function therefore must be everywhere zero.
– Patrick Stevens
Nov 18 '18 at 15:52
Try using Rouche's theorem on a small disk away from $0$.
– anomaly
Nov 18 '18 at 15:58
See math.stackexchange.com/questions/632660/… or math.stackexchange.com/questions/657208/…
– Martin R
Nov 18 '18 at 16:08
@Patrick Stevens: I just saw this question and immediately thought of what you said, then saw what you said (and that it was an hour ago). Oh well. Anyway, in case the OP wants more detail, a slight modification of the usual proof that compact sets in Euclidean space are Bolzano-Weierstrass compact works --- Suppose uncountably many zeros. Then there must be some closed square of side length $1$ containing uncountably many zeros (because $mathbb C$ can be covered by countably many such squares), and in this square there must be a closed square of side length $1/2$ containing uncountably...
– Dave L. Renfro
Nov 18 '18 at 17:38
add a comment |
I have shown that the following transcendental equation has only one real root. But I am looking for the argument to show that it has infinitely many numbers of roots in the complex plane. Moreover, these roots are countable.
$$e^x+x=0$$
And real root dominate all other roots in the sense of absolute value. Or real part of all complex roots is less than the real root.
Would be grateful for the hint and remarks.
real-analysis complex-analysis transcendental-equations
asked Nov 18 '18 at 15:45
JohnJohn
7810
I have shown that the following transcendental equation has only one real root. But I am looking for the argument to show that it has infinitely many numbers of roots in the complex plane. Moreover, these roots are countable.
$$e^x+x=0$$
And real root dominate all other roots in the sense of absolute value. Or real part of all complex roots is less than the real root.
Would be grateful for the hint and remarks.
real-analysis complex-analysis transcendental-equations
real-analysis complex-analysis transcendental-equations
asked Nov 18 '18 at 15:45
JohnJohn
7810
asked Nov 18 '18 at 15:45
JohnJohn
7810
asked Nov 18 '18 at 15:45
JohnJohn
7810
asked Nov 18 '18 at 15:45
JohnJohn
7810
asked Nov 18 '18 at 15:45
JohnJohn
7810
7810
2
An non-constant entire function has at most countably many zeros, since if it had uncountably many then you could find a sequence tending to some point such that $f$ was zero on that sequence. By the identity theorem, that function therefore must be everywhere zero.
– Patrick Stevens
Nov 18 '18 at 15:52
Try using Rouche's theorem on a small disk away from $0$.
– anomaly
Nov 18 '18 at 15:58
See math.stackexchange.com/questions/632660/… or math.stackexchange.com/questions/657208/…
– Martin R
Nov 18 '18 at 16:08
@Patrick Stevens: I just saw this question and immediately thought of what you said, then saw what you said (and that it was an hour ago). Oh well. Anyway, in case the OP wants more detail, a slight modification of the usual proof that compact sets in Euclidean space are Bolzano-Weierstrass compact works --- Suppose uncountably many zeros. Then there must be some closed square of side length $1$ containing uncountably many zeros (because $mathbb C$ can be covered by countably many such squares), and in this square there must be a closed square of side length $1/2$ containing uncountably...
– Dave L. Renfro
Nov 18 '18 at 17:38
add a comment |
2
An non-constant entire function has at most countably many zeros, since if it had uncountably many then you could find a sequence tending to some point such that $f$ was zero on that sequence. By the identity theorem, that function therefore must be everywhere zero.
– Patrick Stevens
Nov 18 '18 at 15:52
Try using Rouche's theorem on a small disk away from $0$.
– anomaly
Nov 18 '18 at 15:58
See math.stackexchange.com/questions/632660/… or math.stackexchange.com/questions/657208/…
– Martin R
Nov 18 '18 at 16:08
@Patrick Stevens: I just saw this question and immediately thought of what you said, then saw what you said (and that it was an hour ago). Oh well. Anyway, in case the OP wants more detail, a slight modification of the usual proof that compact sets in Euclidean space are Bolzano-Weierstrass compact works --- Suppose uncountably many zeros. Then there must be some closed square of side length $1$ containing uncountably many zeros (because $mathbb C$ can be covered by countably many such squares), and in this square there must be a closed square of side length $1/2$ containing uncountably...
– Dave L. Renfro
Nov 18 '18 at 17:38
2
2
An non-constant entire function has at most countably many zeros, since if it had uncountably many then you could find a sequence tending to some point such that $f$ was zero on that sequence. By the identity theorem, that function therefore must be everywhere zero.
– Patrick Stevens
Nov 18 '18 at 15:52
An non-constant entire function has at most countably many zeros, since if it had uncountably many then you could find a sequence tending to some point such that $f$ was zero on that sequence. By the identity theorem, that function therefore must be everywhere zero.
– Patrick Stevens
Nov 18 '18 at 15:52
Try using Rouche's theorem on a small disk away from $0$.
– anomaly
Nov 18 '18 at 15:58
Try using Rouche's theorem on a small disk away from $0$.
– anomaly
Nov 18 '18 at 15:58
See math.stackexchange.com/questions/632660/… or math.stackexchange.com/questions/657208/…
– Martin R
Nov 18 '18 at 16:08
See math.stackexchange.com/questions/632660/… or math.stackexchange.com/questions/657208/…
– Martin R
Nov 18 '18 at 16:08
@Patrick Stevens: I just saw this question and immediately thought of what you said, then saw what you said (and that it was an hour ago). Oh well. Anyway, in case the OP wants more detail, a slight modification of the usual proof that compact sets in Euclidean space are Bolzano-Weierstrass compact works --- Suppose uncountably many zeros. Then there must be some closed square of side length $1$ containing uncountably many zeros (because $mathbb C$ can be covered by countably many such squares), and in this square there must be a closed square of side length $1/2$ containing uncountably...
– Dave L. Renfro
Nov 18 '18 at 17:38
@Patrick Stevens: I just saw this question and immediately thought of what you said, then saw what you said (and that it was an hour ago). Oh well. Anyway, in case the OP wants more detail, a slight modification of the usual proof that compact sets in Euclidean space are Bolzano-Weierstrass compact works --- Suppose uncountably many zeros. Then there must be some closed square of side length $1$ containing uncountably many zeros (because $mathbb C$ can be covered by countably many such squares), and in this square there must be a closed square of side length $1/2$ containing uncountably...
– Dave L. Renfro
Nov 18 '18 at 17:38
add a comment |
2 Answers
2
active
oldest
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$f(z)=e^z+z$ is an entire function, hence it attains every complex value (with at most one exception) countable times, by Picard's theorem. Actually there are no exceptional values, since
$$ f(z+2pi i) = f(z)+2pi i $$
(the presence of an exceptional value would imply the presence of infinite exceptional values) hence there are countable zeroes in the complex plane. The number of zeroes in the region $|z|leq 2pi M$ is given by
$$ frac{1}{2pi i}oint_{|z|=2pi M}frac{f'(z)}{f(z)},dz =frac{1}{2pi i}oint_{|z|=2pi M}frac{1-z}{e^z+z},dz$$
but such roots are probably best accounted by noticing that $f(a+bi)=0$ is equivalent to
$$ cos b = -a e^{-a}, qquad frac{sin b}{b} = -e^{-a} $$
hence they are given by the intersections of the following blue/purple curves:
$hspace{1cm}$
The roots closest to the origin lie at $approx 1.53391 pm 4.37519i$.
answered Nov 18 '18 at 17:24
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
The solutions of $z+e^z=0$ come in complex conjugate pairs. Consider the roots with positive imaginary part. From $|z|=e^{Re(z)}$ one gets for $x=Re(z)gg 1$ that $y=Im(z)sim e^xgg x$. Considering the dominance of the imaginary part, transform the equation to
$$
-iz=ie^z=e^{z+ipi/2}implies z = Ln(-iz)+i2npi-ifracpi2
$$
for some $ninBbb N$. Inserting this into itself gives in the next step
$$
z= ln((2n-tfrac12)pi)+i(2n-tfrac12)pi+w.
$$
for some small $w$. Inserting back gives an equation for $w$,
$$
w = Lnleft(1-ifrac{ln((2n-tfrac12)pi)+w}{(2n-tfrac12)pi}right)
$$
For $n$ large enough this is a contraction on ${w:|w|lefrac12}$, ensuring the existence of a solution. The same also holds for the unshifted iteration for $z$.
Different values of $n$ give different fixed-point iterations resulting in different solutions of the original equation, ensuring a countably infinite set of solutions.
Comparing the above first approximations with later iterates of the fixed-point iteration
$$
z_{k+1} = Ln(-iz_k)+i(2n-tfrac12)pi
$$
shows rapid (numerical) convergence and gives the table
begin{array}{l|lll}
n& z_0 & z_{15} & z_{15}-z_0 \ hline
1 & (1.55019499396+4.71238898038j) & (1.53391331978+4.37518515309j) & (-0.0162816741736-0.337203827291j) \
2 & (2.39749285434+10.9955742876j) & (2.40158510487+10.7762995161j) & (0.00409225052324-0.219274771449j) \
3 & (2.84947797809+17.2787595947j) & (2.85358175541+17.1135355394j) & (0.00410377732121-0.165224055332j) \
4 & (3.15963290639+23.5619449019j) & (3.1629527388+23.4277475038j) & (0.00331983241242-0.134197398168j) \
5 & (3.39602168446+29.8451302091j) & (3.39869219676+29.7313107078j) & (0.00267051230882-0.113819501275j) \
6 & (3.58707692122+36.1283155163j) & (3.58926252453+36.0290217034j) & (0.00218560331097-0.0992938128549j) \
7 & (3.74741957129+42.4115008235j) & (3.74924254122+42.3231453612j) & (0.0018229699232-0.0883554622252j) \
8 & (3.88556990977+48.6946861306j) & (3.88711644955+48.6148985649j) & (0.00154653977456-0.0797875657055j) \
9 & (4.00693076678+54.9778714378j) & (4.00826205311+54.9049971233j) & (0.00133128633039-0.0728743144716j) \
10 & (4.11514435142+61.261056745j) & (4.116304664+61.193891332j) & (0.00116031258267-0.0671654130445j) \
11 & (4.21278282098+67.5442420522j) & (4.21380491472+67.48187952j) & (0.00102209373376-0.0623625321652j) \
12 & (4.301730307+73.8274273594j) & (4.3026389193+73.769167656j) & (0.000908612303842-0.0582597033192j) \
end{array}
It also demonstrates a good fit of the initial approximation even for small $n$.
answered Nov 21 '18 at 17:11
LutzLLutzL
56.5k42054
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
$f(z)=e^z+z$ is an entire function, hence it attains every complex value (with at most one exception) countable times, by Picard's theorem. Actually there are no exceptional values, since
$$ f(z+2pi i) = f(z)+2pi i $$
(the presence of an exceptional value would imply the presence of infinite exceptional values) hence there are countable zeroes in the complex plane. The number of zeroes in the region $|z|leq 2pi M$ is given by
$$ frac{1}{2pi i}oint_{|z|=2pi M}frac{f'(z)}{f(z)},dz =frac{1}{2pi i}oint_{|z|=2pi M}frac{1-z}{e^z+z},dz$$
but such roots are probably best accounted by noticing that $f(a+bi)=0$ is equivalent to
$$ cos b = -a e^{-a}, qquad frac{sin b}{b} = -e^{-a} $$
hence they are given by the intersections of the following blue/purple curves:
$hspace{1cm}$
The roots closest to the origin lie at $approx 1.53391 pm 4.37519i$.
answered Nov 18 '18 at 17:24
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
$f(z)=e^z+z$ is an entire function, hence it attains every complex value (with at most one exception) countable times, by Picard's theorem. Actually there are no exceptional values, since
$$ f(z+2pi i) = f(z)+2pi i $$
(the presence of an exceptional value would imply the presence of infinite exceptional values) hence there are countable zeroes in the complex plane. The number of zeroes in the region $|z|leq 2pi M$ is given by
$$ frac{1}{2pi i}oint_{|z|=2pi M}frac{f'(z)}{f(z)},dz =frac{1}{2pi i}oint_{|z|=2pi M}frac{1-z}{e^z+z},dz$$
but such roots are probably best accounted by noticing that $f(a+bi)=0$ is equivalent to
$$ cos b = -a e^{-a}, qquad frac{sin b}{b} = -e^{-a} $$
hence they are given by the intersections of the following blue/purple curves:
$hspace{1cm}$
The roots closest to the origin lie at $approx 1.53391 pm 4.37519i$.
answered Nov 18 '18 at 17:24
Jack D'AurizioJack D'Aurizio
287k33280658
add a comment |
$f(z)=e^z+z$ is an entire function, hence it attains every complex value (with at most one exception) countable times, by Picard's theorem. Actually there are no exceptional values, since
$$ f(z+2pi i) = f(z)+2pi i $$
(the presence of an exceptional value would imply the presence of infinite exceptional values) hence there are countable zeroes in the complex plane. The number of zeroes in the region $|z|leq 2pi M$ is given by
$$ frac{1}{2pi i}oint_{|z|=2pi M}frac{f'(z)}{f(z)},dz =frac{1}{2pi i}oint_{|z|=2pi M}frac{1-z}{e^z+z},dz$$
but such roots are probably best accounted by noticing that $f(a+bi)=0$ is equivalent to
$$ cos b = -a e^{-a}, qquad frac{sin b}{b} = -e^{-a} $$
hence they are given by the intersections of the following blue/purple curves:
$hspace{1cm}$
The roots closest to the origin lie at $approx 1.53391 pm 4.37519i$.
answered Nov 18 '18 at 17:24
Jack D'AurizioJack D'Aurizio
287k33280658
$f(z)=e^z+z$ is an entire function, hence it attains every complex value (with at most one exception) countable times, by Picard's theorem. Actually there are no exceptional values, since
$$ f(z+2pi i) = f(z)+2pi i $$
(the presence of an exceptional value would imply the presence of infinite exceptional values) hence there are countable zeroes in the complex plane. The number of zeroes in the region $|z|leq 2pi M$ is given by
$$ frac{1}{2pi i}oint_{|z|=2pi M}frac{f'(z)}{f(z)},dz =frac{1}{2pi i}oint_{|z|=2pi M}frac{1-z}{e^z+z},dz$$
but such roots are probably best accounted by noticing that $f(a+bi)=0$ is equivalent to
$$ cos b = -a e^{-a}, qquad frac{sin b}{b} = -e^{-a} $$
hence they are given by the intersections of the following blue/purple curves:
$hspace{1cm}$
The roots closest to the origin lie at $approx 1.53391 pm 4.37519i$.
answered Nov 18 '18 at 17:24
Jack D'AurizioJack D'Aurizio
287k33280658
answered Nov 18 '18 at 17:24
Jack D'AurizioJack D'Aurizio
287k33280658
answered Nov 18 '18 at 17:24
Jack D'AurizioJack D'Aurizio
287k33280658
answered Nov 18 '18 at 17:24
Jack D'AurizioJack D'Aurizio
287k33280658
287k33280658
add a comment |
add a comment |
The solutions of $z+e^z=0$ come in complex conjugate pairs. Consider the roots with positive imaginary part. From $|z|=e^{Re(z)}$ one gets for $x=Re(z)gg 1$ that $y=Im(z)sim e^xgg x$. Considering the dominance of the imaginary part, transform the equation to
$$
-iz=ie^z=e^{z+ipi/2}implies z = Ln(-iz)+i2npi-ifracpi2
$$
for some $ninBbb N$. Inserting this into itself gives in the next step
$$
z= ln((2n-tfrac12)pi)+i(2n-tfrac12)pi+w.
$$
for some small $w$. Inserting back gives an equation for $w$,
$$
w = Lnleft(1-ifrac{ln((2n-tfrac12)pi)+w}{(2n-tfrac12)pi}right)
$$
For $n$ large enough this is a contraction on ${w:|w|lefrac12}$, ensuring the existence of a solution. The same also holds for the unshifted iteration for $z$.
Different values of $n$ give different fixed-point iterations resulting in different solutions of the original equation, ensuring a countably infinite set of solutions.
Comparing the above first approximations with later iterates of the fixed-point iteration
$$
z_{k+1} = Ln(-iz_k)+i(2n-tfrac12)pi
$$
shows rapid (numerical) convergence and gives the table
begin{array}{l|lll}
n& z_0 & z_{15} & z_{15}-z_0 \ hline
1 & (1.55019499396+4.71238898038j) & (1.53391331978+4.37518515309j) & (-0.0162816741736-0.337203827291j) \
2 & (2.39749285434+10.9955742876j) & (2.40158510487+10.7762995161j) & (0.00409225052324-0.219274771449j) \
3 & (2.84947797809+17.2787595947j) & (2.85358175541+17.1135355394j) & (0.00410377732121-0.165224055332j) \
4 & (3.15963290639+23.5619449019j) & (3.1629527388+23.4277475038j) & (0.00331983241242-0.134197398168j) \
5 & (3.39602168446+29.8451302091j) & (3.39869219676+29.7313107078j) & (0.00267051230882-0.113819501275j) \
6 & (3.58707692122+36.1283155163j) & (3.58926252453+36.0290217034j) & (0.00218560331097-0.0992938128549j) \
7 & (3.74741957129+42.4115008235j) & (3.74924254122+42.3231453612j) & (0.0018229699232-0.0883554622252j) \
8 & (3.88556990977+48.6946861306j) & (3.88711644955+48.6148985649j) & (0.00154653977456-0.0797875657055j) \
9 & (4.00693076678+54.9778714378j) & (4.00826205311+54.9049971233j) & (0.00133128633039-0.0728743144716j) \
10 & (4.11514435142+61.261056745j) & (4.116304664+61.193891332j) & (0.00116031258267-0.0671654130445j) \
11 & (4.21278282098+67.5442420522j) & (4.21380491472+67.48187952j) & (0.00102209373376-0.0623625321652j) \
12 & (4.301730307+73.8274273594j) & (4.3026389193+73.769167656j) & (0.000908612303842-0.0582597033192j) \
end{array}
It also demonstrates a good fit of the initial approximation even for small $n$.
answered Nov 21 '18 at 17:11
LutzLLutzL
56.5k42054
add a comment |
The solutions of $z+e^z=0$ come in complex conjugate pairs. Consider the roots with positive imaginary part. From $|z|=e^{Re(z)}$ one gets for $x=Re(z)gg 1$ that $y=Im(z)sim e^xgg x$. Considering the dominance of the imaginary part, transform the equation to
$$
-iz=ie^z=e^{z+ipi/2}implies z = Ln(-iz)+i2npi-ifracpi2
$$
for some $ninBbb N$. Inserting this into itself gives in the next step
$$
z= ln((2n-tfrac12)pi)+i(2n-tfrac12)pi+w.
$$
for some small $w$. Inserting back gives an equation for $w$,
$$
w = Lnleft(1-ifrac{ln((2n-tfrac12)pi)+w}{(2n-tfrac12)pi}right)
$$
For $n$ large enough this is a contraction on ${w:|w|lefrac12}$, ensuring the existence of a solution. The same also holds for the unshifted iteration for $z$.
Different values of $n$ give different fixed-point iterations resulting in different solutions of the original equation, ensuring a countably infinite set of solutions.
Comparing the above first approximations with later iterates of the fixed-point iteration
$$
z_{k+1} = Ln(-iz_k)+i(2n-tfrac12)pi
$$
shows rapid (numerical) convergence and gives the table
begin{array}{l|lll}
n& z_0 & z_{15} & z_{15}-z_0 \ hline
1 & (1.55019499396+4.71238898038j) & (1.53391331978+4.37518515309j) & (-0.0162816741736-0.337203827291j) \
2 & (2.39749285434+10.9955742876j) & (2.40158510487+10.7762995161j) & (0.00409225052324-0.219274771449j) \
3 & (2.84947797809+17.2787595947j) & (2.85358175541+17.1135355394j) & (0.00410377732121-0.165224055332j) \
4 & (3.15963290639+23.5619449019j) & (3.1629527388+23.4277475038j) & (0.00331983241242-0.134197398168j) \
5 & (3.39602168446+29.8451302091j) & (3.39869219676+29.7313107078j) & (0.00267051230882-0.113819501275j) \
6 & (3.58707692122+36.1283155163j) & (3.58926252453+36.0290217034j) & (0.00218560331097-0.0992938128549j) \
7 & (3.74741957129+42.4115008235j) & (3.74924254122+42.3231453612j) & (0.0018229699232-0.0883554622252j) \
8 & (3.88556990977+48.6946861306j) & (3.88711644955+48.6148985649j) & (0.00154653977456-0.0797875657055j) \
9 & (4.00693076678+54.9778714378j) & (4.00826205311+54.9049971233j) & (0.00133128633039-0.0728743144716j) \
10 & (4.11514435142+61.261056745j) & (4.116304664+61.193891332j) & (0.00116031258267-0.0671654130445j) \
11 & (4.21278282098+67.5442420522j) & (4.21380491472+67.48187952j) & (0.00102209373376-0.0623625321652j) \
12 & (4.301730307+73.8274273594j) & (4.3026389193+73.769167656j) & (0.000908612303842-0.0582597033192j) \
end{array}
It also demonstrates a good fit of the initial approximation even for small $n$.
answered Nov 21 '18 at 17:11
LutzLLutzL
56.5k42054
add a comment |
The solutions of $z+e^z=0$ come in complex conjugate pairs. Consider the roots with positive imaginary part. From $|z|=e^{Re(z)}$ one gets for $x=Re(z)gg 1$ that $y=Im(z)sim e^xgg x$. Considering the dominance of the imaginary part, transform the equation to
$$
-iz=ie^z=e^{z+ipi/2}implies z = Ln(-iz)+i2npi-ifracpi2
$$
for some $ninBbb N$. Inserting this into itself gives in the next step
$$
z= ln((2n-tfrac12)pi)+i(2n-tfrac12)pi+w.
$$
for some small $w$. Inserting back gives an equation for $w$,
$$
w = Lnleft(1-ifrac{ln((2n-tfrac12)pi)+w}{(2n-tfrac12)pi}right)
$$
For $n$ large enough this is a contraction on ${w:|w|lefrac12}$, ensuring the existence of a solution. The same also holds for the unshifted iteration for $z$.
Different values of $n$ give different fixed-point iterations resulting in different solutions of the original equation, ensuring a countably infinite set of solutions.
Comparing the above first approximations with later iterates of the fixed-point iteration
$$
z_{k+1} = Ln(-iz_k)+i(2n-tfrac12)pi
$$
shows rapid (numerical) convergence and gives the table
begin{array}{l|lll}
n& z_0 & z_{15} & z_{15}-z_0 \ hline
1 & (1.55019499396+4.71238898038j) & (1.53391331978+4.37518515309j) & (-0.0162816741736-0.337203827291j) \
2 & (2.39749285434+10.9955742876j) & (2.40158510487+10.7762995161j) & (0.00409225052324-0.219274771449j) \
3 & (2.84947797809+17.2787595947j) & (2.85358175541+17.1135355394j) & (0.00410377732121-0.165224055332j) \
4 & (3.15963290639+23.5619449019j) & (3.1629527388+23.4277475038j) & (0.00331983241242-0.134197398168j) \
5 & (3.39602168446+29.8451302091j) & (3.39869219676+29.7313107078j) & (0.00267051230882-0.113819501275j) \
6 & (3.58707692122+36.1283155163j) & (3.58926252453+36.0290217034j) & (0.00218560331097-0.0992938128549j) \
7 & (3.74741957129+42.4115008235j) & (3.74924254122+42.3231453612j) & (0.0018229699232-0.0883554622252j) \
8 & (3.88556990977+48.6946861306j) & (3.88711644955+48.6148985649j) & (0.00154653977456-0.0797875657055j) \
9 & (4.00693076678+54.9778714378j) & (4.00826205311+54.9049971233j) & (0.00133128633039-0.0728743144716j) \
10 & (4.11514435142+61.261056745j) & (4.116304664+61.193891332j) & (0.00116031258267-0.0671654130445j) \
11 & (4.21278282098+67.5442420522j) & (4.21380491472+67.48187952j) & (0.00102209373376-0.0623625321652j) \
12 & (4.301730307+73.8274273594j) & (4.3026389193+73.769167656j) & (0.000908612303842-0.0582597033192j) \
end{array}
It also demonstrates a good fit of the initial approximation even for small $n$.
answered Nov 21 '18 at 17:11
LutzLLutzL
56.5k42054
The solutions of $z+e^z=0$ come in complex conjugate pairs. Consider the roots with positive imaginary part. From $|z|=e^{Re(z)}$ one gets for $x=Re(z)gg 1$ that $y=Im(z)sim e^xgg x$. Considering the dominance of the imaginary part, transform the equation to
$$
-iz=ie^z=e^{z+ipi/2}implies z = Ln(-iz)+i2npi-ifracpi2
$$
for some $ninBbb N$. Inserting this into itself gives in the next step
$$
z= ln((2n-tfrac12)pi)+i(2n-tfrac12)pi+w.
$$
for some small $w$. Inserting back gives an equation for $w$,
$$
w = Lnleft(1-ifrac{ln((2n-tfrac12)pi)+w}{(2n-tfrac12)pi}right)
$$
For $n$ large enough this is a contraction on ${w:|w|lefrac12}$, ensuring the existence of a solution. The same also holds for the unshifted iteration for $z$.
Different values of $n$ give different fixed-point iterations resulting in different solutions of the original equation, ensuring a countably infinite set of solutions.
Comparing the above first approximations with later iterates of the fixed-point iteration
$$
z_{k+1} = Ln(-iz_k)+i(2n-tfrac12)pi
$$
shows rapid (numerical) convergence and gives the table
begin{array}{l|lll}
n& z_0 & z_{15} & z_{15}-z_0 \ hline
1 & (1.55019499396+4.71238898038j) & (1.53391331978+4.37518515309j) & (-0.0162816741736-0.337203827291j) \
2 & (2.39749285434+10.9955742876j) & (2.40158510487+10.7762995161j) & (0.00409225052324-0.219274771449j) \
3 & (2.84947797809+17.2787595947j) & (2.85358175541+17.1135355394j) & (0.00410377732121-0.165224055332j) \
4 & (3.15963290639+23.5619449019j) & (3.1629527388+23.4277475038j) & (0.00331983241242-0.134197398168j) \
5 & (3.39602168446+29.8451302091j) & (3.39869219676+29.7313107078j) & (0.00267051230882-0.113819501275j) \
6 & (3.58707692122+36.1283155163j) & (3.58926252453+36.0290217034j) & (0.00218560331097-0.0992938128549j) \
7 & (3.74741957129+42.4115008235j) & (3.74924254122+42.3231453612j) & (0.0018229699232-0.0883554622252j) \
8 & (3.88556990977+48.6946861306j) & (3.88711644955+48.6148985649j) & (0.00154653977456-0.0797875657055j) \
9 & (4.00693076678+54.9778714378j) & (4.00826205311+54.9049971233j) & (0.00133128633039-0.0728743144716j) \
10 & (4.11514435142+61.261056745j) & (4.116304664+61.193891332j) & (0.00116031258267-0.0671654130445j) \
11 & (4.21278282098+67.5442420522j) & (4.21380491472+67.48187952j) & (0.00102209373376-0.0623625321652j) \
12 & (4.301730307+73.8274273594j) & (4.3026389193+73.769167656j) & (0.000908612303842-0.0582597033192j) \
end{array}
It also demonstrates a good fit of the initial approximation even for small $n$.
answered Nov 21 '18 at 17:11
LutzLLutzL
56.5k42054
answered Nov 21 '18 at 17:11
LutzLLutzL
56.5k42054
answered Nov 21 '18 at 17:11
LutzLLutzL
56.5k42054
answered Nov 21 '18 at 17:11
LutzLLutzL
56.5k42054
56.5k42054
add a comment |
add a comment |
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2
An non-constant entire function has at most countably many zeros, since if it had uncountably many then you could find a sequence tending to some point such that $f$ was zero on that sequence. By the identity theorem, that function therefore must be everywhere zero.
– Patrick Stevens
Nov 18 '18 at 15:52
Try using Rouche's theorem on a small disk away from $0$.
– anomaly
Nov 18 '18 at 15:58
See math.stackexchange.com/questions/632660/… or math.stackexchange.com/questions/657208/…
– Martin R
Nov 18 '18 at 16:08
@Patrick Stevens: I just saw this question and immediately thought of what you said, then saw what you said (and that it was an hour ago). Oh well. Anyway, in case the OP wants more detail, a slight modification of the usual proof that compact sets in Euclidean space are Bolzano-Weierstrass compact works --- Suppose uncountably many zeros. Then there must be some closed square of side length $1$ containing uncountably many zeros (because $mathbb C$ can be covered by countably many such squares), and in this square there must be a closed square of side length $1/2$ containing uncountably...
– Dave L. Renfro
Nov 18 '18 at 17:38