In the definition of partial derivative, why the function must be defined on an open set?
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On the page Partial derivative on Wikipedia, the following formal definition was found:
I am wondering if in this definition, the condition that $U$ being open is always necessary. For example, if in $mathbb{R}^n$, a function $f$ is defined on the set of union of all axises: $A={(x_1,ldots,x_n)mid text{all but one }x_i text{'s are zero}}$, it seems to me that one can still compute all partial derivatives of $f$ at $(0,ldots, 0)$ without requiring the domain of $f$ being an open set.
Should we loosen the condition that $U$ is open to something like "U contains the intersection ${(x_1,ldots,x_n)+(a_1,ldots, a_n)mid text{all but one }x_i text{'s are zero}}cap B$, where $B$ is an open ball containing $(a_1,ldots, a_n)$"? In this case the domain of $f$ is not necessarily an open set but the definition of the partial derivative at $(a_1,ldots, a_n)$ still makes sense to me. Did I miss anything here?
multivariable-calculus definition partial-derivative
asked Dec 1 at 12:42
Zuriel
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up vote
2
down vote
favorite
On the page Partial derivative on Wikipedia, the following formal definition was found:
I am wondering if in this definition, the condition that $U$ being open is always necessary. For example, if in $mathbb{R}^n$, a function $f$ is defined on the set of union of all axises: $A={(x_1,ldots,x_n)mid text{all but one }x_i text{'s are zero}}$, it seems to me that one can still compute all partial derivatives of $f$ at $(0,ldots, 0)$ without requiring the domain of $f$ being an open set.
Should we loosen the condition that $U$ is open to something like "U contains the intersection ${(x_1,ldots,x_n)+(a_1,ldots, a_n)mid text{all but one }x_i text{'s are zero}}cap B$, where $B$ is an open ball containing $(a_1,ldots, a_n)$"? In this case the domain of $f$ is not necessarily an open set but the definition of the partial derivative at $(a_1,ldots, a_n)$ still makes sense to me. Did I miss anything here?
multivariable-calculus definition partial-derivative
asked Dec 1 at 12:42
Zuriel
1,5291028
Related questions: math.stackexchange.com/q/2830875, math.stackexchange.com/q/161910
– Paul Frost
Dec 1 at 13:58
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
On the page Partial derivative on Wikipedia, the following formal definition was found:
I am wondering if in this definition, the condition that $U$ being open is always necessary. For example, if in $mathbb{R}^n$, a function $f$ is defined on the set of union of all axises: $A={(x_1,ldots,x_n)mid text{all but one }x_i text{'s are zero}}$, it seems to me that one can still compute all partial derivatives of $f$ at $(0,ldots, 0)$ without requiring the domain of $f$ being an open set.
Should we loosen the condition that $U$ is open to something like "U contains the intersection ${(x_1,ldots,x_n)+(a_1,ldots, a_n)mid text{all but one }x_i text{'s are zero}}cap B$, where $B$ is an open ball containing $(a_1,ldots, a_n)$"? In this case the domain of $f$ is not necessarily an open set but the definition of the partial derivative at $(a_1,ldots, a_n)$ still makes sense to me. Did I miss anything here?
multivariable-calculus definition partial-derivative
asked Dec 1 at 12:42
Zuriel
1,5291028
On the page Partial derivative on Wikipedia, the following formal definition was found:
I am wondering if in this definition, the condition that $U$ being open is always necessary. For example, if in $mathbb{R}^n$, a function $f$ is defined on the set of union of all axises: $A={(x_1,ldots,x_n)mid text{all but one }x_i text{'s are zero}}$, it seems to me that one can still compute all partial derivatives of $f$ at $(0,ldots, 0)$ without requiring the domain of $f$ being an open set.
Should we loosen the condition that $U$ is open to something like "U contains the intersection ${(x_1,ldots,x_n)+(a_1,ldots, a_n)mid text{all but one }x_i text{'s are zero}}cap B$, where $B$ is an open ball containing $(a_1,ldots, a_n)$"? In this case the domain of $f$ is not necessarily an open set but the definition of the partial derivative at $(a_1,ldots, a_n)$ still makes sense to me. Did I miss anything here?
multivariable-calculus definition partial-derivative
multivariable-calculus definition partial-derivative
asked Dec 1 at 12:42
Zuriel
1,5291028
asked Dec 1 at 12:42
Zuriel
1,5291028
edited Dec 1 at 14:10
asked Dec 1 at 12:42
Zuriel
1,5291028
asked Dec 1 at 12:42
Zuriel
1,5291028
asked Dec 1 at 12:42
Zuriel
1,5291028
1,5291028
Related questions: math.stackexchange.com/q/2830875, math.stackexchange.com/q/161910
– Paul Frost
Dec 1 at 13:58
add a comment |
Related questions: math.stackexchange.com/q/2830875, math.stackexchange.com/q/161910
– Paul Frost
Dec 1 at 13:58
Related questions: math.stackexchange.com/q/2830875, math.stackexchange.com/q/161910
– Paul Frost
Dec 1 at 13:58
Related questions: math.stackexchange.com/q/2830875, math.stackexchange.com/q/161910
– Paul Frost
Dec 1 at 13:58
add a comment |
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Related questions: math.stackexchange.com/q/2830875, math.stackexchange.com/q/161910
– Paul Frost
Dec 1 at 13:58