De Moivres Theorem
$begingroup$
This is the questions I got and I'm stuck as to where to proceed after these few steps
complex-numbers
asked Jan 24 at 2:29
user122343user122343
805
$endgroup$
add a comment |
$begingroup$
This is the questions I got and I'm stuck as to where to proceed after these few steps
complex-numbers
asked Jan 24 at 2:29
user122343user122343
805
$endgroup$
add a comment |
$begingroup$
This is the questions I got and I'm stuck as to where to proceed after these few steps
complex-numbers
asked Jan 24 at 2:29
user122343user122343
805
$endgroup$
This is the questions I got and I'm stuck as to where to proceed after these few steps
complex-numbers
complex-numbers
asked Jan 24 at 2:29
user122343user122343
805
asked Jan 24 at 2:29
user122343user122343
805
asked Jan 24 at 2:29
user122343user122343
805
asked Jan 24 at 2:29
user122343user122343
805
asked Jan 24 at 2:29
user122343user122343
805
805
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
For the second part since we want to solve the equation $$x^4-6x^2+1=0$$ in terms of $tan(frac{npi}{8})$ we assume that:
$$frac{npi}{8}=theta$$
Now $$4theta = frac{npi}{2}$$
Therefore, $$tan(4theta) = tan(frac{npi}{2})$$
RHS is always either $infty$ or $-infty$ so whatever expansion of $tan(4theta)$ is in the denominator that will obviously be $0$.
Now since, denominator is $0$ or $$tan^4theta-6tan^2theta+1=0$$
Now you assume that $$x=tantheta$$ and solve the biquadratic equation to get the answer. The equation will become:
$$x^4-6x^2+1=0$$
How to solve this equation?? Put $$x^2=t$$ and rewrite the biquadratic equation as a quadratic equation in $t$ giving
$$t^2-6t+1=0$$. Solve for $t$ and then put $x^2=t$ to get the roots of x.
Hope this helps ......
answered Jan 24 at 4:02
SNEHIL SANYALSNEHIL SANYAL
614110
$endgroup$
$begingroup$
Thanks it did help
$endgroup$
– user122343
Jan 25 at 10:26
$begingroup$
I am happy to know that, welcome
$endgroup$
– SNEHIL SANYAL
Jan 25 at 13:23
add a comment |
$begingroup$
For the first part of the question:
Notice by DMT:
$$(costheta +isintheta)^4=cos(4theta)+isin(4theta)$$
Since $tan x=frac{sin x}{cos x}$, we have that: $$tan(4theta)=frac{Im((costheta+isintheta)^4)}{Re((costheta+isintheta)^4)}$$
Use the binomial expansion to do the rest.
answered Jan 24 at 3:36
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
$begingroup$
For the second part, set $dfrac1{tan4theta }=0$
$implies4theta=(2n+1)dfracpi2$ for some integer $n$
$theta=dfrac{(2n+1)pi}8$ where $nequiv0,1,2,3pmod4$
answered Jan 24 at 3:55
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
$$sin(4 theta) = 2 sin (2 theta) cos(2 theta)$$
$$cos(4 theta) = cos^2(2 theta)-sin^2(2theta)$$
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
Note that all of these can be derived from de Moivre's theorem $(cos theta + j sin theta)^n = cos(n theta) + jsin(n theta)$.
For example, when $n=2$,
$$(cos theta + j sin theta)^2 = cos^2(theta) - sin^2(theta) + j2 sin theta cos theta = cos(2 theta) + jsin(2 theta)$$
implying
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
You can similarly derive the result for $n=4$.
Anyway, now you can compute
$$tan(4 theta) = {sin(4 theta) over cos(4 theta)} = {4 sin theta cos theta ; (cos^2 theta - sin^2 theta) over (cos^2 theta - sin^2 theta)^2 - 4sin^2 theta cos^2 theta}$$
If you divide the numerator and denominator by $cos^4 theta$, this simplifies to
$$tan(4 theta) = {4 tan theta - 4 tan^3 theta over 1 - 6 tan^2 theta + tan^4 theta}$$
Now, note that when $4 theta = n pi/2$, $tan 4 theta = infty$, i.e. $1 - 6 tan^2 theta + tan^4 theta = 0$. This is exactly the equation you want to solve (with $x = tan theta$), so you get $x = tan (n pi/8)$.
answered Jan 24 at 3:53
Aditya DuaAditya Dua
1,15418
$endgroup$
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%2f3085362%2fde-moivres-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the second part since we want to solve the equation $$x^4-6x^2+1=0$$ in terms of $tan(frac{npi}{8})$ we assume that:
$$frac{npi}{8}=theta$$
Now $$4theta = frac{npi}{2}$$
Therefore, $$tan(4theta) = tan(frac{npi}{2})$$
RHS is always either $infty$ or $-infty$ so whatever expansion of $tan(4theta)$ is in the denominator that will obviously be $0$.
Now since, denominator is $0$ or $$tan^4theta-6tan^2theta+1=0$$
Now you assume that $$x=tantheta$$ and solve the biquadratic equation to get the answer. The equation will become:
$$x^4-6x^2+1=0$$
How to solve this equation?? Put $$x^2=t$$ and rewrite the biquadratic equation as a quadratic equation in $t$ giving
$$t^2-6t+1=0$$. Solve for $t$ and then put $x^2=t$ to get the roots of x.
Hope this helps ......
answered Jan 24 at 4:02
SNEHIL SANYALSNEHIL SANYAL
614110
$endgroup$
$begingroup$
Thanks it did help
$endgroup$
– user122343
Jan 25 at 10:26
$begingroup$
I am happy to know that, welcome
$endgroup$
– SNEHIL SANYAL
Jan 25 at 13:23
add a comment |
$begingroup$
For the second part since we want to solve the equation $$x^4-6x^2+1=0$$ in terms of $tan(frac{npi}{8})$ we assume that:
$$frac{npi}{8}=theta$$
Now $$4theta = frac{npi}{2}$$
Therefore, $$tan(4theta) = tan(frac{npi}{2})$$
RHS is always either $infty$ or $-infty$ so whatever expansion of $tan(4theta)$ is in the denominator that will obviously be $0$.
Now since, denominator is $0$ or $$tan^4theta-6tan^2theta+1=0$$
Now you assume that $$x=tantheta$$ and solve the biquadratic equation to get the answer. The equation will become:
$$x^4-6x^2+1=0$$
How to solve this equation?? Put $$x^2=t$$ and rewrite the biquadratic equation as a quadratic equation in $t$ giving
$$t^2-6t+1=0$$. Solve for $t$ and then put $x^2=t$ to get the roots of x.
Hope this helps ......
answered Jan 24 at 4:02
SNEHIL SANYALSNEHIL SANYAL
614110
$endgroup$
$begingroup$
Thanks it did help
$endgroup$
– user122343
Jan 25 at 10:26
$begingroup$
I am happy to know that, welcome
$endgroup$
– SNEHIL SANYAL
Jan 25 at 13:23
add a comment |
$begingroup$
For the second part since we want to solve the equation $$x^4-6x^2+1=0$$ in terms of $tan(frac{npi}{8})$ we assume that:
$$frac{npi}{8}=theta$$
Now $$4theta = frac{npi}{2}$$
Therefore, $$tan(4theta) = tan(frac{npi}{2})$$
RHS is always either $infty$ or $-infty$ so whatever expansion of $tan(4theta)$ is in the denominator that will obviously be $0$.
Now since, denominator is $0$ or $$tan^4theta-6tan^2theta+1=0$$
Now you assume that $$x=tantheta$$ and solve the biquadratic equation to get the answer. The equation will become:
$$x^4-6x^2+1=0$$
How to solve this equation?? Put $$x^2=t$$ and rewrite the biquadratic equation as a quadratic equation in $t$ giving
$$t^2-6t+1=0$$. Solve for $t$ and then put $x^2=t$ to get the roots of x.
Hope this helps ......
answered Jan 24 at 4:02
SNEHIL SANYALSNEHIL SANYAL
614110
$endgroup$
For the second part since we want to solve the equation $$x^4-6x^2+1=0$$ in terms of $tan(frac{npi}{8})$ we assume that:
$$frac{npi}{8}=theta$$
Now $$4theta = frac{npi}{2}$$
Therefore, $$tan(4theta) = tan(frac{npi}{2})$$
RHS is always either $infty$ or $-infty$ so whatever expansion of $tan(4theta)$ is in the denominator that will obviously be $0$.
Now since, denominator is $0$ or $$tan^4theta-6tan^2theta+1=0$$
Now you assume that $$x=tantheta$$ and solve the biquadratic equation to get the answer. The equation will become:
$$x^4-6x^2+1=0$$
How to solve this equation?? Put $$x^2=t$$ and rewrite the biquadratic equation as a quadratic equation in $t$ giving
$$t^2-6t+1=0$$. Solve for $t$ and then put $x^2=t$ to get the roots of x.
Hope this helps ......
answered Jan 24 at 4:02
SNEHIL SANYALSNEHIL SANYAL
614110
answered Jan 24 at 4:02
SNEHIL SANYALSNEHIL SANYAL
614110
answered Jan 24 at 4:02
SNEHIL SANYALSNEHIL SANYAL
614110
answered Jan 24 at 4:02
SNEHIL SANYALSNEHIL SANYAL
614110
614110
$begingroup$
Thanks it did help
$endgroup$
– user122343
Jan 25 at 10:26
$begingroup$
I am happy to know that, welcome
$endgroup$
– SNEHIL SANYAL
Jan 25 at 13:23
add a comment |
$begingroup$
Thanks it did help
$endgroup$
– user122343
Jan 25 at 10:26
$begingroup$
I am happy to know that, welcome
$endgroup$
– SNEHIL SANYAL
Jan 25 at 13:23
$begingroup$
Thanks it did help
$endgroup$
– user122343
Jan 25 at 10:26
$begingroup$
Thanks it did help
$endgroup$
– user122343
Jan 25 at 10:26
$begingroup$
I am happy to know that, welcome
$endgroup$
– SNEHIL SANYAL
Jan 25 at 13:23
$begingroup$
I am happy to know that, welcome
$endgroup$
– SNEHIL SANYAL
Jan 25 at 13:23
add a comment |
$begingroup$
For the first part of the question:
Notice by DMT:
$$(costheta +isintheta)^4=cos(4theta)+isin(4theta)$$
Since $tan x=frac{sin x}{cos x}$, we have that: $$tan(4theta)=frac{Im((costheta+isintheta)^4)}{Re((costheta+isintheta)^4)}$$
Use the binomial expansion to do the rest.
answered Jan 24 at 3:36
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
$begingroup$
For the first part of the question:
Notice by DMT:
$$(costheta +isintheta)^4=cos(4theta)+isin(4theta)$$
Since $tan x=frac{sin x}{cos x}$, we have that: $$tan(4theta)=frac{Im((costheta+isintheta)^4)}{Re((costheta+isintheta)^4)}$$
Use the binomial expansion to do the rest.
answered Jan 24 at 3:36
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
$begingroup$
For the first part of the question:
Notice by DMT:
$$(costheta +isintheta)^4=cos(4theta)+isin(4theta)$$
Since $tan x=frac{sin x}{cos x}$, we have that: $$tan(4theta)=frac{Im((costheta+isintheta)^4)}{Re((costheta+isintheta)^4)}$$
Use the binomial expansion to do the rest.
answered Jan 24 at 3:36
Rhys HughesRhys Hughes
6,9571530
$endgroup$
For the first part of the question:
Notice by DMT:
$$(costheta +isintheta)^4=cos(4theta)+isin(4theta)$$
Since $tan x=frac{sin x}{cos x}$, we have that: $$tan(4theta)=frac{Im((costheta+isintheta)^4)}{Re((costheta+isintheta)^4)}$$
Use the binomial expansion to do the rest.
answered Jan 24 at 3:36
Rhys HughesRhys Hughes
6,9571530
answered Jan 24 at 3:36
Rhys HughesRhys Hughes
6,9571530
answered Jan 24 at 3:36
Rhys HughesRhys Hughes
6,9571530
answered Jan 24 at 3:36
Rhys HughesRhys Hughes
6,9571530
6,9571530
add a comment |
add a comment |
$begingroup$
For the second part, set $dfrac1{tan4theta }=0$
$implies4theta=(2n+1)dfracpi2$ for some integer $n$
$theta=dfrac{(2n+1)pi}8$ where $nequiv0,1,2,3pmod4$
answered Jan 24 at 3:55
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
For the second part, set $dfrac1{tan4theta }=0$
$implies4theta=(2n+1)dfracpi2$ for some integer $n$
$theta=dfrac{(2n+1)pi}8$ where $nequiv0,1,2,3pmod4$
answered Jan 24 at 3:55
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
For the second part, set $dfrac1{tan4theta }=0$
$implies4theta=(2n+1)dfracpi2$ for some integer $n$
$theta=dfrac{(2n+1)pi}8$ where $nequiv0,1,2,3pmod4$
answered Jan 24 at 3:55
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
For the second part, set $dfrac1{tan4theta }=0$
$implies4theta=(2n+1)dfracpi2$ for some integer $n$
$theta=dfrac{(2n+1)pi}8$ where $nequiv0,1,2,3pmod4$
answered Jan 24 at 3:55
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 24 at 3:55
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 24 at 3:55
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 24 at 3:55
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
$begingroup$
$$sin(4 theta) = 2 sin (2 theta) cos(2 theta)$$
$$cos(4 theta) = cos^2(2 theta)-sin^2(2theta)$$
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
Note that all of these can be derived from de Moivre's theorem $(cos theta + j sin theta)^n = cos(n theta) + jsin(n theta)$.
For example, when $n=2$,
$$(cos theta + j sin theta)^2 = cos^2(theta) - sin^2(theta) + j2 sin theta cos theta = cos(2 theta) + jsin(2 theta)$$
implying
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
You can similarly derive the result for $n=4$.
Anyway, now you can compute
$$tan(4 theta) = {sin(4 theta) over cos(4 theta)} = {4 sin theta cos theta ; (cos^2 theta - sin^2 theta) over (cos^2 theta - sin^2 theta)^2 - 4sin^2 theta cos^2 theta}$$
If you divide the numerator and denominator by $cos^4 theta$, this simplifies to
$$tan(4 theta) = {4 tan theta - 4 tan^3 theta over 1 - 6 tan^2 theta + tan^4 theta}$$
Now, note that when $4 theta = n pi/2$, $tan 4 theta = infty$, i.e. $1 - 6 tan^2 theta + tan^4 theta = 0$. This is exactly the equation you want to solve (with $x = tan theta$), so you get $x = tan (n pi/8)$.
answered Jan 24 at 3:53
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
$$sin(4 theta) = 2 sin (2 theta) cos(2 theta)$$
$$cos(4 theta) = cos^2(2 theta)-sin^2(2theta)$$
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
Note that all of these can be derived from de Moivre's theorem $(cos theta + j sin theta)^n = cos(n theta) + jsin(n theta)$.
For example, when $n=2$,
$$(cos theta + j sin theta)^2 = cos^2(theta) - sin^2(theta) + j2 sin theta cos theta = cos(2 theta) + jsin(2 theta)$$
implying
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
You can similarly derive the result for $n=4$.
Anyway, now you can compute
$$tan(4 theta) = {sin(4 theta) over cos(4 theta)} = {4 sin theta cos theta ; (cos^2 theta - sin^2 theta) over (cos^2 theta - sin^2 theta)^2 - 4sin^2 theta cos^2 theta}$$
If you divide the numerator and denominator by $cos^4 theta$, this simplifies to
$$tan(4 theta) = {4 tan theta - 4 tan^3 theta over 1 - 6 tan^2 theta + tan^4 theta}$$
Now, note that when $4 theta = n pi/2$, $tan 4 theta = infty$, i.e. $1 - 6 tan^2 theta + tan^4 theta = 0$. This is exactly the equation you want to solve (with $x = tan theta$), so you get $x = tan (n pi/8)$.
answered Jan 24 at 3:53
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
$$sin(4 theta) = 2 sin (2 theta) cos(2 theta)$$
$$cos(4 theta) = cos^2(2 theta)-sin^2(2theta)$$
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
Note that all of these can be derived from de Moivre's theorem $(cos theta + j sin theta)^n = cos(n theta) + jsin(n theta)$.
For example, when $n=2$,
$$(cos theta + j sin theta)^2 = cos^2(theta) - sin^2(theta) + j2 sin theta cos theta = cos(2 theta) + jsin(2 theta)$$
implying
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
You can similarly derive the result for $n=4$.
Anyway, now you can compute
$$tan(4 theta) = {sin(4 theta) over cos(4 theta)} = {4 sin theta cos theta ; (cos^2 theta - sin^2 theta) over (cos^2 theta - sin^2 theta)^2 - 4sin^2 theta cos^2 theta}$$
If you divide the numerator and denominator by $cos^4 theta$, this simplifies to
$$tan(4 theta) = {4 tan theta - 4 tan^3 theta over 1 - 6 tan^2 theta + tan^4 theta}$$
Now, note that when $4 theta = n pi/2$, $tan 4 theta = infty$, i.e. $1 - 6 tan^2 theta + tan^4 theta = 0$. This is exactly the equation you want to solve (with $x = tan theta$), so you get $x = tan (n pi/8)$.
answered Jan 24 at 3:53
Aditya DuaAditya Dua
1,15418
$endgroup$
$$sin(4 theta) = 2 sin (2 theta) cos(2 theta)$$
$$cos(4 theta) = cos^2(2 theta)-sin^2(2theta)$$
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
Note that all of these can be derived from de Moivre's theorem $(cos theta + j sin theta)^n = cos(n theta) + jsin(n theta)$.
For example, when $n=2$,
$$(cos theta + j sin theta)^2 = cos^2(theta) - sin^2(theta) + j2 sin theta cos theta = cos(2 theta) + jsin(2 theta)$$
implying
$$sin(2theta) = 2sin(theta)cos(theta)$$
$$cos(2theta) = cos^2(theta)-sin^2(theta)$$
You can similarly derive the result for $n=4$.
Anyway, now you can compute
$$tan(4 theta) = {sin(4 theta) over cos(4 theta)} = {4 sin theta cos theta ; (cos^2 theta - sin^2 theta) over (cos^2 theta - sin^2 theta)^2 - 4sin^2 theta cos^2 theta}$$
If you divide the numerator and denominator by $cos^4 theta$, this simplifies to
$$tan(4 theta) = {4 tan theta - 4 tan^3 theta over 1 - 6 tan^2 theta + tan^4 theta}$$
Now, note that when $4 theta = n pi/2$, $tan 4 theta = infty$, i.e. $1 - 6 tan^2 theta + tan^4 theta = 0$. This is exactly the equation you want to solve (with $x = tan theta$), so you get $x = tan (n pi/8)$.
answered Jan 24 at 3:53
Aditya DuaAditya Dua
1,15418
answered Jan 24 at 3:53
Aditya DuaAditya Dua
1,15418
answered Jan 24 at 3:53
Aditya DuaAditya Dua
1,15418
answered Jan 24 at 3:53
Aditya DuaAditya Dua
1,15418
1,15418
add a comment |
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%2f3085362%2fde-moivres-theorem%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
