Mixing
Prove that a product's limit converges to a sum
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I accidentally discovered this equality, in which I can prove numerically using python.
$$lim_ktoinfty sqrt[n] prod_i=1^n(x_i+k) - k = fracsum_i=1^nx_in$$
But I need to prove this equality in an algebraic way and I could not get anywhere. The right-hand side is nothing more than an arithmetic mean, while the left-hand side is a modification of the geometric mean.
I hope someone can help me at this point.
real-analysis discrete-mathematics means
asked 3 hours ago
Gustavo Ale
213
New contributor
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
I accidentally discovered this equality, in which I can prove numerically using python.
$$lim_ktoinfty sqrt[n] prod_i=1^n(x_i+k) - k = fracsum_i=1^nx_in$$
But I need to prove this equality in an algebraic way and I could not get anywhere. The right-hand side is nothing more than an arithmetic mean, while the left-hand side is a modification of the geometric mean.
I hope someone can help me at this point.
real-analysis discrete-mathematics means
asked 3 hours ago
Gustavo Ale
213
New contributor
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I accidentally discovered this equality, in which I can prove numerically using python.
$$lim_ktoinfty sqrt[n] prod_i=1^n(x_i+k) - k = fracsum_i=1^nx_in$$
But I need to prove this equality in an algebraic way and I could not get anywhere. The right-hand side is nothing more than an arithmetic mean, while the left-hand side is a modification of the geometric mean.
I hope someone can help me at this point.
real-analysis discrete-mathematics means
asked 3 hours ago
Gustavo Ale
213
New contributor
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I accidentally discovered this equality, in which I can prove numerically using python.
$$lim_ktoinfty sqrt[n] prod_i=1^n(x_i+k) - k = fracsum_i=1^nx_in$$
But I need to prove this equality in an algebraic way and I could not get anywhere. The right-hand side is nothing more than an arithmetic mean, while the left-hand side is a modification of the geometric mean.
I hope someone can help me at this point.
real-analysis discrete-mathematics means
real-analysis discrete-mathematics means
asked 3 hours ago
Gustavo Ale
213
New contributor
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Gustavo Ale
213
New contributor
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
asked 3 hours ago
Gustavo Ale
213
New contributor
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Gustavo Ale
213
asked 3 hours ago
Gustavo Ale
213
213
New contributor
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gustavo Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Assume that $k$ is greater than twice $|x_1|+ldots+|x_n|$ and also greater than twice $M=x_1^2+ldots+x_n^2$. Over the interval $left(-frac12,frac12right)$ we have
$$ e^x = 1+x+C(x)x^2, qquad log(1+x)=x+D(x)x^2 $$
with $|C(x)|,|D(x)|leq 1$. It follows that
$$begineqnarray*textGM(x_1+k,ldots,x_n+k)&=&kcdot textGMleft(1+tfracx_1k,ldots,1+tfracx_nkright)\&=&kexpleft[frac1nsum_j=1^nlogleft(1+fracx_jkright)right]\&=&kexpleft[frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]\&=&kleft[1+frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]endeqnarray*$$
and the claim is proved. It looks very reasonable also without a formal proof: the magnitude of the difference between the arithmetic mean and the geometric mean is controlled by $fractextVar(x_1,ldots,x_n)textAM(x_1,ldots,x_n)$. A translation towards the right leaves the variance unchanged and increases the mean.
answered 1 hour ago
Jack D'Aurizio
280k33272651
add a comment |Â
up vote
1
down vote
Your expression under limit can be written as $k(a-b) $ and further $a-b=(a^n-b^n) / c$ where $a, b$ tend to $1$ and $c$ tends to $n$ as $ktoinfty$. And $$a^n-b^n=prod_i=1^nleft(1+fracx_ikright)-1$$ I hope you can continue it from here.
answered 3 hours ago
Paramanand Singh
47.3k555149
add a comment |Â
up vote
1
down vote
You could even get an interesting asymptotics considering $$y_n=sqrt[n] prod_i=1^n(x_i+k)$$ $$ log(y_n)=frac 1n sum_i=1^nlogleft((x_i+k)right)=frac 1n sum_i=1^nleft(log(k)+logleft(1+frac x_ikright)right)$$ Assuming that $x_1<x_2<cdots < x_n ll k$, use the Taylor expansion of $logleft(1+frac x_ikright)$. Then continuing with Taylor series
$$y_n=e^log(y_n)implies
y_n=k+fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ making
$$ sqrt[n] prod_i=1^n(x_i+k) - k =fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$
For example, using $n=10$, $k=1000$ and $x_i=p_i$, the approximation would give $frac2572671200000approx 12.8634$ while the exact calculation would lead to $approx 12.8639$
answered 35 mins ago
Claude Leibovici
115k1155130
1
@GustavoAle. For sure ! I just wanted to see what would be the asymptotics. Cheers.
– Claude Leibovici
8 mins ago
At this point the equality is proven by applying the limit itself,$$lim_ktoinftysqrt[n] prod_i=1^n(x_i+k) - k = lim_ktoinftyfracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ where $$lim_ktoinftyfracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k + Oleft(frac1k^2right) to 0$$ remaining only $$fracsum_i=1^n x_in$$
– Gustavo Ale
6 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Assume that $k$ is greater than twice $|x_1|+ldots+|x_n|$ and also greater than twice $M=x_1^2+ldots+x_n^2$. Over the interval $left(-frac12,frac12right)$ we have
$$ e^x = 1+x+C(x)x^2, qquad log(1+x)=x+D(x)x^2 $$
with $|C(x)|,|D(x)|leq 1$. It follows that
$$begineqnarray*textGM(x_1+k,ldots,x_n+k)&=&kcdot textGMleft(1+tfracx_1k,ldots,1+tfracx_nkright)\&=&kexpleft[frac1nsum_j=1^nlogleft(1+fracx_jkright)right]\&=&kexpleft[frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]\&=&kleft[1+frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]endeqnarray*$$
and the claim is proved. It looks very reasonable also without a formal proof: the magnitude of the difference between the arithmetic mean and the geometric mean is controlled by $fractextVar(x_1,ldots,x_n)textAM(x_1,ldots,x_n)$. A translation towards the right leaves the variance unchanged and increases the mean.
answered 1 hour ago
Jack D'Aurizio
280k33272651
add a comment |Â
up vote
2
down vote
Assume that $k$ is greater than twice $|x_1|+ldots+|x_n|$ and also greater than twice $M=x_1^2+ldots+x_n^2$. Over the interval $left(-frac12,frac12right)$ we have
$$ e^x = 1+x+C(x)x^2, qquad log(1+x)=x+D(x)x^2 $$
with $|C(x)|,|D(x)|leq 1$. It follows that
$$begineqnarray*textGM(x_1+k,ldots,x_n+k)&=&kcdot textGMleft(1+tfracx_1k,ldots,1+tfracx_nkright)\&=&kexpleft[frac1nsum_j=1^nlogleft(1+fracx_jkright)right]\&=&kexpleft[frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]\&=&kleft[1+frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]endeqnarray*$$
and the claim is proved. It looks very reasonable also without a formal proof: the magnitude of the difference between the arithmetic mean and the geometric mean is controlled by $fractextVar(x_1,ldots,x_n)textAM(x_1,ldots,x_n)$. A translation towards the right leaves the variance unchanged and increases the mean.
answered 1 hour ago
Jack D'Aurizio
280k33272651
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Assume that $k$ is greater than twice $|x_1|+ldots+|x_n|$ and also greater than twice $M=x_1^2+ldots+x_n^2$. Over the interval $left(-frac12,frac12right)$ we have
$$ e^x = 1+x+C(x)x^2, qquad log(1+x)=x+D(x)x^2 $$
with $|C(x)|,|D(x)|leq 1$. It follows that
$$begineqnarray*textGM(x_1+k,ldots,x_n+k)&=&kcdot textGMleft(1+tfracx_1k,ldots,1+tfracx_nkright)\&=&kexpleft[frac1nsum_j=1^nlogleft(1+fracx_jkright)right]\&=&kexpleft[frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]\&=&kleft[1+frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]endeqnarray*$$
and the claim is proved. It looks very reasonable also without a formal proof: the magnitude of the difference between the arithmetic mean and the geometric mean is controlled by $fractextVar(x_1,ldots,x_n)textAM(x_1,ldots,x_n)$. A translation towards the right leaves the variance unchanged and increases the mean.
answered 1 hour ago
Jack D'Aurizio
280k33272651
Assume that $k$ is greater than twice $|x_1|+ldots+|x_n|$ and also greater than twice $M=x_1^2+ldots+x_n^2$. Over the interval $left(-frac12,frac12right)$ we have
$$ e^x = 1+x+C(x)x^2, qquad log(1+x)=x+D(x)x^2 $$
with $|C(x)|,|D(x)|leq 1$. It follows that
$$begineqnarray*textGM(x_1+k,ldots,x_n+k)&=&kcdot textGMleft(1+tfracx_1k,ldots,1+tfracx_nkright)\&=&kexpleft[frac1nsum_j=1^nlogleft(1+fracx_jkright)right]\&=&kexpleft[frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]\&=&kleft[1+frac1ksum_j=1^nfracx_jn+Thetaleft(fracMnk^2right)right]endeqnarray*$$
and the claim is proved. It looks very reasonable also without a formal proof: the magnitude of the difference between the arithmetic mean and the geometric mean is controlled by $fractextVar(x_1,ldots,x_n)textAM(x_1,ldots,x_n)$. A translation towards the right leaves the variance unchanged and increases the mean.
answered 1 hour ago
Jack D'Aurizio
280k33272651
answered 1 hour ago
Jack D'Aurizio
280k33272651
answered 1 hour ago
Jack D'Aurizio
280k33272651
answered 1 hour ago
Jack D'Aurizio
280k33272651
280k33272651
add a comment |Â
add a comment |Â
up vote
1
down vote
Your expression under limit can be written as $k(a-b) $ and further $a-b=(a^n-b^n) / c$ where $a, b$ tend to $1$ and $c$ tends to $n$ as $ktoinfty$. And $$a^n-b^n=prod_i=1^nleft(1+fracx_ikright)-1$$ I hope you can continue it from here.
answered 3 hours ago
Paramanand Singh
47.3k555149
add a comment |Â
up vote
1
down vote
Your expression under limit can be written as $k(a-b) $ and further $a-b=(a^n-b^n) / c$ where $a, b$ tend to $1$ and $c$ tends to $n$ as $ktoinfty$. And $$a^n-b^n=prod_i=1^nleft(1+fracx_ikright)-1$$ I hope you can continue it from here.
answered 3 hours ago
Paramanand Singh
47.3k555149
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your expression under limit can be written as $k(a-b) $ and further $a-b=(a^n-b^n) / c$ where $a, b$ tend to $1$ and $c$ tends to $n$ as $ktoinfty$. And $$a^n-b^n=prod_i=1^nleft(1+fracx_ikright)-1$$ I hope you can continue it from here.
answered 3 hours ago
Paramanand Singh
47.3k555149
Your expression under limit can be written as $k(a-b) $ and further $a-b=(a^n-b^n) / c$ where $a, b$ tend to $1$ and $c$ tends to $n$ as $ktoinfty$. And $$a^n-b^n=prod_i=1^nleft(1+fracx_ikright)-1$$ I hope you can continue it from here.
answered 3 hours ago
Paramanand Singh
47.3k555149
answered 3 hours ago
Paramanand Singh
47.3k555149
answered 3 hours ago
Paramanand Singh
47.3k555149
answered 3 hours ago
Paramanand Singh
47.3k555149
47.3k555149
add a comment |Â
add a comment |Â
up vote
1
down vote
You could even get an interesting asymptotics considering $$y_n=sqrt[n] prod_i=1^n(x_i+k)$$ $$ log(y_n)=frac 1n sum_i=1^nlogleft((x_i+k)right)=frac 1n sum_i=1^nleft(log(k)+logleft(1+frac x_ikright)right)$$ Assuming that $x_1<x_2<cdots < x_n ll k$, use the Taylor expansion of $logleft(1+frac x_ikright)$. Then continuing with Taylor series
$$y_n=e^log(y_n)implies
y_n=k+fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ making
$$ sqrt[n] prod_i=1^n(x_i+k) - k =fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$
For example, using $n=10$, $k=1000$ and $x_i=p_i$, the approximation would give $frac2572671200000approx 12.8634$ while the exact calculation would lead to $approx 12.8639$
answered 35 mins ago
Claude Leibovici
115k1155130
1
@GustavoAle. For sure ! I just wanted to see what would be the asymptotics. Cheers.
– Claude Leibovici
8 mins ago
At this point the equality is proven by applying the limit itself,$$lim_ktoinftysqrt[n] prod_i=1^n(x_i+k) - k = lim_ktoinftyfracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ where $$lim_ktoinftyfracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k + Oleft(frac1k^2right) to 0$$ remaining only $$fracsum_i=1^n x_in$$
– Gustavo Ale
6 mins ago
add a comment |Â
up vote
1
down vote
You could even get an interesting asymptotics considering $$y_n=sqrt[n] prod_i=1^n(x_i+k)$$ $$ log(y_n)=frac 1n sum_i=1^nlogleft((x_i+k)right)=frac 1n sum_i=1^nleft(log(k)+logleft(1+frac x_ikright)right)$$ Assuming that $x_1<x_2<cdots < x_n ll k$, use the Taylor expansion of $logleft(1+frac x_ikright)$. Then continuing with Taylor series
$$y_n=e^log(y_n)implies
y_n=k+fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ making
$$ sqrt[n] prod_i=1^n(x_i+k) - k =fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$
For example, using $n=10$, $k=1000$ and $x_i=p_i$, the approximation would give $frac2572671200000approx 12.8634$ while the exact calculation would lead to $approx 12.8639$
answered 35 mins ago
Claude Leibovici
115k1155130
1
@GustavoAle. For sure ! I just wanted to see what would be the asymptotics. Cheers.
– Claude Leibovici
8 mins ago
At this point the equality is proven by applying the limit itself,$$lim_ktoinftysqrt[n] prod_i=1^n(x_i+k) - k = lim_ktoinftyfracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ where $$lim_ktoinftyfracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k + Oleft(frac1k^2right) to 0$$ remaining only $$fracsum_i=1^n x_in$$
– Gustavo Ale
6 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You could even get an interesting asymptotics considering $$y_n=sqrt[n] prod_i=1^n(x_i+k)$$ $$ log(y_n)=frac 1n sum_i=1^nlogleft((x_i+k)right)=frac 1n sum_i=1^nleft(log(k)+logleft(1+frac x_ikright)right)$$ Assuming that $x_1<x_2<cdots < x_n ll k$, use the Taylor expansion of $logleft(1+frac x_ikright)$. Then continuing with Taylor series
$$y_n=e^log(y_n)implies
y_n=k+fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ making
$$ sqrt[n] prod_i=1^n(x_i+k) - k =fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$
For example, using $n=10$, $k=1000$ and $x_i=p_i$, the approximation would give $frac2572671200000approx 12.8634$ while the exact calculation would lead to $approx 12.8639$
answered 35 mins ago
Claude Leibovici
115k1155130
You could even get an interesting asymptotics considering $$y_n=sqrt[n] prod_i=1^n(x_i+k)$$ $$ log(y_n)=frac 1n sum_i=1^nlogleft((x_i+k)right)=frac 1n sum_i=1^nleft(log(k)+logleft(1+frac x_ikright)right)$$ Assuming that $x_1<x_2<cdots < x_n ll k$, use the Taylor expansion of $logleft(1+frac x_ikright)$. Then continuing with Taylor series
$$y_n=e^log(y_n)implies
y_n=k+fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ making
$$ sqrt[n] prod_i=1^n(x_i+k) - k =fracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$
For example, using $n=10$, $k=1000$ and $x_i=p_i$, the approximation would give $frac2572671200000approx 12.8634$ while the exact calculation would lead to $approx 12.8639$
answered 35 mins ago
Claude Leibovici
115k1155130
edited 24 mins ago
answered 35 mins ago
Claude Leibovici
115k1155130
answered 35 mins ago
Claude Leibovici
115k1155130
answered 35 mins ago
Claude Leibovici
115k1155130
115k1155130
1
@GustavoAle. For sure ! I just wanted to see what would be the asymptotics. Cheers.
– Claude Leibovici
8 mins ago
At this point the equality is proven by applying the limit itself,$$lim_ktoinftysqrt[n] prod_i=1^n(x_i+k) - k = lim_ktoinftyfracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ where $$lim_ktoinftyfracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k + Oleft(frac1k^2right) to 0$$ remaining only $$fracsum_i=1^n x_in$$
– Gustavo Ale
6 mins ago
add a comment |Â
1
@GustavoAle. For sure ! I just wanted to see what would be the asymptotics. Cheers.
– Claude Leibovici
8 mins ago
At this point the equality is proven by applying the limit itself,$$lim_ktoinftysqrt[n] prod_i=1^n(x_i+k) - k = lim_ktoinftyfracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ where $$lim_ktoinftyfracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k + Oleft(frac1k^2right) to 0$$ remaining only $$fracsum_i=1^n x_in$$
– Gustavo Ale
6 mins ago
1
1
@GustavoAle. For sure ! I just wanted to see what would be the asymptotics. Cheers.
– Claude Leibovici
8 mins ago
@GustavoAle. For sure ! I just wanted to see what would be the asymptotics. Cheers.
– Claude Leibovici
8 mins ago
At this point the equality is proven by applying the limit itself,$$lim_ktoinftysqrt[n] prod_i=1^n(x_i+k) - k = lim_ktoinftyfracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ where $$lim_ktoinftyfracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k + Oleft(frac1k^2right) to 0$$ remaining only $$fracsum_i=1^n x_in$$
– Gustavo Ale
6 mins ago
At this point the equality is proven by applying the limit itself,$$lim_ktoinftysqrt[n] prod_i=1^n(x_i+k) - k = lim_ktoinftyfracsum_i=1^n x_in+fracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k+Oleft(frac1k^2right)$$ where $$lim_ktoinftyfracleft(sum_i=1^n x_iright)^2-nsum_i=1^n x^2_i2n^2 k + Oleft(frac1k^2right) to 0$$ remaining only $$fracsum_i=1^n x_in$$
– Gustavo Ale
6 mins ago
add a comment |Â
Gustavo Ale is a new contributor. Be nice, and check out our Code of Conduct.
Gustavo Ale is a new contributor. Be nice, and check out our Code of Conduct.
Gustavo Ale is a new contributor. Be nice, and check out our Code of Conduct.
Gustavo Ale is a new contributor. Be nice, and check out our Code of Conduct.
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%2f2983641%2fprove-that-a-products-limit-converges-to-a-sum%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
