Why we can not differentiate functions that have closed ball domain?
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Why we can not differentiate functions that have closed ball domain?
I think this comes from the idea that we can not differentiate functions that have closed intervals domain, as in this case we can study differentiation on the open interval and then study differentiation from the left at the end point of the interval and study differentiation from right at the starting point of the interval. am I correct?
If so how can I explain this for functions having closed ball domain?
Thanks!
general-topology derivatives differential-geometry differential-topology
asked Aug 26 at 20:01
Idonotknow
582214
add a comment |Â
up vote
0
down vote
favorite
Why we can not differentiate functions that have closed ball domain?
I think this comes from the idea that we can not differentiate functions that have closed intervals domain, as in this case we can study differentiation on the open interval and then study differentiation from the left at the end point of the interval and study differentiation from right at the starting point of the interval. am I correct?
If so how can I explain this for functions having closed ball domain?
Thanks!
general-topology derivatives differential-geometry differential-topology
asked Aug 26 at 20:01
Idonotknow
582214
You are correct. Consider a point on the boundary of the ball. As in the closed interval, we cannot take the derivative in each direction.
– John Douma
Aug 26 at 20:13
@JohnDouma why? and in the case of the ball there are several directions not justleft and right ..... sohow we differentiate a closed ball?
– Idonotknow
Aug 26 at 20:31
1
The function is not defined there. We are on the boundary of the ball and the function is only defined on the ball. Imagine taking a ball around this boundary point.
– John Douma
Aug 26 at 20:32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Why we can not differentiate functions that have closed ball domain?
I think this comes from the idea that we can not differentiate functions that have closed intervals domain, as in this case we can study differentiation on the open interval and then study differentiation from the left at the end point of the interval and study differentiation from right at the starting point of the interval. am I correct?
If so how can I explain this for functions having closed ball domain?
Thanks!
general-topology derivatives differential-geometry differential-topology
asked Aug 26 at 20:01
Idonotknow
582214
Why we can not differentiate functions that have closed ball domain?
I think this comes from the idea that we can not differentiate functions that have closed intervals domain, as in this case we can study differentiation on the open interval and then study differentiation from the left at the end point of the interval and study differentiation from right at the starting point of the interval. am I correct?
If so how can I explain this for functions having closed ball domain?
Thanks!
general-topology derivatives differential-geometry differential-topology
asked Aug 26 at 20:01
Idonotknow
582214
asked Aug 26 at 20:01
Idonotknow
582214
asked Aug 26 at 20:01
Idonotknow
582214
asked Aug 26 at 20:01
Idonotknow
582214
582214
You are correct. Consider a point on the boundary of the ball. As in the closed interval, we cannot take the derivative in each direction.
– John Douma
Aug 26 at 20:13
@JohnDouma why? and in the case of the ball there are several directions not justleft and right ..... sohow we differentiate a closed ball?
– Idonotknow
Aug 26 at 20:31
1
The function is not defined there. We are on the boundary of the ball and the function is only defined on the ball. Imagine taking a ball around this boundary point.
– John Douma
Aug 26 at 20:32
add a comment |Â
You are correct. Consider a point on the boundary of the ball. As in the closed interval, we cannot take the derivative in each direction.
– John Douma
Aug 26 at 20:13
@JohnDouma why? and in the case of the ball there are several directions not justleft and right ..... sohow we differentiate a closed ball?
– Idonotknow
Aug 26 at 20:31
1
The function is not defined there. We are on the boundary of the ball and the function is only defined on the ball. Imagine taking a ball around this boundary point.
– John Douma
Aug 26 at 20:32
You are correct. Consider a point on the boundary of the ball. As in the closed interval, we cannot take the derivative in each direction.
– John Douma
Aug 26 at 20:13
You are correct. Consider a point on the boundary of the ball. As in the closed interval, we cannot take the derivative in each direction.
– John Douma
Aug 26 at 20:13
@JohnDouma why? and in the case of the ball there are several directions not justleft and right ..... sohow we differentiate a closed ball?
– Idonotknow
Aug 26 at 20:31
@JohnDouma why? and in the case of the ball there are several directions not justleft and right ..... sohow we differentiate a closed ball?
– Idonotknow
Aug 26 at 20:31
1
1
The function is not defined there. We are on the boundary of the ball and the function is only defined on the ball. Imagine taking a ball around this boundary point.
– John Douma
Aug 26 at 20:32
The function is not defined there. We are on the boundary of the ball and the function is only defined on the ball. Imagine taking a ball around this boundary point.
– John Douma
Aug 26 at 20:32
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
There's no intrinsic reason not to define a derivative of a function defined on a closed ball, at a point on the boundary on that ball.
In more detail, if $B = xinmathbb R^n: $ and $f:Bto mathbb R^m$, then for any point $x_0$ domain of $f$ we can define that the derivative of $f$ at $x_0$ means a linear mapping $T$ such that
$$ f(x) = f(x_0) + T(x-x_0) + o(x-x_0) $$
for $xto x_0$ and $xin B$.
Even at the boundary points of $B$, there are neighborhoods of the ball that are rich enough that the derivative is unique if it exists. (All that's needed for this is that every neighborhood of $x_0$ relative to $B$ contains points whose difference from $x_0$ span $mathbb R^n$).
The reason why textbooks often don't do this is that most of what they want to say can be said quite fine while restricting our attention to open domains, and doing so is quite a bit simpler -- there are (literally!) fewer edge cases that need to be considered when stating the results.
For one example of one such corner case, consider the function
$$ f(x,y) = begincases 0 & textif y < x \
0 & textif y > 4x \
0 & textif x = y = 0 \
1 & textif 2x < y < 3x \
textundefined & textotherwise endcases $$
whose partial derivatives exist and are continuous (indeed identically $0$) everywhere in the domain of $f$, but $f$ is not even continuous at $(0,0)$.
answered Aug 26 at 21:20
Henning Makholm
230k16296527
But if the domain of the function $f$ is not open we can not speak about partial derivatives of it ..... correct? if so why?
– Idonotknow
Sep 3 at 14:54
1
@Idonotknow, correct: Partial deriviatives need more than I describe here. A most points on the boundary of a closed ball we could speak of partial derivitives well enough, but there are some points that are exceptions -- for example, for $f$ defined on the closed unit disk, $partial f/partial x$ would not make sense at $(0,1)$.
– Henning Makholm
Sep 3 at 14:59
why partial derivative at that point would not make sense?
– Idonotknow
Sep 3 at 15:14
1
@Idonotknow: Because $xmapsto f(x,1)$ is only defined at $x=0$, and even with the definition above you can't meaningfully take a derivative at an isolated point. See new addition to answer for an even more dramatic example of things going "wrong".
– Henning Makholm
Sep 3 at 15:33
add a comment |Â
up vote
2
down vote
This really depends on what type of function and derivative you are thinking of. I assume you are thinking of the usual definition of the total derivative of a function between real vector spaces. In this case, your intuition is correct!
Consider some function f defined on some ball B(0,R) of radius R sitting inside Real n-space. Since any derivative( given it exists ) at a point p
is a measure of change in the output of f coming from an infinitesimal change in the input around p, you have to be able to change the input approaching p from the direction you are measuring the change.
Suppose p is on the boundary of the ball. At p it makes sense to talk about directional derivatives in directions pointing into the ball, but not out of it. However since there are directions in which we cannot change the input ( or cannot approach p from ) we cannot express the rate of change at p in all directions, which is the data of the total derivative.
Another way of thinking about total derivatives is as a linearization( I.e. linear map )of the function, which approximates f around p, thus one needs to have a “tiny vector space†i.e an open ball, consisting of tangent vectors around p in order for f to have a chance to exist.
I hope that this comment was useful.
answered Aug 26 at 20:47
user587399
1015
1
Actually, directional derivatives (ordinarily) are interpreted as two-sided, not single-sided, so when you compute a directional derivative of the direction of $vec v$ pointing into the ball, the definition still requires that you look at $f(p+tvec v)$ for $t<0$, and there you're out of luck.
– Ted Shifrin
Aug 26 at 22:59
1
Thank you for clarifying! My overview of the literature is still lacking, so I am thankful that you took the time to improve my answer.
– user587399
Aug 27 at 12:26
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There's no intrinsic reason not to define a derivative of a function defined on a closed ball, at a point on the boundary on that ball.
In more detail, if $B = xinmathbb R^n: $ and $f:Bto mathbb R^m$, then for any point $x_0$ domain of $f$ we can define that the derivative of $f$ at $x_0$ means a linear mapping $T$ such that
$$ f(x) = f(x_0) + T(x-x_0) + o(x-x_0) $$
for $xto x_0$ and $xin B$.
Even at the boundary points of $B$, there are neighborhoods of the ball that are rich enough that the derivative is unique if it exists. (All that's needed for this is that every neighborhood of $x_0$ relative to $B$ contains points whose difference from $x_0$ span $mathbb R^n$).
The reason why textbooks often don't do this is that most of what they want to say can be said quite fine while restricting our attention to open domains, and doing so is quite a bit simpler -- there are (literally!) fewer edge cases that need to be considered when stating the results.
For one example of one such corner case, consider the function
$$ f(x,y) = begincases 0 & textif y < x \
0 & textif y > 4x \
0 & textif x = y = 0 \
1 & textif 2x < y < 3x \
textundefined & textotherwise endcases $$
whose partial derivatives exist and are continuous (indeed identically $0$) everywhere in the domain of $f$, but $f$ is not even continuous at $(0,0)$.
answered Aug 26 at 21:20
Henning Makholm
230k16296527
But if the domain of the function $f$ is not open we can not speak about partial derivatives of it ..... correct? if so why?
– Idonotknow
Sep 3 at 14:54
1
@Idonotknow, correct: Partial deriviatives need more than I describe here. A most points on the boundary of a closed ball we could speak of partial derivitives well enough, but there are some points that are exceptions -- for example, for $f$ defined on the closed unit disk, $partial f/partial x$ would not make sense at $(0,1)$.
– Henning Makholm
Sep 3 at 14:59
why partial derivative at that point would not make sense?
– Idonotknow
Sep 3 at 15:14
1
@Idonotknow: Because $xmapsto f(x,1)$ is only defined at $x=0$, and even with the definition above you can't meaningfully take a derivative at an isolated point. See new addition to answer for an even more dramatic example of things going "wrong".
– Henning Makholm
Sep 3 at 15:33
add a comment |Â
up vote
2
down vote
accepted
There's no intrinsic reason not to define a derivative of a function defined on a closed ball, at a point on the boundary on that ball.
In more detail, if $B = xinmathbb R^n: $ and $f:Bto mathbb R^m$, then for any point $x_0$ domain of $f$ we can define that the derivative of $f$ at $x_0$ means a linear mapping $T$ such that
$$ f(x) = f(x_0) + T(x-x_0) + o(x-x_0) $$
for $xto x_0$ and $xin B$.
Even at the boundary points of $B$, there are neighborhoods of the ball that are rich enough that the derivative is unique if it exists. (All that's needed for this is that every neighborhood of $x_0$ relative to $B$ contains points whose difference from $x_0$ span $mathbb R^n$).
The reason why textbooks often don't do this is that most of what they want to say can be said quite fine while restricting our attention to open domains, and doing so is quite a bit simpler -- there are (literally!) fewer edge cases that need to be considered when stating the results.
For one example of one such corner case, consider the function
$$ f(x,y) = begincases 0 & textif y < x \
0 & textif y > 4x \
0 & textif x = y = 0 \
1 & textif 2x < y < 3x \
textundefined & textotherwise endcases $$
whose partial derivatives exist and are continuous (indeed identically $0$) everywhere in the domain of $f$, but $f$ is not even continuous at $(0,0)$.
answered Aug 26 at 21:20
Henning Makholm
230k16296527
But if the domain of the function $f$ is not open we can not speak about partial derivatives of it ..... correct? if so why?
– Idonotknow
Sep 3 at 14:54
1
@Idonotknow, correct: Partial deriviatives need more than I describe here. A most points on the boundary of a closed ball we could speak of partial derivitives well enough, but there are some points that are exceptions -- for example, for $f$ defined on the closed unit disk, $partial f/partial x$ would not make sense at $(0,1)$.
– Henning Makholm
Sep 3 at 14:59
why partial derivative at that point would not make sense?
– Idonotknow
Sep 3 at 15:14
1
@Idonotknow: Because $xmapsto f(x,1)$ is only defined at $x=0$, and even with the definition above you can't meaningfully take a derivative at an isolated point. See new addition to answer for an even more dramatic example of things going "wrong".
– Henning Makholm
Sep 3 at 15:33
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There's no intrinsic reason not to define a derivative of a function defined on a closed ball, at a point on the boundary on that ball.
In more detail, if $B = xinmathbb R^n: $ and $f:Bto mathbb R^m$, then for any point $x_0$ domain of $f$ we can define that the derivative of $f$ at $x_0$ means a linear mapping $T$ such that
$$ f(x) = f(x_0) + T(x-x_0) + o(x-x_0) $$
for $xto x_0$ and $xin B$.
Even at the boundary points of $B$, there are neighborhoods of the ball that are rich enough that the derivative is unique if it exists. (All that's needed for this is that every neighborhood of $x_0$ relative to $B$ contains points whose difference from $x_0$ span $mathbb R^n$).
The reason why textbooks often don't do this is that most of what they want to say can be said quite fine while restricting our attention to open domains, and doing so is quite a bit simpler -- there are (literally!) fewer edge cases that need to be considered when stating the results.
For one example of one such corner case, consider the function
$$ f(x,y) = begincases 0 & textif y < x \
0 & textif y > 4x \
0 & textif x = y = 0 \
1 & textif 2x < y < 3x \
textundefined & textotherwise endcases $$
whose partial derivatives exist and are continuous (indeed identically $0$) everywhere in the domain of $f$, but $f$ is not even continuous at $(0,0)$.
answered Aug 26 at 21:20
Henning Makholm
230k16296527
There's no intrinsic reason not to define a derivative of a function defined on a closed ball, at a point on the boundary on that ball.
In more detail, if $B = xinmathbb R^n: $ and $f:Bto mathbb R^m$, then for any point $x_0$ domain of $f$ we can define that the derivative of $f$ at $x_0$ means a linear mapping $T$ such that
$$ f(x) = f(x_0) + T(x-x_0) + o(x-x_0) $$
for $xto x_0$ and $xin B$.
Even at the boundary points of $B$, there are neighborhoods of the ball that are rich enough that the derivative is unique if it exists. (All that's needed for this is that every neighborhood of $x_0$ relative to $B$ contains points whose difference from $x_0$ span $mathbb R^n$).
The reason why textbooks often don't do this is that most of what they want to say can be said quite fine while restricting our attention to open domains, and doing so is quite a bit simpler -- there are (literally!) fewer edge cases that need to be considered when stating the results.
For one example of one such corner case, consider the function
$$ f(x,y) = begincases 0 & textif y < x \
0 & textif y > 4x \
0 & textif x = y = 0 \
1 & textif 2x < y < 3x \
textundefined & textotherwise endcases $$
whose partial derivatives exist and are continuous (indeed identically $0$) everywhere in the domain of $f$, but $f$ is not even continuous at $(0,0)$.
answered Aug 26 at 21:20
Henning Makholm
230k16296527
edited Sep 3 at 15:34
answered Aug 26 at 21:20
Henning Makholm
230k16296527
answered Aug 26 at 21:20
Henning Makholm
230k16296527
answered Aug 26 at 21:20
Henning Makholm
230k16296527
230k16296527
But if the domain of the function $f$ is not open we can not speak about partial derivatives of it ..... correct? if so why?
– Idonotknow
Sep 3 at 14:54
1
@Idonotknow, correct: Partial deriviatives need more than I describe here. A most points on the boundary of a closed ball we could speak of partial derivitives well enough, but there are some points that are exceptions -- for example, for $f$ defined on the closed unit disk, $partial f/partial x$ would not make sense at $(0,1)$.
– Henning Makholm
Sep 3 at 14:59
why partial derivative at that point would not make sense?
– Idonotknow
Sep 3 at 15:14
1
@Idonotknow: Because $xmapsto f(x,1)$ is only defined at $x=0$, and even with the definition above you can't meaningfully take a derivative at an isolated point. See new addition to answer for an even more dramatic example of things going "wrong".
– Henning Makholm
Sep 3 at 15:33
add a comment |Â
But if the domain of the function $f$ is not open we can not speak about partial derivatives of it ..... correct? if so why?
– Idonotknow
Sep 3 at 14:54
1
@Idonotknow, correct: Partial deriviatives need more than I describe here. A most points on the boundary of a closed ball we could speak of partial derivitives well enough, but there are some points that are exceptions -- for example, for $f$ defined on the closed unit disk, $partial f/partial x$ would not make sense at $(0,1)$.
– Henning Makholm
Sep 3 at 14:59
why partial derivative at that point would not make sense?
– Idonotknow
Sep 3 at 15:14
1
@Idonotknow: Because $xmapsto f(x,1)$ is only defined at $x=0$, and even with the definition above you can't meaningfully take a derivative at an isolated point. See new addition to answer for an even more dramatic example of things going "wrong".
– Henning Makholm
Sep 3 at 15:33
But if the domain of the function $f$ is not open we can not speak about partial derivatives of it ..... correct? if so why?
– Idonotknow
Sep 3 at 14:54
But if the domain of the function $f$ is not open we can not speak about partial derivatives of it ..... correct? if so why?
– Idonotknow
Sep 3 at 14:54
1
1
@Idonotknow, correct: Partial deriviatives need more than I describe here. A most points on the boundary of a closed ball we could speak of partial derivitives well enough, but there are some points that are exceptions -- for example, for $f$ defined on the closed unit disk, $partial f/partial x$ would not make sense at $(0,1)$.
– Henning Makholm
Sep 3 at 14:59
@Idonotknow, correct: Partial deriviatives need more than I describe here. A most points on the boundary of a closed ball we could speak of partial derivitives well enough, but there are some points that are exceptions -- for example, for $f$ defined on the closed unit disk, $partial f/partial x$ would not make sense at $(0,1)$.
– Henning Makholm
Sep 3 at 14:59
why partial derivative at that point would not make sense?
– Idonotknow
Sep 3 at 15:14
why partial derivative at that point would not make sense?
– Idonotknow
Sep 3 at 15:14
1
1
@Idonotknow: Because $xmapsto f(x,1)$ is only defined at $x=0$, and even with the definition above you can't meaningfully take a derivative at an isolated point. See new addition to answer for an even more dramatic example of things going "wrong".
– Henning Makholm
Sep 3 at 15:33
@Idonotknow: Because $xmapsto f(x,1)$ is only defined at $x=0$, and even with the definition above you can't meaningfully take a derivative at an isolated point. See new addition to answer for an even more dramatic example of things going "wrong".
– Henning Makholm
Sep 3 at 15:33
add a comment |Â
up vote
2
down vote
This really depends on what type of function and derivative you are thinking of. I assume you are thinking of the usual definition of the total derivative of a function between real vector spaces. In this case, your intuition is correct!
Consider some function f defined on some ball B(0,R) of radius R sitting inside Real n-space. Since any derivative( given it exists ) at a point p
is a measure of change in the output of f coming from an infinitesimal change in the input around p, you have to be able to change the input approaching p from the direction you are measuring the change.
Suppose p is on the boundary of the ball. At p it makes sense to talk about directional derivatives in directions pointing into the ball, but not out of it. However since there are directions in which we cannot change the input ( or cannot approach p from ) we cannot express the rate of change at p in all directions, which is the data of the total derivative.
Another way of thinking about total derivatives is as a linearization( I.e. linear map )of the function, which approximates f around p, thus one needs to have a “tiny vector space†i.e an open ball, consisting of tangent vectors around p in order for f to have a chance to exist.
I hope that this comment was useful.
answered Aug 26 at 20:47
user587399
1015
1
Actually, directional derivatives (ordinarily) are interpreted as two-sided, not single-sided, so when you compute a directional derivative of the direction of $vec v$ pointing into the ball, the definition still requires that you look at $f(p+tvec v)$ for $t<0$, and there you're out of luck.
– Ted Shifrin
Aug 26 at 22:59
1
Thank you for clarifying! My overview of the literature is still lacking, so I am thankful that you took the time to improve my answer.
– user587399
Aug 27 at 12:26
add a comment |Â
up vote
2
down vote
This really depends on what type of function and derivative you are thinking of. I assume you are thinking of the usual definition of the total derivative of a function between real vector spaces. In this case, your intuition is correct!
Consider some function f defined on some ball B(0,R) of radius R sitting inside Real n-space. Since any derivative( given it exists ) at a point p
is a measure of change in the output of f coming from an infinitesimal change in the input around p, you have to be able to change the input approaching p from the direction you are measuring the change.
Suppose p is on the boundary of the ball. At p it makes sense to talk about directional derivatives in directions pointing into the ball, but not out of it. However since there are directions in which we cannot change the input ( or cannot approach p from ) we cannot express the rate of change at p in all directions, which is the data of the total derivative.
Another way of thinking about total derivatives is as a linearization( I.e. linear map )of the function, which approximates f around p, thus one needs to have a “tiny vector space†i.e an open ball, consisting of tangent vectors around p in order for f to have a chance to exist.
I hope that this comment was useful.
answered Aug 26 at 20:47
user587399
1015
1
Actually, directional derivatives (ordinarily) are interpreted as two-sided, not single-sided, so when you compute a directional derivative of the direction of $vec v$ pointing into the ball, the definition still requires that you look at $f(p+tvec v)$ for $t<0$, and there you're out of luck.
– Ted Shifrin
Aug 26 at 22:59
1
Thank you for clarifying! My overview of the literature is still lacking, so I am thankful that you took the time to improve my answer.
– user587399
Aug 27 at 12:26
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This really depends on what type of function and derivative you are thinking of. I assume you are thinking of the usual definition of the total derivative of a function between real vector spaces. In this case, your intuition is correct!
Consider some function f defined on some ball B(0,R) of radius R sitting inside Real n-space. Since any derivative( given it exists ) at a point p
is a measure of change in the output of f coming from an infinitesimal change in the input around p, you have to be able to change the input approaching p from the direction you are measuring the change.
Suppose p is on the boundary of the ball. At p it makes sense to talk about directional derivatives in directions pointing into the ball, but not out of it. However since there are directions in which we cannot change the input ( or cannot approach p from ) we cannot express the rate of change at p in all directions, which is the data of the total derivative.
Another way of thinking about total derivatives is as a linearization( I.e. linear map )of the function, which approximates f around p, thus one needs to have a “tiny vector space†i.e an open ball, consisting of tangent vectors around p in order for f to have a chance to exist.
I hope that this comment was useful.
answered Aug 26 at 20:47
user587399
1015
This really depends on what type of function and derivative you are thinking of. I assume you are thinking of the usual definition of the total derivative of a function between real vector spaces. In this case, your intuition is correct!
Consider some function f defined on some ball B(0,R) of radius R sitting inside Real n-space. Since any derivative( given it exists ) at a point p
is a measure of change in the output of f coming from an infinitesimal change in the input around p, you have to be able to change the input approaching p from the direction you are measuring the change.
Suppose p is on the boundary of the ball. At p it makes sense to talk about directional derivatives in directions pointing into the ball, but not out of it. However since there are directions in which we cannot change the input ( or cannot approach p from ) we cannot express the rate of change at p in all directions, which is the data of the total derivative.
Another way of thinking about total derivatives is as a linearization( I.e. linear map )of the function, which approximates f around p, thus one needs to have a “tiny vector space†i.e an open ball, consisting of tangent vectors around p in order for f to have a chance to exist.
I hope that this comment was useful.
answered Aug 26 at 20:47
user587399
1015
answered Aug 26 at 20:47
user587399
1015
answered Aug 26 at 20:47
user587399
1015
answered Aug 26 at 20:47
user587399
1015
1015
1
Actually, directional derivatives (ordinarily) are interpreted as two-sided, not single-sided, so when you compute a directional derivative of the direction of $vec v$ pointing into the ball, the definition still requires that you look at $f(p+tvec v)$ for $t<0$, and there you're out of luck.
– Ted Shifrin
Aug 26 at 22:59
1
Thank you for clarifying! My overview of the literature is still lacking, so I am thankful that you took the time to improve my answer.
– user587399
Aug 27 at 12:26
add a comment |Â
1
Actually, directional derivatives (ordinarily) are interpreted as two-sided, not single-sided, so when you compute a directional derivative of the direction of $vec v$ pointing into the ball, the definition still requires that you look at $f(p+tvec v)$ for $t<0$, and there you're out of luck.
– Ted Shifrin
Aug 26 at 22:59
1
Thank you for clarifying! My overview of the literature is still lacking, so I am thankful that you took the time to improve my answer.
– user587399
Aug 27 at 12:26
1
1
Actually, directional derivatives (ordinarily) are interpreted as two-sided, not single-sided, so when you compute a directional derivative of the direction of $vec v$ pointing into the ball, the definition still requires that you look at $f(p+tvec v)$ for $t<0$, and there you're out of luck.
– Ted Shifrin
Aug 26 at 22:59
Actually, directional derivatives (ordinarily) are interpreted as two-sided, not single-sided, so when you compute a directional derivative of the direction of $vec v$ pointing into the ball, the definition still requires that you look at $f(p+tvec v)$ for $t<0$, and there you're out of luck.
– Ted Shifrin
Aug 26 at 22:59
1
1
Thank you for clarifying! My overview of the literature is still lacking, so I am thankful that you took the time to improve my answer.
– user587399
Aug 27 at 12:26
Thank you for clarifying! My overview of the literature is still lacking, so I am thankful that you took the time to improve my answer.
– user587399
Aug 27 at 12:26
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You are correct. Consider a point on the boundary of the ball. As in the closed interval, we cannot take the derivative in each direction.
– John Douma
Aug 26 at 20:13
@JohnDouma why? and in the case of the ball there are several directions not justleft and right ..... sohow we differentiate a closed ball?
– Idonotknow
Aug 26 at 20:31
1
The function is not defined there. We are on the boundary of the ball and the function is only defined on the ball. Imagine taking a ball around this boundary point.
– John Douma
Aug 26 at 20:32