Dimensions of an algebraic expression [closed]
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From section 24 of the book "Integral calculus for beginners " by Joseph Edwards.
How to find the dimension of the algebraic expression, that the author is talking about? According to me the dimensions of 1/ $sqrt{a^2 - x^2}$ should be $-1$ but the author states that is is $0$. Please explain because if x and a are linear then dx must be constant hence the degree of 1/ $sqrt{a^2-x^2}$ should be $-1$ only
integration
asked Dec 16 '18 at 6:38
SLADESLADE
112
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closed as unclear what you're asking by Eevee Trainer, Did, BigbearZzz, Paul Frost, José Carlos Santos Dec 16 '18 at 15:31
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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$begingroup$
From section 24 of the book "Integral calculus for beginners " by Joseph Edwards.
How to find the dimension of the algebraic expression, that the author is talking about? According to me the dimensions of 1/ $sqrt{a^2 - x^2}$ should be $-1$ but the author states that is is $0$. Please explain because if x and a are linear then dx must be constant hence the degree of 1/ $sqrt{a^2-x^2}$ should be $-1$ only
integration
asked Dec 16 '18 at 6:38
SLADESLADE
112
$endgroup$
closed as unclear what you're asking by Eevee Trainer, Did, BigbearZzz, Paul Frost, José Carlos Santos Dec 16 '18 at 15:31
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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What do you mean by "dimensions" of $sqrt{a^2 - x^2}$? Who is the author? Any link to understand what you are talking about?
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– Robert Z
Dec 16 '18 at 6:40
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Actually i dont know ...this was mentioned in article 24 of the book "Integral calculus for beginners " by joseph edwards
$endgroup$
– SLADE
Dec 16 '18 at 6:43
1
$begingroup$
"but the author states that is is 0" No the author does not state that, and the dimension of $sqrt{a^2-x^2}$ is $+1$, obviously.
$endgroup$
– Did
Dec 16 '18 at 12:39
$begingroup$
"How to find the dimension of the algebraic expression, that the author is talking about?" In the example? As the author says, by assuming $a$ and $x$ and $dx$ are linear hence $a^2-x^2$ is quadratic, $sqrt{a^2-x^2}$ is linear and $dx/sqrt{a^2-x^2}$ has dimension $0$.
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– Did
Dec 16 '18 at 12:41
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@Did i have edited the question !!
$endgroup$
– SLADE
Dec 17 '18 at 6:23
|
show 2 more comments
$begingroup$
From section 24 of the book "Integral calculus for beginners " by Joseph Edwards.
How to find the dimension of the algebraic expression, that the author is talking about? According to me the dimensions of 1/ $sqrt{a^2 - x^2}$ should be $-1$ but the author states that is is $0$. Please explain because if x and a are linear then dx must be constant hence the degree of 1/ $sqrt{a^2-x^2}$ should be $-1$ only
integration
asked Dec 16 '18 at 6:38
SLADESLADE
112
$endgroup$
From section 24 of the book "Integral calculus for beginners " by Joseph Edwards.
How to find the dimension of the algebraic expression, that the author is talking about? According to me the dimensions of 1/ $sqrt{a^2 - x^2}$ should be $-1$ but the author states that is is $0$. Please explain because if x and a are linear then dx must be constant hence the degree of 1/ $sqrt{a^2-x^2}$ should be $-1$ only
integration
integration
asked Dec 16 '18 at 6:38
SLADESLADE
112
asked Dec 16 '18 at 6:38
SLADESLADE
112
edited Dec 17 '18 at 6:22
SLADE
asked Dec 16 '18 at 6:38
SLADESLADE
112
asked Dec 16 '18 at 6:38
SLADESLADE
112
asked Dec 16 '18 at 6:38
SLADESLADE
112
112
closed as unclear what you're asking by Eevee Trainer, Did, BigbearZzz, Paul Frost, José Carlos Santos Dec 16 '18 at 15:31
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 Eevee Trainer, Did, BigbearZzz, Paul Frost, José Carlos Santos Dec 16 '18 at 15:31
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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What do you mean by "dimensions" of $sqrt{a^2 - x^2}$? Who is the author? Any link to understand what you are talking about?
$endgroup$
– Robert Z
Dec 16 '18 at 6:40
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Actually i dont know ...this was mentioned in article 24 of the book "Integral calculus for beginners " by joseph edwards
$endgroup$
– SLADE
Dec 16 '18 at 6:43
1
$begingroup$
"but the author states that is is 0" No the author does not state that, and the dimension of $sqrt{a^2-x^2}$ is $+1$, obviously.
$endgroup$
– Did
Dec 16 '18 at 12:39
$begingroup$
"How to find the dimension of the algebraic expression, that the author is talking about?" In the example? As the author says, by assuming $a$ and $x$ and $dx$ are linear hence $a^2-x^2$ is quadratic, $sqrt{a^2-x^2}$ is linear and $dx/sqrt{a^2-x^2}$ has dimension $0$.
$endgroup$
– Did
Dec 16 '18 at 12:41
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@Did i have edited the question !!
$endgroup$
– SLADE
Dec 17 '18 at 6:23
|
show 2 more comments
$begingroup$
What do you mean by "dimensions" of $sqrt{a^2 - x^2}$? Who is the author? Any link to understand what you are talking about?
$endgroup$
– Robert Z
Dec 16 '18 at 6:40
$begingroup$
Actually i dont know ...this was mentioned in article 24 of the book "Integral calculus for beginners " by joseph edwards
$endgroup$
– SLADE
Dec 16 '18 at 6:43
1
$begingroup$
"but the author states that is is 0" No the author does not state that, and the dimension of $sqrt{a^2-x^2}$ is $+1$, obviously.
$endgroup$
– Did
Dec 16 '18 at 12:39
$begingroup$
"How to find the dimension of the algebraic expression, that the author is talking about?" In the example? As the author says, by assuming $a$ and $x$ and $dx$ are linear hence $a^2-x^2$ is quadratic, $sqrt{a^2-x^2}$ is linear and $dx/sqrt{a^2-x^2}$ has dimension $0$.
$endgroup$
– Did
Dec 16 '18 at 12:41
$begingroup$
@Did i have edited the question !!
$endgroup$
– SLADE
Dec 17 '18 at 6:23
$begingroup$
What do you mean by "dimensions" of $sqrt{a^2 - x^2}$? Who is the author? Any link to understand what you are talking about?
$endgroup$
– Robert Z
Dec 16 '18 at 6:40
$begingroup$
What do you mean by "dimensions" of $sqrt{a^2 - x^2}$? Who is the author? Any link to understand what you are talking about?
$endgroup$
– Robert Z
Dec 16 '18 at 6:40
$begingroup$
Actually i dont know ...this was mentioned in article 24 of the book "Integral calculus for beginners " by joseph edwards
$endgroup$
– SLADE
Dec 16 '18 at 6:43
$begingroup$
Actually i dont know ...this was mentioned in article 24 of the book "Integral calculus for beginners " by joseph edwards
$endgroup$
– SLADE
Dec 16 '18 at 6:43
1
1
$begingroup$
"but the author states that is is 0" No the author does not state that, and the dimension of $sqrt{a^2-x^2}$ is $+1$, obviously.
$endgroup$
– Did
Dec 16 '18 at 12:39
$begingroup$
"but the author states that is is 0" No the author does not state that, and the dimension of $sqrt{a^2-x^2}$ is $+1$, obviously.
$endgroup$
– Did
Dec 16 '18 at 12:39
$begingroup$
"How to find the dimension of the algebraic expression, that the author is talking about?" In the example? As the author says, by assuming $a$ and $x$ and $dx$ are linear hence $a^2-x^2$ is quadratic, $sqrt{a^2-x^2}$ is linear and $dx/sqrt{a^2-x^2}$ has dimension $0$.
$endgroup$
– Did
Dec 16 '18 at 12:41
$begingroup$
"How to find the dimension of the algebraic expression, that the author is talking about?" In the example? As the author says, by assuming $a$ and $x$ and $dx$ are linear hence $a^2-x^2$ is quadratic, $sqrt{a^2-x^2}$ is linear and $dx/sqrt{a^2-x^2}$ has dimension $0$.
$endgroup$
– Did
Dec 16 '18 at 12:41
$begingroup$
@Did i have edited the question !!
$endgroup$
– SLADE
Dec 17 '18 at 6:23
$begingroup$
@Did i have edited the question !!
$endgroup$
– SLADE
Dec 17 '18 at 6:23
|
show 2 more comments
1 Answer
1
active
oldest
votes
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The author is saying that
$$intfrac{1}{sqrt{a^2-x^2}},dx $$
has dimensions $0$ because $sqrt{a^2-x^2}$ at the denominator is of dimensions $2cdot (1/2)=1$ and $dx$ at the numerator is of dimensions $1$. So $1-1=0$.
By using the same argument,
$$intfrac{1}{a^2+x^2},dx $$
is of dimensions $1-2=-1$.
answered Dec 16 '18 at 6:50
Robert ZRobert Z
100k1068140
$endgroup$
$begingroup$
Thank you , i never thought about dx
$endgroup$
– SLADE
Dec 16 '18 at 6:53
$begingroup$
But if x is linear ,isn't dx supposed to be of 0 degree then?
$endgroup$
– SLADE
Dec 17 '18 at 6:18
$begingroup$
$x=x^1$ is of $1$ degree. $dx$ is a variation of $x$ (a difference of $x$) not a derivative.
$endgroup$
– Robert Z
Dec 17 '18 at 6:23
$begingroup$
hence dx is of 0 degree?
$endgroup$
– SLADE
Dec 17 '18 at 6:24
$begingroup$
$dx$ is a variation of $x$ (a difference of $x$) not a derivative. So $dx$ has the same dimensions of $x$.
$endgroup$
– Robert Z
Dec 17 '18 at 6:24
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The author is saying that
$$intfrac{1}{sqrt{a^2-x^2}},dx $$
has dimensions $0$ because $sqrt{a^2-x^2}$ at the denominator is of dimensions $2cdot (1/2)=1$ and $dx$ at the numerator is of dimensions $1$. So $1-1=0$.
By using the same argument,
$$intfrac{1}{a^2+x^2},dx $$
is of dimensions $1-2=-1$.
answered Dec 16 '18 at 6:50
Robert ZRobert Z
100k1068140
$endgroup$
$begingroup$
Thank you , i never thought about dx
$endgroup$
– SLADE
Dec 16 '18 at 6:53
$begingroup$
But if x is linear ,isn't dx supposed to be of 0 degree then?
$endgroup$
– SLADE
Dec 17 '18 at 6:18
$begingroup$
$x=x^1$ is of $1$ degree. $dx$ is a variation of $x$ (a difference of $x$) not a derivative.
$endgroup$
– Robert Z
Dec 17 '18 at 6:23
$begingroup$
hence dx is of 0 degree?
$endgroup$
– SLADE
Dec 17 '18 at 6:24
$begingroup$
$dx$ is a variation of $x$ (a difference of $x$) not a derivative. So $dx$ has the same dimensions of $x$.
$endgroup$
– Robert Z
Dec 17 '18 at 6:24
add a comment |
$begingroup$
The author is saying that
$$intfrac{1}{sqrt{a^2-x^2}},dx $$
has dimensions $0$ because $sqrt{a^2-x^2}$ at the denominator is of dimensions $2cdot (1/2)=1$ and $dx$ at the numerator is of dimensions $1$. So $1-1=0$.
By using the same argument,
$$intfrac{1}{a^2+x^2},dx $$
is of dimensions $1-2=-1$.
answered Dec 16 '18 at 6:50
Robert ZRobert Z
100k1068140
$endgroup$
$begingroup$
Thank you , i never thought about dx
$endgroup$
– SLADE
Dec 16 '18 at 6:53
$begingroup$
But if x is linear ,isn't dx supposed to be of 0 degree then?
$endgroup$
– SLADE
Dec 17 '18 at 6:18
$begingroup$
$x=x^1$ is of $1$ degree. $dx$ is a variation of $x$ (a difference of $x$) not a derivative.
$endgroup$
– Robert Z
Dec 17 '18 at 6:23
$begingroup$
hence dx is of 0 degree?
$endgroup$
– SLADE
Dec 17 '18 at 6:24
$begingroup$
$dx$ is a variation of $x$ (a difference of $x$) not a derivative. So $dx$ has the same dimensions of $x$.
$endgroup$
– Robert Z
Dec 17 '18 at 6:24
add a comment |
$begingroup$
The author is saying that
$$intfrac{1}{sqrt{a^2-x^2}},dx $$
has dimensions $0$ because $sqrt{a^2-x^2}$ at the denominator is of dimensions $2cdot (1/2)=1$ and $dx$ at the numerator is of dimensions $1$. So $1-1=0$.
By using the same argument,
$$intfrac{1}{a^2+x^2},dx $$
is of dimensions $1-2=-1$.
answered Dec 16 '18 at 6:50
Robert ZRobert Z
100k1068140
$endgroup$
The author is saying that
$$intfrac{1}{sqrt{a^2-x^2}},dx $$
has dimensions $0$ because $sqrt{a^2-x^2}$ at the denominator is of dimensions $2cdot (1/2)=1$ and $dx$ at the numerator is of dimensions $1$. So $1-1=0$.
By using the same argument,
$$intfrac{1}{a^2+x^2},dx $$
is of dimensions $1-2=-1$.
answered Dec 16 '18 at 6:50
Robert ZRobert Z
100k1068140
edited Dec 16 '18 at 6:55
answered Dec 16 '18 at 6:50
Robert ZRobert Z
100k1068140
answered Dec 16 '18 at 6:50
Robert ZRobert Z
100k1068140
answered Dec 16 '18 at 6:50
Robert ZRobert Z
100k1068140
100k1068140
$begingroup$
Thank you , i never thought about dx
$endgroup$
– SLADE
Dec 16 '18 at 6:53
$begingroup$
But if x is linear ,isn't dx supposed to be of 0 degree then?
$endgroup$
– SLADE
Dec 17 '18 at 6:18
$begingroup$
$x=x^1$ is of $1$ degree. $dx$ is a variation of $x$ (a difference of $x$) not a derivative.
$endgroup$
– Robert Z
Dec 17 '18 at 6:23
$begingroup$
hence dx is of 0 degree?
$endgroup$
– SLADE
Dec 17 '18 at 6:24
$begingroup$
$dx$ is a variation of $x$ (a difference of $x$) not a derivative. So $dx$ has the same dimensions of $x$.
$endgroup$
– Robert Z
Dec 17 '18 at 6:24
add a comment |
$begingroup$
Thank you , i never thought about dx
$endgroup$
– SLADE
Dec 16 '18 at 6:53
$begingroup$
But if x is linear ,isn't dx supposed to be of 0 degree then?
$endgroup$
– SLADE
Dec 17 '18 at 6:18
$begingroup$
$x=x^1$ is of $1$ degree. $dx$ is a variation of $x$ (a difference of $x$) not a derivative.
$endgroup$
– Robert Z
Dec 17 '18 at 6:23
$begingroup$
hence dx is of 0 degree?
$endgroup$
– SLADE
Dec 17 '18 at 6:24
$begingroup$
$dx$ is a variation of $x$ (a difference of $x$) not a derivative. So $dx$ has the same dimensions of $x$.
$endgroup$
– Robert Z
Dec 17 '18 at 6:24
$begingroup$
Thank you , i never thought about dx
$endgroup$
– SLADE
Dec 16 '18 at 6:53
$begingroup$
Thank you , i never thought about dx
$endgroup$
– SLADE
Dec 16 '18 at 6:53
$begingroup$
But if x is linear ,isn't dx supposed to be of 0 degree then?
$endgroup$
– SLADE
Dec 17 '18 at 6:18
$begingroup$
But if x is linear ,isn't dx supposed to be of 0 degree then?
$endgroup$
– SLADE
Dec 17 '18 at 6:18
$begingroup$
$x=x^1$ is of $1$ degree. $dx$ is a variation of $x$ (a difference of $x$) not a derivative.
$endgroup$
– Robert Z
Dec 17 '18 at 6:23
$begingroup$
$x=x^1$ is of $1$ degree. $dx$ is a variation of $x$ (a difference of $x$) not a derivative.
$endgroup$
– Robert Z
Dec 17 '18 at 6:23
$begingroup$
hence dx is of 0 degree?
$endgroup$
– SLADE
Dec 17 '18 at 6:24
$begingroup$
hence dx is of 0 degree?
$endgroup$
– SLADE
Dec 17 '18 at 6:24
$begingroup$
$dx$ is a variation of $x$ (a difference of $x$) not a derivative. So $dx$ has the same dimensions of $x$.
$endgroup$
– Robert Z
Dec 17 '18 at 6:24
$begingroup$
$dx$ is a variation of $x$ (a difference of $x$) not a derivative. So $dx$ has the same dimensions of $x$.
$endgroup$
– Robert Z
Dec 17 '18 at 6:24
add a comment |
$begingroup$
What do you mean by "dimensions" of $sqrt{a^2 - x^2}$? Who is the author? Any link to understand what you are talking about?
$endgroup$
– Robert Z
Dec 16 '18 at 6:40
$begingroup$
Actually i dont know ...this was mentioned in article 24 of the book "Integral calculus for beginners " by joseph edwards
$endgroup$
– SLADE
Dec 16 '18 at 6:43
1
$begingroup$
"but the author states that is is 0" No the author does not state that, and the dimension of $sqrt{a^2-x^2}$ is $+1$, obviously.
$endgroup$
– Did
Dec 16 '18 at 12:39
$begingroup$
"How to find the dimension of the algebraic expression, that the author is talking about?" In the example? As the author says, by assuming $a$ and $x$ and $dx$ are linear hence $a^2-x^2$ is quadratic, $sqrt{a^2-x^2}$ is linear and $dx/sqrt{a^2-x^2}$ has dimension $0$.
$endgroup$
– Did
Dec 16 '18 at 12:41
$begingroup$
@Did i have edited the question !!
$endgroup$
– SLADE
Dec 17 '18 at 6:23