If $f_nrightarrow f$ uniformly on compact subsets $Omega$ and $f$ is not constant, prove $f(Omega)subseteq...
$begingroup$
Let $Omega$ be open and connected, and let ${f_n}$ be a sequence of holomorphic functions on $Omega$ such that $f_n(Omega)subseteq Omega$. If $f_nrightarrow f$ uniformly on compact subsets $Omega$ and $f$ is not constant, prove $f(Omega)subseteq Omega$
My attempt:
Let $z_0in Omega$ and let $g(z)=f(z)-f(z_0)$ and $h(z)=f_n(z)-f(z_0)$. Clearly, $g(z)$ has a zero in a small neighborhood of $z_0$, which is $z_0$.
If I'm able to show that $h(z)$ also has a zero in that neighborhood, say $z_1$, this implies that $f(z_0)=f_n(z_1)in Omega$ and this completes the proof.
But I'm having trouble showing that $h(z)$ has a zero in the neighborhood of $z_0$. I'm trying to use Rouche's theorem by showing if $|g(z)-h(z)|<|g(z)|$ in the boundary of some neighborhood of $z_0$, $g(z)$ and $h(z)$ has the same number of roots in the neighborhood, hence, there exists some $z_1$ that satisfies my claim.
Can anyone show me how to get the inequality?
complex-analysis
asked Dec 19 '18 at 18:44
Ya GYa G
529211
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be open and connected, and let ${f_n}$ be a sequence of holomorphic functions on $Omega$ such that $f_n(Omega)subseteq Omega$. If $f_nrightarrow f$ uniformly on compact subsets $Omega$ and $f$ is not constant, prove $f(Omega)subseteq Omega$
My attempt:
Let $z_0in Omega$ and let $g(z)=f(z)-f(z_0)$ and $h(z)=f_n(z)-f(z_0)$. Clearly, $g(z)$ has a zero in a small neighborhood of $z_0$, which is $z_0$.
If I'm able to show that $h(z)$ also has a zero in that neighborhood, say $z_1$, this implies that $f(z_0)=f_n(z_1)in Omega$ and this completes the proof.
But I'm having trouble showing that $h(z)$ has a zero in the neighborhood of $z_0$. I'm trying to use Rouche's theorem by showing if $|g(z)-h(z)|<|g(z)|$ in the boundary of some neighborhood of $z_0$, $g(z)$ and $h(z)$ has the same number of roots in the neighborhood, hence, there exists some $z_1$ that satisfies my claim.
Can anyone show me how to get the inequality?
complex-analysis
asked Dec 19 '18 at 18:44
Ya GYa G
529211
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be open and connected, and let ${f_n}$ be a sequence of holomorphic functions on $Omega$ such that $f_n(Omega)subseteq Omega$. If $f_nrightarrow f$ uniformly on compact subsets $Omega$ and $f$ is not constant, prove $f(Omega)subseteq Omega$
My attempt:
Let $z_0in Omega$ and let $g(z)=f(z)-f(z_0)$ and $h(z)=f_n(z)-f(z_0)$. Clearly, $g(z)$ has a zero in a small neighborhood of $z_0$, which is $z_0$.
If I'm able to show that $h(z)$ also has a zero in that neighborhood, say $z_1$, this implies that $f(z_0)=f_n(z_1)in Omega$ and this completes the proof.
But I'm having trouble showing that $h(z)$ has a zero in the neighborhood of $z_0$. I'm trying to use Rouche's theorem by showing if $|g(z)-h(z)|<|g(z)|$ in the boundary of some neighborhood of $z_0$, $g(z)$ and $h(z)$ has the same number of roots in the neighborhood, hence, there exists some $z_1$ that satisfies my claim.
Can anyone show me how to get the inequality?
complex-analysis
asked Dec 19 '18 at 18:44
Ya GYa G
529211
$endgroup$
Let $Omega$ be open and connected, and let ${f_n}$ be a sequence of holomorphic functions on $Omega$ such that $f_n(Omega)subseteq Omega$. If $f_nrightarrow f$ uniformly on compact subsets $Omega$ and $f$ is not constant, prove $f(Omega)subseteq Omega$
My attempt:
Let $z_0in Omega$ and let $g(z)=f(z)-f(z_0)$ and $h(z)=f_n(z)-f(z_0)$. Clearly, $g(z)$ has a zero in a small neighborhood of $z_0$, which is $z_0$.
If I'm able to show that $h(z)$ also has a zero in that neighborhood, say $z_1$, this implies that $f(z_0)=f_n(z_1)in Omega$ and this completes the proof.
But I'm having trouble showing that $h(z)$ has a zero in the neighborhood of $z_0$. I'm trying to use Rouche's theorem by showing if $|g(z)-h(z)|<|g(z)|$ in the boundary of some neighborhood of $z_0$, $g(z)$ and $h(z)$ has the same number of roots in the neighborhood, hence, there exists some $z_1$ that satisfies my claim.
Can anyone show me how to get the inequality?
complex-analysis
complex-analysis
asked Dec 19 '18 at 18:44
Ya GYa G
529211
asked Dec 19 '18 at 18:44
Ya GYa G
529211
asked Dec 19 '18 at 18:44
Ya GYa G
529211
asked Dec 19 '18 at 18:44
Ya GYa G
529211
asked Dec 19 '18 at 18:44
Ya GYa G
529211
529211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose for the sake of contradiction there exists a point $z_0$ such that $z_0 notin Omega$, such that $f(w_0) = z_0$. WLOG $z_0 = 0$, then by the argument principle begin{equation} int_{gamma}
frac{f'}{f} geq 1 end{equation} where $gamma$ is a small contour around zero so that it doesn't intersect $Omega$; this is possible since $Omega$ is open. But observe begin{equation} int_{gamma} frac{f_n'}{f_n} = 0 end{equation} for any $n$ since $f_n(Omega) subset Omega$, so $f_n neq 0$ for any $z in Omega$. So by uniform convergence we get begin{equation} int_{gamma}
frac{f'}{f} = 0 end{equation} which is our contradiction. In general uniform convergence + argument principle gives you many strong results such as if $f_n$ is $1-1$ and converges uniformly to $f$, then either $f$ is constant or $1-1$ and similarly if $f_n neq 0$ for any $z$. (Uniform convergence allows you to pass limit to integrals, and this integral tells you how many zeros there are.)
answered Dec 19 '18 at 18:55
Story123Story123
23718
$endgroup$
$begingroup$
So the argument principle states that if $f$ is meromorphic with roots and poles not on the contour $gamma$, we have $frac{1}{2pi i}int_{gamma}frac{f'}{f}=$ number of zeros in $gamma$ - number of poles in $gamma$. How do we know that $int_{gamma}frac{f'}{f}geq1$? I guess I just need some more explanation on your reasoning.
$endgroup$
– Ya G
Dec 19 '18 at 19:22
$begingroup$
As $f_n rightarrow f$ uniformly on compact sets, $f_n$ holomorphic, we know by Morerra Theorem and uniform convergence that $f$ is holomorphic. So it has no poles, but $f(z_0) = 0$, so it has at least one zero in $gamma$ namely $z_0$. So the integral is greater than or equal to 1.
$endgroup$
– Story123
Dec 19 '18 at 19:26
1
$begingroup$
Got it! So you are assuming that there is a root inside the contour such that the root itself is not in $Omega$. But since it's inside the contour, the argument principle suggests that there exists at least one zero in the contour.
$endgroup$
– Ya G
Dec 19 '18 at 19:31
$begingroup$
Yes, you can assume WLOG it's a root since if it isn't consider $g(z) := f(z) - w_0$, then $g'(z) = f'(z)$ and you can apply the argument principle to $g$.
$endgroup$
– Story123
Dec 19 '18 at 19:36
$begingroup$
What if $z_0in partial Omega?$ Then every nbhd of $z_0$ intersects $Omega$
$endgroup$
– Matematleta
Dec 19 '18 at 22:06
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046732%2fif-f-n-rightarrow-f-uniformly-on-compact-subsets-omega-and-f-is-not-const%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose for the sake of contradiction there exists a point $z_0$ such that $z_0 notin Omega$, such that $f(w_0) = z_0$. WLOG $z_0 = 0$, then by the argument principle begin{equation} int_{gamma}
frac{f'}{f} geq 1 end{equation} where $gamma$ is a small contour around zero so that it doesn't intersect $Omega$; this is possible since $Omega$ is open. But observe begin{equation} int_{gamma} frac{f_n'}{f_n} = 0 end{equation} for any $n$ since $f_n(Omega) subset Omega$, so $f_n neq 0$ for any $z in Omega$. So by uniform convergence we get begin{equation} int_{gamma}
frac{f'}{f} = 0 end{equation} which is our contradiction. In general uniform convergence + argument principle gives you many strong results such as if $f_n$ is $1-1$ and converges uniformly to $f$, then either $f$ is constant or $1-1$ and similarly if $f_n neq 0$ for any $z$. (Uniform convergence allows you to pass limit to integrals, and this integral tells you how many zeros there are.)
answered Dec 19 '18 at 18:55
Story123Story123
23718
$endgroup$
$begingroup$
So the argument principle states that if $f$ is meromorphic with roots and poles not on the contour $gamma$, we have $frac{1}{2pi i}int_{gamma}frac{f'}{f}=$ number of zeros in $gamma$ - number of poles in $gamma$. How do we know that $int_{gamma}frac{f'}{f}geq1$? I guess I just need some more explanation on your reasoning.
$endgroup$
– Ya G
Dec 19 '18 at 19:22
$begingroup$
As $f_n rightarrow f$ uniformly on compact sets, $f_n$ holomorphic, we know by Morerra Theorem and uniform convergence that $f$ is holomorphic. So it has no poles, but $f(z_0) = 0$, so it has at least one zero in $gamma$ namely $z_0$. So the integral is greater than or equal to 1.
$endgroup$
– Story123
Dec 19 '18 at 19:26
1
$begingroup$
Got it! So you are assuming that there is a root inside the contour such that the root itself is not in $Omega$. But since it's inside the contour, the argument principle suggests that there exists at least one zero in the contour.
$endgroup$
– Ya G
Dec 19 '18 at 19:31
$begingroup$
Yes, you can assume WLOG it's a root since if it isn't consider $g(z) := f(z) - w_0$, then $g'(z) = f'(z)$ and you can apply the argument principle to $g$.
$endgroup$
– Story123
Dec 19 '18 at 19:36
$begingroup$
What if $z_0in partial Omega?$ Then every nbhd of $z_0$ intersects $Omega$
$endgroup$
– Matematleta
Dec 19 '18 at 22:06
add a comment |
$begingroup$
Suppose for the sake of contradiction there exists a point $z_0$ such that $z_0 notin Omega$, such that $f(w_0) = z_0$. WLOG $z_0 = 0$, then by the argument principle begin{equation} int_{gamma}
frac{f'}{f} geq 1 end{equation} where $gamma$ is a small contour around zero so that it doesn't intersect $Omega$; this is possible since $Omega$ is open. But observe begin{equation} int_{gamma} frac{f_n'}{f_n} = 0 end{equation} for any $n$ since $f_n(Omega) subset Omega$, so $f_n neq 0$ for any $z in Omega$. So by uniform convergence we get begin{equation} int_{gamma}
frac{f'}{f} = 0 end{equation} which is our contradiction. In general uniform convergence + argument principle gives you many strong results such as if $f_n$ is $1-1$ and converges uniformly to $f$, then either $f$ is constant or $1-1$ and similarly if $f_n neq 0$ for any $z$. (Uniform convergence allows you to pass limit to integrals, and this integral tells you how many zeros there are.)
answered Dec 19 '18 at 18:55
Story123Story123
23718
$endgroup$
$begingroup$
So the argument principle states that if $f$ is meromorphic with roots and poles not on the contour $gamma$, we have $frac{1}{2pi i}int_{gamma}frac{f'}{f}=$ number of zeros in $gamma$ - number of poles in $gamma$. How do we know that $int_{gamma}frac{f'}{f}geq1$? I guess I just need some more explanation on your reasoning.
$endgroup$
– Ya G
Dec 19 '18 at 19:22
$begingroup$
As $f_n rightarrow f$ uniformly on compact sets, $f_n$ holomorphic, we know by Morerra Theorem and uniform convergence that $f$ is holomorphic. So it has no poles, but $f(z_0) = 0$, so it has at least one zero in $gamma$ namely $z_0$. So the integral is greater than or equal to 1.
$endgroup$
– Story123
Dec 19 '18 at 19:26
1
$begingroup$
Got it! So you are assuming that there is a root inside the contour such that the root itself is not in $Omega$. But since it's inside the contour, the argument principle suggests that there exists at least one zero in the contour.
$endgroup$
– Ya G
Dec 19 '18 at 19:31
$begingroup$
Yes, you can assume WLOG it's a root since if it isn't consider $g(z) := f(z) - w_0$, then $g'(z) = f'(z)$ and you can apply the argument principle to $g$.
$endgroup$
– Story123
Dec 19 '18 at 19:36
$begingroup$
What if $z_0in partial Omega?$ Then every nbhd of $z_0$ intersects $Omega$
$endgroup$
– Matematleta
Dec 19 '18 at 22:06
add a comment |
$begingroup$
Suppose for the sake of contradiction there exists a point $z_0$ such that $z_0 notin Omega$, such that $f(w_0) = z_0$. WLOG $z_0 = 0$, then by the argument principle begin{equation} int_{gamma}
frac{f'}{f} geq 1 end{equation} where $gamma$ is a small contour around zero so that it doesn't intersect $Omega$; this is possible since $Omega$ is open. But observe begin{equation} int_{gamma} frac{f_n'}{f_n} = 0 end{equation} for any $n$ since $f_n(Omega) subset Omega$, so $f_n neq 0$ for any $z in Omega$. So by uniform convergence we get begin{equation} int_{gamma}
frac{f'}{f} = 0 end{equation} which is our contradiction. In general uniform convergence + argument principle gives you many strong results such as if $f_n$ is $1-1$ and converges uniformly to $f$, then either $f$ is constant or $1-1$ and similarly if $f_n neq 0$ for any $z$. (Uniform convergence allows you to pass limit to integrals, and this integral tells you how many zeros there are.)
answered Dec 19 '18 at 18:55
Story123Story123
23718
$endgroup$
Suppose for the sake of contradiction there exists a point $z_0$ such that $z_0 notin Omega$, such that $f(w_0) = z_0$. WLOG $z_0 = 0$, then by the argument principle begin{equation} int_{gamma}
frac{f'}{f} geq 1 end{equation} where $gamma$ is a small contour around zero so that it doesn't intersect $Omega$; this is possible since $Omega$ is open. But observe begin{equation} int_{gamma} frac{f_n'}{f_n} = 0 end{equation} for any $n$ since $f_n(Omega) subset Omega$, so $f_n neq 0$ for any $z in Omega$. So by uniform convergence we get begin{equation} int_{gamma}
frac{f'}{f} = 0 end{equation} which is our contradiction. In general uniform convergence + argument principle gives you many strong results such as if $f_n$ is $1-1$ and converges uniformly to $f$, then either $f$ is constant or $1-1$ and similarly if $f_n neq 0$ for any $z$. (Uniform convergence allows you to pass limit to integrals, and this integral tells you how many zeros there are.)
answered Dec 19 '18 at 18:55
Story123Story123
23718
edited Dec 19 '18 at 19:01
answered Dec 19 '18 at 18:55
Story123Story123
23718
answered Dec 19 '18 at 18:55
Story123Story123
23718
answered Dec 19 '18 at 18:55
Story123Story123
23718
23718
$begingroup$
So the argument principle states that if $f$ is meromorphic with roots and poles not on the contour $gamma$, we have $frac{1}{2pi i}int_{gamma}frac{f'}{f}=$ number of zeros in $gamma$ - number of poles in $gamma$. How do we know that $int_{gamma}frac{f'}{f}geq1$? I guess I just need some more explanation on your reasoning.
$endgroup$
– Ya G
Dec 19 '18 at 19:22
$begingroup$
As $f_n rightarrow f$ uniformly on compact sets, $f_n$ holomorphic, we know by Morerra Theorem and uniform convergence that $f$ is holomorphic. So it has no poles, but $f(z_0) = 0$, so it has at least one zero in $gamma$ namely $z_0$. So the integral is greater than or equal to 1.
$endgroup$
– Story123
Dec 19 '18 at 19:26
1
$begingroup$
Got it! So you are assuming that there is a root inside the contour such that the root itself is not in $Omega$. But since it's inside the contour, the argument principle suggests that there exists at least one zero in the contour.
$endgroup$
– Ya G
Dec 19 '18 at 19:31
$begingroup$
Yes, you can assume WLOG it's a root since if it isn't consider $g(z) := f(z) - w_0$, then $g'(z) = f'(z)$ and you can apply the argument principle to $g$.
$endgroup$
– Story123
Dec 19 '18 at 19:36
$begingroup$
What if $z_0in partial Omega?$ Then every nbhd of $z_0$ intersects $Omega$
$endgroup$
– Matematleta
Dec 19 '18 at 22:06
add a comment |
$begingroup$
So the argument principle states that if $f$ is meromorphic with roots and poles not on the contour $gamma$, we have $frac{1}{2pi i}int_{gamma}frac{f'}{f}=$ number of zeros in $gamma$ - number of poles in $gamma$. How do we know that $int_{gamma}frac{f'}{f}geq1$? I guess I just need some more explanation on your reasoning.
$endgroup$
– Ya G
Dec 19 '18 at 19:22
$begingroup$
As $f_n rightarrow f$ uniformly on compact sets, $f_n$ holomorphic, we know by Morerra Theorem and uniform convergence that $f$ is holomorphic. So it has no poles, but $f(z_0) = 0$, so it has at least one zero in $gamma$ namely $z_0$. So the integral is greater than or equal to 1.
$endgroup$
– Story123
Dec 19 '18 at 19:26
1
$begingroup$
Got it! So you are assuming that there is a root inside the contour such that the root itself is not in $Omega$. But since it's inside the contour, the argument principle suggests that there exists at least one zero in the contour.
$endgroup$
– Ya G
Dec 19 '18 at 19:31
$begingroup$
Yes, you can assume WLOG it's a root since if it isn't consider $g(z) := f(z) - w_0$, then $g'(z) = f'(z)$ and you can apply the argument principle to $g$.
$endgroup$
– Story123
Dec 19 '18 at 19:36
$begingroup$
What if $z_0in partial Omega?$ Then every nbhd of $z_0$ intersects $Omega$
$endgroup$
– Matematleta
Dec 19 '18 at 22:06
$begingroup$
So the argument principle states that if $f$ is meromorphic with roots and poles not on the contour $gamma$, we have $frac{1}{2pi i}int_{gamma}frac{f'}{f}=$ number of zeros in $gamma$ - number of poles in $gamma$. How do we know that $int_{gamma}frac{f'}{f}geq1$? I guess I just need some more explanation on your reasoning.
$endgroup$
– Ya G
Dec 19 '18 at 19:22
$begingroup$
So the argument principle states that if $f$ is meromorphic with roots and poles not on the contour $gamma$, we have $frac{1}{2pi i}int_{gamma}frac{f'}{f}=$ number of zeros in $gamma$ - number of poles in $gamma$. How do we know that $int_{gamma}frac{f'}{f}geq1$? I guess I just need some more explanation on your reasoning.
$endgroup$
– Ya G
Dec 19 '18 at 19:22
$begingroup$
As $f_n rightarrow f$ uniformly on compact sets, $f_n$ holomorphic, we know by Morerra Theorem and uniform convergence that $f$ is holomorphic. So it has no poles, but $f(z_0) = 0$, so it has at least one zero in $gamma$ namely $z_0$. So the integral is greater than or equal to 1.
$endgroup$
– Story123
Dec 19 '18 at 19:26
$begingroup$
As $f_n rightarrow f$ uniformly on compact sets, $f_n$ holomorphic, we know by Morerra Theorem and uniform convergence that $f$ is holomorphic. So it has no poles, but $f(z_0) = 0$, so it has at least one zero in $gamma$ namely $z_0$. So the integral is greater than or equal to 1.
$endgroup$
– Story123
Dec 19 '18 at 19:26
1
1
$begingroup$
Got it! So you are assuming that there is a root inside the contour such that the root itself is not in $Omega$. But since it's inside the contour, the argument principle suggests that there exists at least one zero in the contour.
$endgroup$
– Ya G
Dec 19 '18 at 19:31
$begingroup$
Got it! So you are assuming that there is a root inside the contour such that the root itself is not in $Omega$. But since it's inside the contour, the argument principle suggests that there exists at least one zero in the contour.
$endgroup$
– Ya G
Dec 19 '18 at 19:31
$begingroup$
Yes, you can assume WLOG it's a root since if it isn't consider $g(z) := f(z) - w_0$, then $g'(z) = f'(z)$ and you can apply the argument principle to $g$.
$endgroup$
– Story123
Dec 19 '18 at 19:36
$begingroup$
Yes, you can assume WLOG it's a root since if it isn't consider $g(z) := f(z) - w_0$, then $g'(z) = f'(z)$ and you can apply the argument principle to $g$.
$endgroup$
– Story123
Dec 19 '18 at 19:36
$begingroup$
What if $z_0in partial Omega?$ Then every nbhd of $z_0$ intersects $Omega$
$endgroup$
– Matematleta
Dec 19 '18 at 22:06
$begingroup$
What if $z_0in partial Omega?$ Then every nbhd of $z_0$ intersects $Omega$
$endgroup$
– Matematleta
Dec 19 '18 at 22:06
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046732%2fif-f-n-rightarrow-f-uniformly-on-compact-subsets-omega-and-f-is-not-const%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
