Why do we need the absolute value signs when integrating the square of a function?
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Why do we need the absolute value signs in the definition of square-integrable function? As seen below:
$$ int_-infty^infty lvert f(x) rvert^2 , dx < infty $$
integration notation lp-spaces
edited Sep 9 at 5:04
Michael Hardy
206k23187466
asked Aug 23 '14 at 23:19
BCLC
1
add a comment |Â
up vote
13
down vote
favorite
Why do we need the absolute value signs in the definition of square-integrable function? As seen below:
$$ int_-infty^infty lvert f(x) rvert^2 , dx < infty $$
integration notation lp-spaces
edited Sep 9 at 5:04
Michael Hardy
206k23187466
asked Aug 23 '14 at 23:19
BCLC
1
The absolute value signs ensure that the integrand is a non-negative real function.
– David H
Aug 23 '14 at 23:22
2
The other answers are right, but I'll add that it also looks best (to me) among the options $$int (f(x))^2,dxquad int |f(x)|^2,dxquad int f(x)^2,dxquad int f^2(x),dx$$
– user147263
Aug 23 '14 at 23:42
add a comment |Â
up vote
13
down vote
favorite
up vote
13
down vote
favorite
Why do we need the absolute value signs in the definition of square-integrable function? As seen below:
$$ int_-infty^infty lvert f(x) rvert^2 , dx < infty $$
integration notation lp-spaces
edited Sep 9 at 5:04
Michael Hardy
206k23187466
asked Aug 23 '14 at 23:19
BCLC
1
Why do we need the absolute value signs in the definition of square-integrable function? As seen below:
$$ int_-infty^infty lvert f(x) rvert^2 , dx < infty $$
integration notation lp-spaces
integration notation lp-spaces
edited Sep 9 at 5:04
Michael Hardy
206k23187466
asked Aug 23 '14 at 23:19
BCLC
1
edited Sep 9 at 5:04
Michael Hardy
206k23187466
asked Aug 23 '14 at 23:19
BCLC
1
edited Sep 9 at 5:04
Michael Hardy
206k23187466
edited Sep 9 at 5:04
Michael Hardy
206k23187466
edited Sep 9 at 5:04
Michael Hardy
206k23187466
206k23187466
asked Aug 23 '14 at 23:19
BCLC
1
asked Aug 23 '14 at 23:19
BCLC
1
asked Aug 23 '14 at 23:19
BCLC
1
1
The absolute value signs ensure that the integrand is a non-negative real function.
– David H
Aug 23 '14 at 23:22
2
The other answers are right, but I'll add that it also looks best (to me) among the options $$int (f(x))^2,dxquad int |f(x)|^2,dxquad int f(x)^2,dxquad int f^2(x),dx$$
– user147263
Aug 23 '14 at 23:42
add a comment |Â
The absolute value signs ensure that the integrand is a non-negative real function.
– David H
Aug 23 '14 at 23:22
2
The other answers are right, but I'll add that it also looks best (to me) among the options $$int (f(x))^2,dxquad int |f(x)|^2,dxquad int f(x)^2,dxquad int f^2(x),dx$$
– user147263
Aug 23 '14 at 23:42
The absolute value signs ensure that the integrand is a non-negative real function.
– David H
Aug 23 '14 at 23:22
The absolute value signs ensure that the integrand is a non-negative real function.
– David H
Aug 23 '14 at 23:22
2
2
The other answers are right, but I'll add that it also looks best (to me) among the options $$int (f(x))^2,dxquad int |f(x)|^2,dxquad int f(x)^2,dxquad int f^2(x),dx$$
– user147263
Aug 23 '14 at 23:42
The other answers are right, but I'll add that it also looks best (to me) among the options $$int (f(x))^2,dxquad int |f(x)|^2,dxquad int f(x)^2,dxquad int f^2(x),dx$$
– user147263
Aug 23 '14 at 23:42
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
33
down vote
accepted
Because complex-valued functions are used. The square of a complex number need not be non-negative.
answered Aug 23 '14 at 23:23
Michael Hardy
206k23187466
Why do it need to be non-negative? Just simply a subtraction on the integration, I think.
– Ooker
Sep 1 '14 at 10:20
3
Because one is trying to define a metric with properties similar to those of the metric in Euclidean space.
– Michael Hardy
Sep 1 '14 at 17:50
add a comment |Â
up vote
14
down vote
Besides the complex-valued case, I suspect it also has to do with the existence of other Lp spaces; Wikipedia gives the general definition as
$$
|f|_p = left(int_S |f|^p dmuright)^1/p
$$
Since the absolute value symbols are redundant only for even integral $p$, omitting them disrupts the uniformity of the notation without buying a whole lot.
answered Aug 24 '14 at 3:55
Jonathan Cast
725410
In other words, tradition?
– BCLC
Aug 24 '14 at 3:56
7
@BCLC: not really. Tradition would be "we write it that way because we always have". Which might be true, but it's not an alternative way of phrasing this explanation, which is "we write it that way because we consider $2$ to be a special case of $p$, and it would be weird to write the special case differently".
– Steve Jessop
Aug 24 '14 at 15:21
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
33
down vote
accepted
Because complex-valued functions are used. The square of a complex number need not be non-negative.
answered Aug 23 '14 at 23:23
Michael Hardy
206k23187466
Why do it need to be non-negative? Just simply a subtraction on the integration, I think.
– Ooker
Sep 1 '14 at 10:20
3
Because one is trying to define a metric with properties similar to those of the metric in Euclidean space.
– Michael Hardy
Sep 1 '14 at 17:50
add a comment |Â
up vote
33
down vote
accepted
Because complex-valued functions are used. The square of a complex number need not be non-negative.
answered Aug 23 '14 at 23:23
Michael Hardy
206k23187466
Why do it need to be non-negative? Just simply a subtraction on the integration, I think.
– Ooker
Sep 1 '14 at 10:20
3
Because one is trying to define a metric with properties similar to those of the metric in Euclidean space.
– Michael Hardy
Sep 1 '14 at 17:50
add a comment |Â
up vote
33
down vote
accepted
up vote
33
down vote
accepted
Because complex-valued functions are used. The square of a complex number need not be non-negative.
answered Aug 23 '14 at 23:23
Michael Hardy
206k23187466
Because complex-valued functions are used. The square of a complex number need not be non-negative.
answered Aug 23 '14 at 23:23
Michael Hardy
206k23187466
answered Aug 23 '14 at 23:23
Michael Hardy
206k23187466
answered Aug 23 '14 at 23:23
Michael Hardy
206k23187466
answered Aug 23 '14 at 23:23
Michael Hardy
206k23187466
206k23187466
Why do it need to be non-negative? Just simply a subtraction on the integration, I think.
– Ooker
Sep 1 '14 at 10:20
3
Because one is trying to define a metric with properties similar to those of the metric in Euclidean space.
– Michael Hardy
Sep 1 '14 at 17:50
add a comment |Â
Why do it need to be non-negative? Just simply a subtraction on the integration, I think.
– Ooker
Sep 1 '14 at 10:20
3
Because one is trying to define a metric with properties similar to those of the metric in Euclidean space.
– Michael Hardy
Sep 1 '14 at 17:50
Why do it need to be non-negative? Just simply a subtraction on the integration, I think.
– Ooker
Sep 1 '14 at 10:20
Why do it need to be non-negative? Just simply a subtraction on the integration, I think.
– Ooker
Sep 1 '14 at 10:20
3
3
Because one is trying to define a metric with properties similar to those of the metric in Euclidean space.
– Michael Hardy
Sep 1 '14 at 17:50
Because one is trying to define a metric with properties similar to those of the metric in Euclidean space.
– Michael Hardy
Sep 1 '14 at 17:50
add a comment |Â
up vote
14
down vote
Besides the complex-valued case, I suspect it also has to do with the existence of other Lp spaces; Wikipedia gives the general definition as
$$
|f|_p = left(int_S |f|^p dmuright)^1/p
$$
Since the absolute value symbols are redundant only for even integral $p$, omitting them disrupts the uniformity of the notation without buying a whole lot.
answered Aug 24 '14 at 3:55
Jonathan Cast
725410
In other words, tradition?
– BCLC
Aug 24 '14 at 3:56
7
@BCLC: not really. Tradition would be "we write it that way because we always have". Which might be true, but it's not an alternative way of phrasing this explanation, which is "we write it that way because we consider $2$ to be a special case of $p$, and it would be weird to write the special case differently".
– Steve Jessop
Aug 24 '14 at 15:21
add a comment |Â
up vote
14
down vote
Besides the complex-valued case, I suspect it also has to do with the existence of other Lp spaces; Wikipedia gives the general definition as
$$
|f|_p = left(int_S |f|^p dmuright)^1/p
$$
Since the absolute value symbols are redundant only for even integral $p$, omitting them disrupts the uniformity of the notation without buying a whole lot.
answered Aug 24 '14 at 3:55
Jonathan Cast
725410
In other words, tradition?
– BCLC
Aug 24 '14 at 3:56
7
@BCLC: not really. Tradition would be "we write it that way because we always have". Which might be true, but it's not an alternative way of phrasing this explanation, which is "we write it that way because we consider $2$ to be a special case of $p$, and it would be weird to write the special case differently".
– Steve Jessop
Aug 24 '14 at 15:21
add a comment |Â
up vote
14
down vote
up vote
14
down vote
Besides the complex-valued case, I suspect it also has to do with the existence of other Lp spaces; Wikipedia gives the general definition as
$$
|f|_p = left(int_S |f|^p dmuright)^1/p
$$
Since the absolute value symbols are redundant only for even integral $p$, omitting them disrupts the uniformity of the notation without buying a whole lot.
answered Aug 24 '14 at 3:55
Jonathan Cast
725410
Besides the complex-valued case, I suspect it also has to do with the existence of other Lp spaces; Wikipedia gives the general definition as
$$
|f|_p = left(int_S |f|^p dmuright)^1/p
$$
Since the absolute value symbols are redundant only for even integral $p$, omitting them disrupts the uniformity of the notation without buying a whole lot.
answered Aug 24 '14 at 3:55
Jonathan Cast
725410
answered Aug 24 '14 at 3:55
Jonathan Cast
725410
answered Aug 24 '14 at 3:55
Jonathan Cast
725410
answered Aug 24 '14 at 3:55
Jonathan Cast
725410
725410
In other words, tradition?
– BCLC
Aug 24 '14 at 3:56
7
@BCLC: not really. Tradition would be "we write it that way because we always have". Which might be true, but it's not an alternative way of phrasing this explanation, which is "we write it that way because we consider $2$ to be a special case of $p$, and it would be weird to write the special case differently".
– Steve Jessop
Aug 24 '14 at 15:21
add a comment |Â
In other words, tradition?
– BCLC
Aug 24 '14 at 3:56
7
@BCLC: not really. Tradition would be "we write it that way because we always have". Which might be true, but it's not an alternative way of phrasing this explanation, which is "we write it that way because we consider $2$ to be a special case of $p$, and it would be weird to write the special case differently".
– Steve Jessop
Aug 24 '14 at 15:21
In other words, tradition?
– BCLC
Aug 24 '14 at 3:56
In other words, tradition?
– BCLC
Aug 24 '14 at 3:56
7
7
@BCLC: not really. Tradition would be "we write it that way because we always have". Which might be true, but it's not an alternative way of phrasing this explanation, which is "we write it that way because we consider $2$ to be a special case of $p$, and it would be weird to write the special case differently".
– Steve Jessop
Aug 24 '14 at 15:21
@BCLC: not really. Tradition would be "we write it that way because we always have". Which might be true, but it's not an alternative way of phrasing this explanation, which is "we write it that way because we consider $2$ to be a special case of $p$, and it would be weird to write the special case differently".
– Steve Jessop
Aug 24 '14 at 15:21
add a comment |Â
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The absolute value signs ensure that the integrand is a non-negative real function.
– David H
Aug 23 '14 at 23:22
2
The other answers are right, but I'll add that it also looks best (to me) among the options $$int (f(x))^2,dxquad int |f(x)|^2,dxquad int f(x)^2,dxquad int f^2(x),dx$$
– user147263
Aug 23 '14 at 23:42