Mixing
Prob. 12, Sec. 4.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: If $f(x+y)=f(x)+f(y)$ for all $x, y$ and if $lim_xto 0 f(x)$ exists, . . .
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Here is Prob. 12, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition:
Let $f colon mathbbR to mathbbR$ be such that $f(x+y) = f(x) + f(y)$ for all $x, y$ in $mathbbR$. Assume that $lim_x to 0 f = L$ exists. Prove that $L = 0$, and then prove that $f$ has a limit at every point $c in mathbbR$. [Hint: First note that $f(2x) = f(x) + f(x) = 2 f(x)$ for $x in mathbbR$. Also note that $f(x) = f(x-c) + f(c)$ for $x, c$ in $mathbbR$.]
My Attempt:
As $lim_x to 0 f (x) = L$, so, given any real number $varepsilon > 0$, we can find a real number $delta > 0$ such that
$$ lvert f(x) - L rvert < varepsilon tag1 $$
for all $x in mathbbR$ for which
$$ 0 < lvert x - 0 rvert < delta; tag2 $$
therefore, we can also conclude that, for all $x in mathbbR$ for which
$$ 0 < lvert x - 0 rvert < fracdelta2 $$
holds, for those $x$ we also have
$$ 0 < lvert 2x - 0 rvert < delta, $$
and so
$$ lvert f(2x) - L rvert < varepsilon, $$
and thus it follows that $$ lim_x to 0 f(2x) = L $$
also. But as $f(2x) = 2 f(x)$ for all $x in mathbbR$, so we must have
$$ lim_x to 0 f(2x) = 2 lim_x to 0 f(x), $$
that is we must have
$$ L = 2L,$$
which in turn implies that $L = 0$.
Hence $$ lim_x to 0 f(x) = 0 $$
holds.
Am I right? Is this part of the proof correct and clear enough?
Now for any real numbers $c, x$, we find that
$$ f(x) = f(x-c + c) = f(x-c) + f(c), $$
and so
$$ f(x) - f(c) = f(x-c). $$
Now if $x= c$, then we obtain
$$ f(c) - f(c) = f(c-c) $$
or $$ f(0) = 0. $$
Then using (1) and (2) above, we find that, for every real number $varepsilon > 0$, there exists a real number $delta > 0$ such that
$$ lvert f(x) - f(c) rvert = lvert f(x-c) rvert < varepsilon $$
for all $x in mathbbR$ for which
$$ 0 < lvert x-c rvert < delta. $$
Therefore we can conclude that
$$ lim_x to c f(x) = f(c). $$
Is this proof sound enough in each and every detail hereof?
calculus real-analysis analysis limits proof-verification
edited Aug 3 at 7:35
Mason
1,1271223
asked Aug 3 at 7:00
Saaqib Mahmood
7,09542169
add a comment |Â
up vote
1
down vote
favorite
Here is Prob. 12, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition:
Let $f colon mathbbR to mathbbR$ be such that $f(x+y) = f(x) + f(y)$ for all $x, y$ in $mathbbR$. Assume that $lim_x to 0 f = L$ exists. Prove that $L = 0$, and then prove that $f$ has a limit at every point $c in mathbbR$. [Hint: First note that $f(2x) = f(x) + f(x) = 2 f(x)$ for $x in mathbbR$. Also note that $f(x) = f(x-c) + f(c)$ for $x, c$ in $mathbbR$.]
My Attempt:
As $lim_x to 0 f (x) = L$, so, given any real number $varepsilon > 0$, we can find a real number $delta > 0$ such that
$$ lvert f(x) - L rvert < varepsilon tag1 $$
for all $x in mathbbR$ for which
$$ 0 < lvert x - 0 rvert < delta; tag2 $$
therefore, we can also conclude that, for all $x in mathbbR$ for which
$$ 0 < lvert x - 0 rvert < fracdelta2 $$
holds, for those $x$ we also have
$$ 0 < lvert 2x - 0 rvert < delta, $$
and so
$$ lvert f(2x) - L rvert < varepsilon, $$
and thus it follows that $$ lim_x to 0 f(2x) = L $$
also. But as $f(2x) = 2 f(x)$ for all $x in mathbbR$, so we must have
$$ lim_x to 0 f(2x) = 2 lim_x to 0 f(x), $$
that is we must have
$$ L = 2L,$$
which in turn implies that $L = 0$.
Hence $$ lim_x to 0 f(x) = 0 $$
holds.
Am I right? Is this part of the proof correct and clear enough?
Now for any real numbers $c, x$, we find that
$$ f(x) = f(x-c + c) = f(x-c) + f(c), $$
and so
$$ f(x) - f(c) = f(x-c). $$
Now if $x= c$, then we obtain
$$ f(c) - f(c) = f(c-c) $$
or $$ f(0) = 0. $$
Then using (1) and (2) above, we find that, for every real number $varepsilon > 0$, there exists a real number $delta > 0$ such that
$$ lvert f(x) - f(c) rvert = lvert f(x-c) rvert < varepsilon $$
for all $x in mathbbR$ for which
$$ 0 < lvert x-c rvert < delta. $$
Therefore we can conclude that
$$ lim_x to c f(x) = f(c). $$
Is this proof sound enough in each and every detail hereof?
calculus real-analysis analysis limits proof-verification
edited Aug 3 at 7:35
Mason
1,1271223
asked Aug 3 at 7:00
Saaqib Mahmood
7,09542169
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is Prob. 12, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition:
Let $f colon mathbbR to mathbbR$ be such that $f(x+y) = f(x) + f(y)$ for all $x, y$ in $mathbbR$. Assume that $lim_x to 0 f = L$ exists. Prove that $L = 0$, and then prove that $f$ has a limit at every point $c in mathbbR$. [Hint: First note that $f(2x) = f(x) + f(x) = 2 f(x)$ for $x in mathbbR$. Also note that $f(x) = f(x-c) + f(c)$ for $x, c$ in $mathbbR$.]
My Attempt:
As $lim_x to 0 f (x) = L$, so, given any real number $varepsilon > 0$, we can find a real number $delta > 0$ such that
$$ lvert f(x) - L rvert < varepsilon tag1 $$
for all $x in mathbbR$ for which
$$ 0 < lvert x - 0 rvert < delta; tag2 $$
therefore, we can also conclude that, for all $x in mathbbR$ for which
$$ 0 < lvert x - 0 rvert < fracdelta2 $$
holds, for those $x$ we also have
$$ 0 < lvert 2x - 0 rvert < delta, $$
and so
$$ lvert f(2x) - L rvert < varepsilon, $$
and thus it follows that $$ lim_x to 0 f(2x) = L $$
also. But as $f(2x) = 2 f(x)$ for all $x in mathbbR$, so we must have
$$ lim_x to 0 f(2x) = 2 lim_x to 0 f(x), $$
that is we must have
$$ L = 2L,$$
which in turn implies that $L = 0$.
Hence $$ lim_x to 0 f(x) = 0 $$
holds.
Am I right? Is this part of the proof correct and clear enough?
Now for any real numbers $c, x$, we find that
$$ f(x) = f(x-c + c) = f(x-c) + f(c), $$
and so
$$ f(x) - f(c) = f(x-c). $$
Now if $x= c$, then we obtain
$$ f(c) - f(c) = f(c-c) $$
or $$ f(0) = 0. $$
Then using (1) and (2) above, we find that, for every real number $varepsilon > 0$, there exists a real number $delta > 0$ such that
$$ lvert f(x) - f(c) rvert = lvert f(x-c) rvert < varepsilon $$
for all $x in mathbbR$ for which
$$ 0 < lvert x-c rvert < delta. $$
Therefore we can conclude that
$$ lim_x to c f(x) = f(c). $$
Is this proof sound enough in each and every detail hereof?
calculus real-analysis analysis limits proof-verification
edited Aug 3 at 7:35
Mason
1,1271223
asked Aug 3 at 7:00
Saaqib Mahmood
7,09542169
Here is Prob. 12, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition:
Let $f colon mathbbR to mathbbR$ be such that $f(x+y) = f(x) + f(y)$ for all $x, y$ in $mathbbR$. Assume that $lim_x to 0 f = L$ exists. Prove that $L = 0$, and then prove that $f$ has a limit at every point $c in mathbbR$. [Hint: First note that $f(2x) = f(x) + f(x) = 2 f(x)$ for $x in mathbbR$. Also note that $f(x) = f(x-c) + f(c)$ for $x, c$ in $mathbbR$.]
My Attempt:
As $lim_x to 0 f (x) = L$, so, given any real number $varepsilon > 0$, we can find a real number $delta > 0$ such that
$$ lvert f(x) - L rvert < varepsilon tag1 $$
for all $x in mathbbR$ for which
$$ 0 < lvert x - 0 rvert < delta; tag2 $$
therefore, we can also conclude that, for all $x in mathbbR$ for which
$$ 0 < lvert x - 0 rvert < fracdelta2 $$
holds, for those $x$ we also have
$$ 0 < lvert 2x - 0 rvert < delta, $$
and so
$$ lvert f(2x) - L rvert < varepsilon, $$
and thus it follows that $$ lim_x to 0 f(2x) = L $$
also. But as $f(2x) = 2 f(x)$ for all $x in mathbbR$, so we must have
$$ lim_x to 0 f(2x) = 2 lim_x to 0 f(x), $$
that is we must have
$$ L = 2L,$$
which in turn implies that $L = 0$.
Hence $$ lim_x to 0 f(x) = 0 $$
holds.
Am I right? Is this part of the proof correct and clear enough?
Now for any real numbers $c, x$, we find that
$$ f(x) = f(x-c + c) = f(x-c) + f(c), $$
and so
$$ f(x) - f(c) = f(x-c). $$
Now if $x= c$, then we obtain
$$ f(c) - f(c) = f(c-c) $$
or $$ f(0) = 0. $$
Then using (1) and (2) above, we find that, for every real number $varepsilon > 0$, there exists a real number $delta > 0$ such that
$$ lvert f(x) - f(c) rvert = lvert f(x-c) rvert < varepsilon $$
for all $x in mathbbR$ for which
$$ 0 < lvert x-c rvert < delta. $$
Therefore we can conclude that
$$ lim_x to c f(x) = f(c). $$
Is this proof sound enough in each and every detail hereof?
calculus real-analysis analysis limits proof-verification
edited Aug 3 at 7:35
Mason
1,1271223
asked Aug 3 at 7:00
Saaqib Mahmood
7,09542169
edited Aug 3 at 7:35
Mason
1,1271223
edited Aug 3 at 7:35
Mason
1,1271223
edited Aug 3 at 7:35
Mason
1,1271223
1,1271223
asked Aug 3 at 7:00
Saaqib Mahmood
7,09542169
asked Aug 3 at 7:00
Saaqib Mahmood
7,09542169
asked Aug 3 at 7:00
Saaqib Mahmood
7,09542169
7,09542169
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Proof
Since $$f(x)+f(y)=f(x+y),~~~forall x,y in mathbbR,tag1$$
hence $$f(x)+f(x)=f(x+x),~~~forall xin mathbbR,tag2$$
namely $$2f(x)=f(2x),~~~forall xin mathbbR.tag3$$
Let $x to 0$ and take the limits of the both sides of $(3)$. We obtain $$2L=2lim_x to 0f(x)=lim_x to 02f(x)=lim_x to 0f(2x)=L,$$which shows that $$L=0.$$
Moreover, from $(1)$ we may obtain $$f(c)+f(x-c)=f(x),~~~forall x,c in mathbbR,tag4$$where $c$ is any a constant. Likewise, let $x to c$ and take the limits of both sides of $(4).$ We obtain $$f(c)+0=lim_x to cf(c)+lim_x to cf(x-c)=lim_x to c[f(c)+f(x-c)]=lim_x to cf(x),$$namely $$lim_x to cf(x)=f(c),$$which shows that $f(x)$ has a limit $f(c)$ at $x=c$, in another word, it is continuous at $x=c.$ We are done!
answered Aug 3 at 7:37
mengdie1982
2,840216
This proof I can offer a genuine critique! Where did you use the fact that $L$ exists (is finite)? This proof doesn't use epsilon deltas so it's not really clear where you use this but we know that we have to use it somewhere... otherwise why would the authors have wrote it in the assumptions? It is tucked into $L=2L implies L=0$. Note that $L=2L$ really implies that $L$ is either zero or infinity.
– Mason
Aug 3 at 7:49
@mason Don't you know the rules of limits？ If not， I would not post any more comments.
– mengdie1982
Aug 3 at 7:59
I don't have to post any comments if you prefer it. But I am not sure what limit rule you are referring to... maybe you can teach me...
– Mason
Aug 3 at 8:04
@Mason Considering the CONTEXT of the problem, we should assume that $L$ is a finite limit. If not, can you prove $L=0$? which teacher told you that?
– mengdie1982
Aug 3 at 8:09
Indeed! Nice proof.
– Mason
Aug 3 at 8:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Proof
Since $$f(x)+f(y)=f(x+y),~~~forall x,y in mathbbR,tag1$$
hence $$f(x)+f(x)=f(x+x),~~~forall xin mathbbR,tag2$$
namely $$2f(x)=f(2x),~~~forall xin mathbbR.tag3$$
Let $x to 0$ and take the limits of the both sides of $(3)$. We obtain $$2L=2lim_x to 0f(x)=lim_x to 02f(x)=lim_x to 0f(2x)=L,$$which shows that $$L=0.$$
Moreover, from $(1)$ we may obtain $$f(c)+f(x-c)=f(x),~~~forall x,c in mathbbR,tag4$$where $c$ is any a constant. Likewise, let $x to c$ and take the limits of both sides of $(4).$ We obtain $$f(c)+0=lim_x to cf(c)+lim_x to cf(x-c)=lim_x to c[f(c)+f(x-c)]=lim_x to cf(x),$$namely $$lim_x to cf(x)=f(c),$$which shows that $f(x)$ has a limit $f(c)$ at $x=c$, in another word, it is continuous at $x=c.$ We are done!
answered Aug 3 at 7:37
mengdie1982
2,840216
This proof I can offer a genuine critique! Where did you use the fact that $L$ exists (is finite)? This proof doesn't use epsilon deltas so it's not really clear where you use this but we know that we have to use it somewhere... otherwise why would the authors have wrote it in the assumptions? It is tucked into $L=2L implies L=0$. Note that $L=2L$ really implies that $L$ is either zero or infinity.
– Mason
Aug 3 at 7:49
@mason Don't you know the rules of limits？ If not， I would not post any more comments.
– mengdie1982
Aug 3 at 7:59
I don't have to post any comments if you prefer it. But I am not sure what limit rule you are referring to... maybe you can teach me...
– Mason
Aug 3 at 8:04
@Mason Considering the CONTEXT of the problem, we should assume that $L$ is a finite limit. If not, can you prove $L=0$? which teacher told you that?
– mengdie1982
Aug 3 at 8:09
Indeed! Nice proof.
– Mason
Aug 3 at 8:12
add a comment |Â
up vote
0
down vote
Proof
Since $$f(x)+f(y)=f(x+y),~~~forall x,y in mathbbR,tag1$$
hence $$f(x)+f(x)=f(x+x),~~~forall xin mathbbR,tag2$$
namely $$2f(x)=f(2x),~~~forall xin mathbbR.tag3$$
Let $x to 0$ and take the limits of the both sides of $(3)$. We obtain $$2L=2lim_x to 0f(x)=lim_x to 02f(x)=lim_x to 0f(2x)=L,$$which shows that $$L=0.$$
Moreover, from $(1)$ we may obtain $$f(c)+f(x-c)=f(x),~~~forall x,c in mathbbR,tag4$$where $c$ is any a constant. Likewise, let $x to c$ and take the limits of both sides of $(4).$ We obtain $$f(c)+0=lim_x to cf(c)+lim_x to cf(x-c)=lim_x to c[f(c)+f(x-c)]=lim_x to cf(x),$$namely $$lim_x to cf(x)=f(c),$$which shows that $f(x)$ has a limit $f(c)$ at $x=c$, in another word, it is continuous at $x=c.$ We are done!
answered Aug 3 at 7:37
mengdie1982
2,840216
This proof I can offer a genuine critique! Where did you use the fact that $L$ exists (is finite)? This proof doesn't use epsilon deltas so it's not really clear where you use this but we know that we have to use it somewhere... otherwise why would the authors have wrote it in the assumptions? It is tucked into $L=2L implies L=0$. Note that $L=2L$ really implies that $L$ is either zero or infinity.
– Mason
Aug 3 at 7:49
@mason Don't you know the rules of limits？ If not， I would not post any more comments.
– mengdie1982
Aug 3 at 7:59
I don't have to post any comments if you prefer it. But I am not sure what limit rule you are referring to... maybe you can teach me...
– Mason
Aug 3 at 8:04
@Mason Considering the CONTEXT of the problem, we should assume that $L$ is a finite limit. If not, can you prove $L=0$? which teacher told you that?
– mengdie1982
Aug 3 at 8:09
Indeed! Nice proof.
– Mason
Aug 3 at 8:12
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Proof
Since $$f(x)+f(y)=f(x+y),~~~forall x,y in mathbbR,tag1$$
hence $$f(x)+f(x)=f(x+x),~~~forall xin mathbbR,tag2$$
namely $$2f(x)=f(2x),~~~forall xin mathbbR.tag3$$
Let $x to 0$ and take the limits of the both sides of $(3)$. We obtain $$2L=2lim_x to 0f(x)=lim_x to 02f(x)=lim_x to 0f(2x)=L,$$which shows that $$L=0.$$
Moreover, from $(1)$ we may obtain $$f(c)+f(x-c)=f(x),~~~forall x,c in mathbbR,tag4$$where $c$ is any a constant. Likewise, let $x to c$ and take the limits of both sides of $(4).$ We obtain $$f(c)+0=lim_x to cf(c)+lim_x to cf(x-c)=lim_x to c[f(c)+f(x-c)]=lim_x to cf(x),$$namely $$lim_x to cf(x)=f(c),$$which shows that $f(x)$ has a limit $f(c)$ at $x=c$, in another word, it is continuous at $x=c.$ We are done!
answered Aug 3 at 7:37
mengdie1982
2,840216
Proof
Since $$f(x)+f(y)=f(x+y),~~~forall x,y in mathbbR,tag1$$
hence $$f(x)+f(x)=f(x+x),~~~forall xin mathbbR,tag2$$
namely $$2f(x)=f(2x),~~~forall xin mathbbR.tag3$$
Let $x to 0$ and take the limits of the both sides of $(3)$. We obtain $$2L=2lim_x to 0f(x)=lim_x to 02f(x)=lim_x to 0f(2x)=L,$$which shows that $$L=0.$$
Moreover, from $(1)$ we may obtain $$f(c)+f(x-c)=f(x),~~~forall x,c in mathbbR,tag4$$where $c$ is any a constant. Likewise, let $x to c$ and take the limits of both sides of $(4).$ We obtain $$f(c)+0=lim_x to cf(c)+lim_x to cf(x-c)=lim_x to c[f(c)+f(x-c)]=lim_x to cf(x),$$namely $$lim_x to cf(x)=f(c),$$which shows that $f(x)$ has a limit $f(c)$ at $x=c$, in another word, it is continuous at $x=c.$ We are done!
answered Aug 3 at 7:37
mengdie1982
2,840216
edited Aug 3 at 7:43
answered Aug 3 at 7:37
mengdie1982
2,840216
answered Aug 3 at 7:37
mengdie1982
2,840216
answered Aug 3 at 7:37
mengdie1982
2,840216
2,840216
This proof I can offer a genuine critique! Where did you use the fact that $L$ exists (is finite)? This proof doesn't use epsilon deltas so it's not really clear where you use this but we know that we have to use it somewhere... otherwise why would the authors have wrote it in the assumptions? It is tucked into $L=2L implies L=0$. Note that $L=2L$ really implies that $L$ is either zero or infinity.
– Mason
Aug 3 at 7:49
@mason Don't you know the rules of limits？ If not， I would not post any more comments.
– mengdie1982
Aug 3 at 7:59
I don't have to post any comments if you prefer it. But I am not sure what limit rule you are referring to... maybe you can teach me...
– Mason
Aug 3 at 8:04
@Mason Considering the CONTEXT of the problem, we should assume that $L$ is a finite limit. If not, can you prove $L=0$? which teacher told you that?
– mengdie1982
Aug 3 at 8:09
Indeed! Nice proof.
– Mason
Aug 3 at 8:12
add a comment |Â
This proof I can offer a genuine critique! Where did you use the fact that $L$ exists (is finite)? This proof doesn't use epsilon deltas so it's not really clear where you use this but we know that we have to use it somewhere... otherwise why would the authors have wrote it in the assumptions? It is tucked into $L=2L implies L=0$. Note that $L=2L$ really implies that $L$ is either zero or infinity.
– Mason
Aug 3 at 7:49
@mason Don't you know the rules of limits？ If not， I would not post any more comments.
– mengdie1982
Aug 3 at 7:59
I don't have to post any comments if you prefer it. But I am not sure what limit rule you are referring to... maybe you can teach me...
– Mason
Aug 3 at 8:04
@Mason Considering the CONTEXT of the problem, we should assume that $L$ is a finite limit. If not, can you prove $L=0$? which teacher told you that?
– mengdie1982
Aug 3 at 8:09
Indeed! Nice proof.
– Mason
Aug 3 at 8:12
This proof I can offer a genuine critique! Where did you use the fact that $L$ exists (is finite)? This proof doesn't use epsilon deltas so it's not really clear where you use this but we know that we have to use it somewhere... otherwise why would the authors have wrote it in the assumptions? It is tucked into $L=2L implies L=0$. Note that $L=2L$ really implies that $L$ is either zero or infinity.
– Mason
Aug 3 at 7:49
This proof I can offer a genuine critique! Where did you use the fact that $L$ exists (is finite)? This proof doesn't use epsilon deltas so it's not really clear where you use this but we know that we have to use it somewhere... otherwise why would the authors have wrote it in the assumptions? It is tucked into $L=2L implies L=0$. Note that $L=2L$ really implies that $L$ is either zero or infinity.
– Mason
Aug 3 at 7:49
@mason Don't you know the rules of limits？ If not， I would not post any more comments.
– mengdie1982
Aug 3 at 7:59
@mason Don't you know the rules of limits？ If not， I would not post any more comments.
– mengdie1982
Aug 3 at 7:59
I don't have to post any comments if you prefer it. But I am not sure what limit rule you are referring to... maybe you can teach me...
– Mason
Aug 3 at 8:04
I don't have to post any comments if you prefer it. But I am not sure what limit rule you are referring to... maybe you can teach me...
– Mason
Aug 3 at 8:04
@Mason Considering the CONTEXT of the problem, we should assume that $L$ is a finite limit. If not, can you prove $L=0$? which teacher told you that?
– mengdie1982
Aug 3 at 8:09
@Mason Considering the CONTEXT of the problem, we should assume that $L$ is a finite limit. If not, can you prove $L=0$? which teacher told you that?
– mengdie1982
Aug 3 at 8:09
Indeed! Nice proof.
– Mason
Aug 3 at 8:12
Indeed! Nice proof.
– Mason
Aug 3 at 8:12
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870823%2fprob-12-sec-4-2-in-bartle-sherberts-intro-to-real-analysis-if-fxy-f%23new-answer', 'question_page');
);
Post as a guest
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
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
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