What does $| |^2_2$ mean? [closed]
$begingroup$
$$|X_m-X_n|^2_2.$$ To be clear, I am not asking about the $X_m - X_n$ part. I am asking what the $| |^2_2$ thing means.
notation
edited Dec 2 '18 at 9:03
Saad
19.7k92352
asked Dec 2 '18 at 7:22
chiechie
32
$endgroup$
closed as unclear what you're asking by Shailesh, Saad, Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Rebellos Dec 2 '18 at 9:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$|X_m-X_n|^2_2.$$ To be clear, I am not asking about the $X_m - X_n$ part. I am asking what the $| |^2_2$ thing means.
notation
edited Dec 2 '18 at 9:03
Saad
19.7k92352
asked Dec 2 '18 at 7:22
chiechie
32
$endgroup$
closed as unclear what you're asking by Shailesh, Saad, Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Rebellos Dec 2 '18 at 9:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Welcome to Math.SE. Please use MathJax.
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– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 7:23
2
$begingroup$
It depends on the context. People only exposed to metric spaces will certainly think of it as the square of the $L^2$-norm. In that part of math tat would be the expectation. But, in another context it could mean something else. Basically whatever the author meant! That's why textbooks often include a page or two where notational conventions are spelled out (possibly including pointers to the pages where they are introduced).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 9:07
add a comment |
$begingroup$
$$|X_m-X_n|^2_2.$$ To be clear, I am not asking about the $X_m - X_n$ part. I am asking what the $| |^2_2$ thing means.
notation
edited Dec 2 '18 at 9:03
Saad
19.7k92352
asked Dec 2 '18 at 7:22
chiechie
32
$endgroup$
$$|X_m-X_n|^2_2.$$ To be clear, I am not asking about the $X_m - X_n$ part. I am asking what the $| |^2_2$ thing means.
notation
notation
edited Dec 2 '18 at 9:03
Saad
19.7k92352
asked Dec 2 '18 at 7:22
chiechie
32
edited Dec 2 '18 at 9:03
Saad
19.7k92352
asked Dec 2 '18 at 7:22
chiechie
32
edited Dec 2 '18 at 9:03
Saad
19.7k92352
edited Dec 2 '18 at 9:03
Saad
19.7k92352
edited Dec 2 '18 at 9:03
Saad
19.7k92352
19.7k92352
asked Dec 2 '18 at 7:22
chiechie
32
asked Dec 2 '18 at 7:22
chiechie
32
asked Dec 2 '18 at 7:22
chiechie
32
32
closed as unclear what you're asking by Shailesh, Saad, Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Rebellos Dec 2 '18 at 9:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Shailesh, Saad, Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Rebellos Dec 2 '18 at 9:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to Math.SE. Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 7:23
2
$begingroup$
It depends on the context. People only exposed to metric spaces will certainly think of it as the square of the $L^2$-norm. In that part of math tat would be the expectation. But, in another context it could mean something else. Basically whatever the author meant! That's why textbooks often include a page or two where notational conventions are spelled out (possibly including pointers to the pages where they are introduced).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 9:07
add a comment |
$begingroup$
Welcome to Math.SE. Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 7:23
2
$begingroup$
It depends on the context. People only exposed to metric spaces will certainly think of it as the square of the $L^2$-norm. In that part of math tat would be the expectation. But, in another context it could mean something else. Basically whatever the author meant! That's why textbooks often include a page or two where notational conventions are spelled out (possibly including pointers to the pages where they are introduced).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 9:07
$begingroup$
Welcome to Math.SE. Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 7:23
$begingroup$
Welcome to Math.SE. Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 7:23
2
2
$begingroup$
It depends on the context. People only exposed to metric spaces will certainly think of it as the square of the $L^2$-norm. In that part of math tat would be the expectation. But, in another context it could mean something else. Basically whatever the author meant! That's why textbooks often include a page or two where notational conventions are spelled out (possibly including pointers to the pages where they are introduced).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 9:07
$begingroup$
It depends on the context. People only exposed to metric spaces will certainly think of it as the square of the $L^2$-norm. In that part of math tat would be the expectation. But, in another context it could mean something else. Basically whatever the author meant! That's why textbooks often include a page or two where notational conventions are spelled out (possibly including pointers to the pages where they are introduced).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 9:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$|x|_2$ means the $2$-norm, which is $$sqrt{sum_{i=1}^n x_i^2}$$
while the $2$ on top, is its square
$$|x|^2=sum_{i=1}^n x_i^2.$$
Overall, it means the sum of square of the components.
answered Dec 2 '18 at 7:32
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
So in this case, it would mean: for all m, for all n, take the squared difference between xm and xn.. and then sum everything?
$endgroup$
– chie
Dec 2 '18 at 10:30
$begingroup$
First, are $x_m$ and $x_n$ vectors? If so , when we evaluate $|x_m-x_n|^2$, the $m$ and $n$ are fixed. We compute $x_m-x_n$, then we sum up the square of each component and we sum them up.
$endgroup$
– Siong Thye Goh
Dec 2 '18 at 10:34
add a comment |
$begingroup$
In general, $|| cdot ||_2$ is called the $L^2$-norm, which is a functional from the function space $L^2(Omega,{cal A}, mu)$ to $Bbb R$ defined by $f mapsto (int |f|^2 ,{rm d}mu)^{1/2}$, provided that the integral exists. This is probably the case when you see $$||X_m - X_n||_2$$ since random variables are usually denoted with capital letters in probability-theory.
In particular, when $Omega = Bbb{N}$, $mu$ is the counting measure, we have $||f||_2 = (int |f|^2 ,{rm d}mu)^{1/2} = (sum_n |f_n|^2)^{1/2}$. This is the case if you actually mean
$$||x_m-x_n||_2$$ a norm of vector.
Remark: It's hard to distinguish $X$ from $x$ in your drawing. That's the reason why I promote the use of $rm LaTeX$ in all levels of math writing.
answered Dec 2 '18 at 7:34
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$|x|_2$ means the $2$-norm, which is $$sqrt{sum_{i=1}^n x_i^2}$$
while the $2$ on top, is its square
$$|x|^2=sum_{i=1}^n x_i^2.$$
Overall, it means the sum of square of the components.
answered Dec 2 '18 at 7:32
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
So in this case, it would mean: for all m, for all n, take the squared difference between xm and xn.. and then sum everything?
$endgroup$
– chie
Dec 2 '18 at 10:30
$begingroup$
First, are $x_m$ and $x_n$ vectors? If so , when we evaluate $|x_m-x_n|^2$, the $m$ and $n$ are fixed. We compute $x_m-x_n$, then we sum up the square of each component and we sum them up.
$endgroup$
– Siong Thye Goh
Dec 2 '18 at 10:34
add a comment |
$begingroup$
$|x|_2$ means the $2$-norm, which is $$sqrt{sum_{i=1}^n x_i^2}$$
while the $2$ on top, is its square
$$|x|^2=sum_{i=1}^n x_i^2.$$
Overall, it means the sum of square of the components.
answered Dec 2 '18 at 7:32
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
So in this case, it would mean: for all m, for all n, take the squared difference between xm and xn.. and then sum everything?
$endgroup$
– chie
Dec 2 '18 at 10:30
$begingroup$
First, are $x_m$ and $x_n$ vectors? If so , when we evaluate $|x_m-x_n|^2$, the $m$ and $n$ are fixed. We compute $x_m-x_n$, then we sum up the square of each component and we sum them up.
$endgroup$
– Siong Thye Goh
Dec 2 '18 at 10:34
add a comment |
$begingroup$
$|x|_2$ means the $2$-norm, which is $$sqrt{sum_{i=1}^n x_i^2}$$
while the $2$ on top, is its square
$$|x|^2=sum_{i=1}^n x_i^2.$$
Overall, it means the sum of square of the components.
answered Dec 2 '18 at 7:32
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$|x|_2$ means the $2$-norm, which is $$sqrt{sum_{i=1}^n x_i^2}$$
while the $2$ on top, is its square
$$|x|^2=sum_{i=1}^n x_i^2.$$
Overall, it means the sum of square of the components.
answered Dec 2 '18 at 7:32
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 2 '18 at 7:32
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 2 '18 at 7:32
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 2 '18 at 7:32
Siong Thye GohSiong Thye Goh
100k1466117
100k1466117
$begingroup$
So in this case, it would mean: for all m, for all n, take the squared difference between xm and xn.. and then sum everything?
$endgroup$
– chie
Dec 2 '18 at 10:30
$begingroup$
First, are $x_m$ and $x_n$ vectors? If so , when we evaluate $|x_m-x_n|^2$, the $m$ and $n$ are fixed. We compute $x_m-x_n$, then we sum up the square of each component and we sum them up.
$endgroup$
– Siong Thye Goh
Dec 2 '18 at 10:34
add a comment |
$begingroup$
So in this case, it would mean: for all m, for all n, take the squared difference between xm and xn.. and then sum everything?
$endgroup$
– chie
Dec 2 '18 at 10:30
$begingroup$
First, are $x_m$ and $x_n$ vectors? If so , when we evaluate $|x_m-x_n|^2$, the $m$ and $n$ are fixed. We compute $x_m-x_n$, then we sum up the square of each component and we sum them up.
$endgroup$
– Siong Thye Goh
Dec 2 '18 at 10:34
$begingroup$
So in this case, it would mean: for all m, for all n, take the squared difference between xm and xn.. and then sum everything?
$endgroup$
– chie
Dec 2 '18 at 10:30
$begingroup$
So in this case, it would mean: for all m, for all n, take the squared difference between xm and xn.. and then sum everything?
$endgroup$
– chie
Dec 2 '18 at 10:30
$begingroup$
First, are $x_m$ and $x_n$ vectors? If so , when we evaluate $|x_m-x_n|^2$, the $m$ and $n$ are fixed. We compute $x_m-x_n$, then we sum up the square of each component and we sum them up.
$endgroup$
– Siong Thye Goh
Dec 2 '18 at 10:34
$begingroup$
First, are $x_m$ and $x_n$ vectors? If so , when we evaluate $|x_m-x_n|^2$, the $m$ and $n$ are fixed. We compute $x_m-x_n$, then we sum up the square of each component and we sum them up.
$endgroup$
– Siong Thye Goh
Dec 2 '18 at 10:34
add a comment |
$begingroup$
In general, $|| cdot ||_2$ is called the $L^2$-norm, which is a functional from the function space $L^2(Omega,{cal A}, mu)$ to $Bbb R$ defined by $f mapsto (int |f|^2 ,{rm d}mu)^{1/2}$, provided that the integral exists. This is probably the case when you see $$||X_m - X_n||_2$$ since random variables are usually denoted with capital letters in probability-theory.
In particular, when $Omega = Bbb{N}$, $mu$ is the counting measure, we have $||f||_2 = (int |f|^2 ,{rm d}mu)^{1/2} = (sum_n |f_n|^2)^{1/2}$. This is the case if you actually mean
$$||x_m-x_n||_2$$ a norm of vector.
Remark: It's hard to distinguish $X$ from $x$ in your drawing. That's the reason why I promote the use of $rm LaTeX$ in all levels of math writing.
answered Dec 2 '18 at 7:34
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
$endgroup$
add a comment |
$begingroup$
In general, $|| cdot ||_2$ is called the $L^2$-norm, which is a functional from the function space $L^2(Omega,{cal A}, mu)$ to $Bbb R$ defined by $f mapsto (int |f|^2 ,{rm d}mu)^{1/2}$, provided that the integral exists. This is probably the case when you see $$||X_m - X_n||_2$$ since random variables are usually denoted with capital letters in probability-theory.
In particular, when $Omega = Bbb{N}$, $mu$ is the counting measure, we have $||f||_2 = (int |f|^2 ,{rm d}mu)^{1/2} = (sum_n |f_n|^2)^{1/2}$. This is the case if you actually mean
$$||x_m-x_n||_2$$ a norm of vector.
Remark: It's hard to distinguish $X$ from $x$ in your drawing. That's the reason why I promote the use of $rm LaTeX$ in all levels of math writing.
answered Dec 2 '18 at 7:34
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
$endgroup$
add a comment |
$begingroup$
In general, $|| cdot ||_2$ is called the $L^2$-norm, which is a functional from the function space $L^2(Omega,{cal A}, mu)$ to $Bbb R$ defined by $f mapsto (int |f|^2 ,{rm d}mu)^{1/2}$, provided that the integral exists. This is probably the case when you see $$||X_m - X_n||_2$$ since random variables are usually denoted with capital letters in probability-theory.
In particular, when $Omega = Bbb{N}$, $mu$ is the counting measure, we have $||f||_2 = (int |f|^2 ,{rm d}mu)^{1/2} = (sum_n |f_n|^2)^{1/2}$. This is the case if you actually mean
$$||x_m-x_n||_2$$ a norm of vector.
Remark: It's hard to distinguish $X$ from $x$ in your drawing. That's the reason why I promote the use of $rm LaTeX$ in all levels of math writing.
answered Dec 2 '18 at 7:34
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
$endgroup$
In general, $|| cdot ||_2$ is called the $L^2$-norm, which is a functional from the function space $L^2(Omega,{cal A}, mu)$ to $Bbb R$ defined by $f mapsto (int |f|^2 ,{rm d}mu)^{1/2}$, provided that the integral exists. This is probably the case when you see $$||X_m - X_n||_2$$ since random variables are usually denoted with capital letters in probability-theory.
In particular, when $Omega = Bbb{N}$, $mu$ is the counting measure, we have $||f||_2 = (int |f|^2 ,{rm d}mu)^{1/2} = (sum_n |f_n|^2)^{1/2}$. This is the case if you actually mean
$$||x_m-x_n||_2$$ a norm of vector.
Remark: It's hard to distinguish $X$ from $x$ in your drawing. That's the reason why I promote the use of $rm LaTeX$ in all levels of math writing.
answered Dec 2 '18 at 7:34
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Dec 2 '18 at 7:34
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Dec 2 '18 at 7:34
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Dec 2 '18 at 7:34
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
add a comment |
add a comment |
$begingroup$
Welcome to Math.SE. Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 2 '18 at 7:23
2
$begingroup$
It depends on the context. People only exposed to metric spaces will certainly think of it as the square of the $L^2$-norm. In that part of math tat would be the expectation. But, in another context it could mean something else. Basically whatever the author meant! That's why textbooks often include a page or two where notational conventions are spelled out (possibly including pointers to the pages where they are introduced).
$endgroup$
– Jyrki Lahtonen
Dec 2 '18 at 9:07