Mixing
Why tangent surface is a plane
$begingroup$
Given function $f(vec x) in mathbb R, vec x in mathbb R^n $, We know that for all unit direction $hat u$:
$$
partial_{hat u} f = hat u cdot nabla f
$$
Consider function $f$:
$$
f(left<x,y,zright>) = z - min(lvert x rvert, lvert y rvert)timesfrac{lvert y rvert}{y}
$$
At the origin, along direction $hat u=left<frac{1}{sqrt 3}, frac{1}{sqrt 3}, frac{1}{sqrt 3} right>$:
$$
partial_{hat u} f(vec 0) = lim_{t to 0} frac{f(t hat u) - f(vec 0)}{t} = lim_{t to 0} frac{left(frac{1}{sqrt 3} - minleft(frac{1}{sqrt 3},frac{1}{sqrt 3}right)times 1right) - 0}{t} = lim_{t to 0} frac{0 - 0}{t} = 0\
hat u cdot nabla f(vec 0) = hat u cdot left<f_x,f_y,f_z right> = left<frac{1}{sqrt 3}, frac{1}{sqrt 3}, frac{1}{sqrt 3} right> cdot left<0,0,1right> = frac{1}{sqrt 3}\
partial_{hat u} f(vec 0) ne hat u cdot nabla f(vec 0)
$$
So there must be other requirements for:
$$
partial_{hat u} f = hat u cdot nabla f
$$
From the graph of the function $f$, we can see that even the directional derivative is defined for all direction at the origin, the tangent surface at the origin is not a plane.
Here are my questions:
- does $nabla f(vec 0)$ exist ?
- what's the requirement for $partial_{hat u} f = hat u cdot nabla f$
Edit: Proof that $f$ is not differentiable at the origin
For $f$ to be differentiable, there is a linear transformation $T$ such that:
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T_a(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
Since $mathbf h$ can approach to $mathbf 0$ follow any trajectory, assume $mathbf h$ is following direction $hat u$, $mathbf h = t hat u$ and $lVert hat u rVert = 1$, then we have:
$$
lim_{t to 0}
frac{f(mathbf a+t hat u)-f(mathbf a)-T_a(t hat u)}{t} = 0\
lim_{t to 0}frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = lim_{t to 0}frac{T_a(t hat u)}{t}\
$$
To the left we have:
$$
lim_{t to 0}frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = partial_{hat u} f(a)
$$
To the right, since $T$ is linear, we have:
$$
lim_{t to 0}frac{T_a(t hat u)}{t} = lim_{t to 0}frac{t T_a(hat u)}{t} = T_a(hat u)
$$
So
$$
partial_{hat u} f(a) = T_a(hat u)
$$
$T$ is linear and I have already proofed that $partial_{hat u} f(0)$ is not linear (above in the question). So $f$ is not differentiable at the origin.
What have surprised me is that I thought linearity of $partial_{hat u} f$ is a property of differentiable function, but in reality, it is a requirement. Although, technically there is no difference between property and requirement, they are both $Rightarrow$ in math world.
multivariable-calculus
asked Jan 18 at 11:49
Zang MingJieZang MingJie
1356
$endgroup$
add a comment |
$begingroup$
Given function $f(vec x) in mathbb R, vec x in mathbb R^n $, We know that for all unit direction $hat u$:
$$
partial_{hat u} f = hat u cdot nabla f
$$
Consider function $f$:
$$
f(left<x,y,zright>) = z - min(lvert x rvert, lvert y rvert)timesfrac{lvert y rvert}{y}
$$
At the origin, along direction $hat u=left<frac{1}{sqrt 3}, frac{1}{sqrt 3}, frac{1}{sqrt 3} right>$:
$$
partial_{hat u} f(vec 0) = lim_{t to 0} frac{f(t hat u) - f(vec 0)}{t} = lim_{t to 0} frac{left(frac{1}{sqrt 3} - minleft(frac{1}{sqrt 3},frac{1}{sqrt 3}right)times 1right) - 0}{t} = lim_{t to 0} frac{0 - 0}{t} = 0\
hat u cdot nabla f(vec 0) = hat u cdot left<f_x,f_y,f_z right> = left<frac{1}{sqrt 3}, frac{1}{sqrt 3}, frac{1}{sqrt 3} right> cdot left<0,0,1right> = frac{1}{sqrt 3}\
partial_{hat u} f(vec 0) ne hat u cdot nabla f(vec 0)
$$
So there must be other requirements for:
$$
partial_{hat u} f = hat u cdot nabla f
$$
From the graph of the function $f$, we can see that even the directional derivative is defined for all direction at the origin, the tangent surface at the origin is not a plane.
Here are my questions:
- does $nabla f(vec 0)$ exist ?
- what's the requirement for $partial_{hat u} f = hat u cdot nabla f$
Edit: Proof that $f$ is not differentiable at the origin
For $f$ to be differentiable, there is a linear transformation $T$ such that:
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T_a(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
Since $mathbf h$ can approach to $mathbf 0$ follow any trajectory, assume $mathbf h$ is following direction $hat u$, $mathbf h = t hat u$ and $lVert hat u rVert = 1$, then we have:
$$
lim_{t to 0}
frac{f(mathbf a+t hat u)-f(mathbf a)-T_a(t hat u)}{t} = 0\
lim_{t to 0}frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = lim_{t to 0}frac{T_a(t hat u)}{t}\
$$
To the left we have:
$$
lim_{t to 0}frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = partial_{hat u} f(a)
$$
To the right, since $T$ is linear, we have:
$$
lim_{t to 0}frac{T_a(t hat u)}{t} = lim_{t to 0}frac{t T_a(hat u)}{t} = T_a(hat u)
$$
So
$$
partial_{hat u} f(a) = T_a(hat u)
$$
$T$ is linear and I have already proofed that $partial_{hat u} f(0)$ is not linear (above in the question). So $f$ is not differentiable at the origin.
What have surprised me is that I thought linearity of $partial_{hat u} f$ is a property of differentiable function, but in reality, it is a requirement. Although, technically there is no difference between property and requirement, they are both $Rightarrow$ in math world.
multivariable-calculus
asked Jan 18 at 11:49
Zang MingJieZang MingJie
1356
$endgroup$
add a comment |
$begingroup$
Given function $f(vec x) in mathbb R, vec x in mathbb R^n $, We know that for all unit direction $hat u$:
$$
partial_{hat u} f = hat u cdot nabla f
$$
Consider function $f$:
$$
f(left<x,y,zright>) = z - min(lvert x rvert, lvert y rvert)timesfrac{lvert y rvert}{y}
$$
At the origin, along direction $hat u=left<frac{1}{sqrt 3}, frac{1}{sqrt 3}, frac{1}{sqrt 3} right>$:
$$
partial_{hat u} f(vec 0) = lim_{t to 0} frac{f(t hat u) - f(vec 0)}{t} = lim_{t to 0} frac{left(frac{1}{sqrt 3} - minleft(frac{1}{sqrt 3},frac{1}{sqrt 3}right)times 1right) - 0}{t} = lim_{t to 0} frac{0 - 0}{t} = 0\
hat u cdot nabla f(vec 0) = hat u cdot left<f_x,f_y,f_z right> = left<frac{1}{sqrt 3}, frac{1}{sqrt 3}, frac{1}{sqrt 3} right> cdot left<0,0,1right> = frac{1}{sqrt 3}\
partial_{hat u} f(vec 0) ne hat u cdot nabla f(vec 0)
$$
So there must be other requirements for:
$$
partial_{hat u} f = hat u cdot nabla f
$$
From the graph of the function $f$, we can see that even the directional derivative is defined for all direction at the origin, the tangent surface at the origin is not a plane.
Here are my questions:
- does $nabla f(vec 0)$ exist ?
- what's the requirement for $partial_{hat u} f = hat u cdot nabla f$
Edit: Proof that $f$ is not differentiable at the origin
For $f$ to be differentiable, there is a linear transformation $T$ such that:
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T_a(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
Since $mathbf h$ can approach to $mathbf 0$ follow any trajectory, assume $mathbf h$ is following direction $hat u$, $mathbf h = t hat u$ and $lVert hat u rVert = 1$, then we have:
$$
lim_{t to 0}
frac{f(mathbf a+t hat u)-f(mathbf a)-T_a(t hat u)}{t} = 0\
lim_{t to 0}frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = lim_{t to 0}frac{T_a(t hat u)}{t}\
$$
To the left we have:
$$
lim_{t to 0}frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = partial_{hat u} f(a)
$$
To the right, since $T$ is linear, we have:
$$
lim_{t to 0}frac{T_a(t hat u)}{t} = lim_{t to 0}frac{t T_a(hat u)}{t} = T_a(hat u)
$$
So
$$
partial_{hat u} f(a) = T_a(hat u)
$$
$T$ is linear and I have already proofed that $partial_{hat u} f(0)$ is not linear (above in the question). So $f$ is not differentiable at the origin.
What have surprised me is that I thought linearity of $partial_{hat u} f$ is a property of differentiable function, but in reality, it is a requirement. Although, technically there is no difference between property and requirement, they are both $Rightarrow$ in math world.
multivariable-calculus
asked Jan 18 at 11:49
Zang MingJieZang MingJie
1356
$endgroup$
Given function $f(vec x) in mathbb R, vec x in mathbb R^n $, We know that for all unit direction $hat u$:
$$
partial_{hat u} f = hat u cdot nabla f
$$
Consider function $f$:
$$
f(left<x,y,zright>) = z - min(lvert x rvert, lvert y rvert)timesfrac{lvert y rvert}{y}
$$
At the origin, along direction $hat u=left<frac{1}{sqrt 3}, frac{1}{sqrt 3}, frac{1}{sqrt 3} right>$:
$$
partial_{hat u} f(vec 0) = lim_{t to 0} frac{f(t hat u) - f(vec 0)}{t} = lim_{t to 0} frac{left(frac{1}{sqrt 3} - minleft(frac{1}{sqrt 3},frac{1}{sqrt 3}right)times 1right) - 0}{t} = lim_{t to 0} frac{0 - 0}{t} = 0\
hat u cdot nabla f(vec 0) = hat u cdot left<f_x,f_y,f_z right> = left<frac{1}{sqrt 3}, frac{1}{sqrt 3}, frac{1}{sqrt 3} right> cdot left<0,0,1right> = frac{1}{sqrt 3}\
partial_{hat u} f(vec 0) ne hat u cdot nabla f(vec 0)
$$
So there must be other requirements for:
$$
partial_{hat u} f = hat u cdot nabla f
$$
From the graph of the function $f$, we can see that even the directional derivative is defined for all direction at the origin, the tangent surface at the origin is not a plane.
Here are my questions:
- does $nabla f(vec 0)$ exist ?
- what's the requirement for $partial_{hat u} f = hat u cdot nabla f$
Edit: Proof that $f$ is not differentiable at the origin
For $f$ to be differentiable, there is a linear transformation $T$ such that:
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T_a(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
Since $mathbf h$ can approach to $mathbf 0$ follow any trajectory, assume $mathbf h$ is following direction $hat u$, $mathbf h = t hat u$ and $lVert hat u rVert = 1$, then we have:
$$
lim_{t to 0}
frac{f(mathbf a+t hat u)-f(mathbf a)-T_a(t hat u)}{t} = 0\
lim_{t to 0}frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = lim_{t to 0}frac{T_a(t hat u)}{t}\
$$
To the left we have:
$$
lim_{t to 0}frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = partial_{hat u} f(a)
$$
To the right, since $T$ is linear, we have:
$$
lim_{t to 0}frac{T_a(t hat u)}{t} = lim_{t to 0}frac{t T_a(hat u)}{t} = T_a(hat u)
$$
So
$$
partial_{hat u} f(a) = T_a(hat u)
$$
$T$ is linear and I have already proofed that $partial_{hat u} f(0)$ is not linear (above in the question). So $f$ is not differentiable at the origin.
What have surprised me is that I thought linearity of $partial_{hat u} f$ is a property of differentiable function, but in reality, it is a requirement. Although, technically there is no difference between property and requirement, they are both $Rightarrow$ in math world.
multivariable-calculus
multivariable-calculus
asked Jan 18 at 11:49
Zang MingJieZang MingJie
1356
asked Jan 18 at 11:49
Zang MingJieZang MingJie
1356
edited Jan 19 at 20:05
Zang MingJie
asked Jan 18 at 11:49
Zang MingJieZang MingJie
1356
asked Jan 18 at 11:49
Zang MingJieZang MingJie
1356
asked Jan 18 at 11:49
Zang MingJieZang MingJie
1356
1356
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In order to ensure that $partial_{hat u} f = hat u cdot nabla f$
at a given point $mathbf a in mathbb R^n,$
you can simply require that the function $f$ is differentiable at $mathbf a.$
There may be weaker conditions that ensure that $partial_{hat u} f = hat u cdot nabla f$, but I do not know what they are. Differentiability seems to be the one usually used.
Of course you must also define what it means for a function over $mathbb R^n$
to be differentiable at $mathbf a.$
Here is a typical definition:
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^n,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
A good definition of the gradient $nabla f$, in turn, would require that $f$ be differentiable at every point where $nabla f$ is defined.
Your function does not have such a derivative at $mathbf 0.$
There is more that can be said about this, but that is the fundamental issue.
answered Jan 19 at 15:03
David KDavid K
54.6k343120
$endgroup$
$begingroup$
Thanks, I have written a proof that $f$ is not differentiable at the origin in my question. And for linearity of partial derivative, I have an already answered question here math.stackexchange.com/questions/3079820
$endgroup$
– Zang MingJie
Jan 19 at 21:35
add a comment |
$begingroup$
You are taking a derivative where your function doesn't have one. $f(x) = |x|$ does not have a derivative at 0. However it is convex and has the subdifferential $[-1,1]$ at $0$.
Edit:
Your partial derivative wrt $x$ is not smooth. Consider $|y|>0$, then the function $g(x) = max(|x|,|y|)frac{|y|}{y}$ does not have a derivative at $x=0$. So even if your function has partial derivatives at $(0,0,0)$, the partial derivative wrt $x$ is not continuous.
answered Jan 18 at 11:56
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
My $f$ is already carefully chosen that the derivative exist for all directions. you can try derivative it using limitation.
$endgroup$
– Zang MingJie
Jan 18 at 12:02
$begingroup$
@Zang MingJie It is discontinuous along $x=0$ and also along $y=0$. Verify that the left and right derivatives there do not match.
$endgroup$
– lightxbulb
Jan 18 at 12:15
$begingroup$
@ZangMingJie having a partial derivative in all directions does not necessarily make your function differentiable.
$endgroup$
– Randall
Jan 18 at 12:18
$begingroup$
@lightxbulb $f vert_{x=0} = z - min(0, lvert y rvert)times frac{lvert y rvert}{y} = z$ can you explain why it is discontinuous.
$endgroup$
– Zang MingJie
Jan 18 at 12:33
$begingroup$
@Randall this maybe the answer I need, I want to know exactly the requirements for a function be differentiable. Any why my $f$ is not differentiable.
$endgroup$
– Zang MingJie
Jan 18 at 12:35
|
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2 Answers
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2 Answers
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$begingroup$
In order to ensure that $partial_{hat u} f = hat u cdot nabla f$
at a given point $mathbf a in mathbb R^n,$
you can simply require that the function $f$ is differentiable at $mathbf a.$
There may be weaker conditions that ensure that $partial_{hat u} f = hat u cdot nabla f$, but I do not know what they are. Differentiability seems to be the one usually used.
Of course you must also define what it means for a function over $mathbb R^n$
to be differentiable at $mathbf a.$
Here is a typical definition:
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^n,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
A good definition of the gradient $nabla f$, in turn, would require that $f$ be differentiable at every point where $nabla f$ is defined.
Your function does not have such a derivative at $mathbf 0.$
There is more that can be said about this, but that is the fundamental issue.
answered Jan 19 at 15:03
David KDavid K
54.6k343120
$endgroup$
$begingroup$
Thanks, I have written a proof that $f$ is not differentiable at the origin in my question. And for linearity of partial derivative, I have an already answered question here math.stackexchange.com/questions/3079820
$endgroup$
– Zang MingJie
Jan 19 at 21:35
add a comment |
$begingroup$
In order to ensure that $partial_{hat u} f = hat u cdot nabla f$
at a given point $mathbf a in mathbb R^n,$
you can simply require that the function $f$ is differentiable at $mathbf a.$
There may be weaker conditions that ensure that $partial_{hat u} f = hat u cdot nabla f$, but I do not know what they are. Differentiability seems to be the one usually used.
Of course you must also define what it means for a function over $mathbb R^n$
to be differentiable at $mathbf a.$
Here is a typical definition:
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^n,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
A good definition of the gradient $nabla f$, in turn, would require that $f$ be differentiable at every point where $nabla f$ is defined.
Your function does not have such a derivative at $mathbf 0.$
There is more that can be said about this, but that is the fundamental issue.
answered Jan 19 at 15:03
David KDavid K
54.6k343120
$endgroup$
$begingroup$
Thanks, I have written a proof that $f$ is not differentiable at the origin in my question. And for linearity of partial derivative, I have an already answered question here math.stackexchange.com/questions/3079820
$endgroup$
– Zang MingJie
Jan 19 at 21:35
add a comment |
$begingroup$
In order to ensure that $partial_{hat u} f = hat u cdot nabla f$
at a given point $mathbf a in mathbb R^n,$
you can simply require that the function $f$ is differentiable at $mathbf a.$
There may be weaker conditions that ensure that $partial_{hat u} f = hat u cdot nabla f$, but I do not know what they are. Differentiability seems to be the one usually used.
Of course you must also define what it means for a function over $mathbb R^n$
to be differentiable at $mathbf a.$
Here is a typical definition:
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^n,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
A good definition of the gradient $nabla f$, in turn, would require that $f$ be differentiable at every point where $nabla f$ is defined.
Your function does not have such a derivative at $mathbf 0.$
There is more that can be said about this, but that is the fundamental issue.
answered Jan 19 at 15:03
David KDavid K
54.6k343120
$endgroup$
In order to ensure that $partial_{hat u} f = hat u cdot nabla f$
at a given point $mathbf a in mathbb R^n,$
you can simply require that the function $f$ is differentiable at $mathbf a.$
There may be weaker conditions that ensure that $partial_{hat u} f = hat u cdot nabla f$, but I do not know what they are. Differentiability seems to be the one usually used.
Of course you must also define what it means for a function over $mathbb R^n$
to be differentiable at $mathbf a.$
Here is a typical definition:
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^n,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
A good definition of the gradient $nabla f$, in turn, would require that $f$ be differentiable at every point where $nabla f$ is defined.
Your function does not have such a derivative at $mathbf 0.$
There is more that can be said about this, but that is the fundamental issue.
answered Jan 19 at 15:03
David KDavid K
54.6k343120
answered Jan 19 at 15:03
David KDavid K
54.6k343120
answered Jan 19 at 15:03
David KDavid K
54.6k343120
answered Jan 19 at 15:03
David KDavid K
54.6k343120
54.6k343120
$begingroup$
Thanks, I have written a proof that $f$ is not differentiable at the origin in my question. And for linearity of partial derivative, I have an already answered question here math.stackexchange.com/questions/3079820
$endgroup$
– Zang MingJie
Jan 19 at 21:35
add a comment |
$begingroup$
Thanks, I have written a proof that $f$ is not differentiable at the origin in my question. And for linearity of partial derivative, I have an already answered question here math.stackexchange.com/questions/3079820
$endgroup$
– Zang MingJie
Jan 19 at 21:35
$begingroup$
Thanks, I have written a proof that $f$ is not differentiable at the origin in my question. And for linearity of partial derivative, I have an already answered question here math.stackexchange.com/questions/3079820
$endgroup$
– Zang MingJie
Jan 19 at 21:35
$begingroup$
Thanks, I have written a proof that $f$ is not differentiable at the origin in my question. And for linearity of partial derivative, I have an already answered question here math.stackexchange.com/questions/3079820
$endgroup$
– Zang MingJie
Jan 19 at 21:35
add a comment |
$begingroup$
You are taking a derivative where your function doesn't have one. $f(x) = |x|$ does not have a derivative at 0. However it is convex and has the subdifferential $[-1,1]$ at $0$.
Edit:
Your partial derivative wrt $x$ is not smooth. Consider $|y|>0$, then the function $g(x) = max(|x|,|y|)frac{|y|}{y}$ does not have a derivative at $x=0$. So even if your function has partial derivatives at $(0,0,0)$, the partial derivative wrt $x$ is not continuous.
answered Jan 18 at 11:56
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
My $f$ is already carefully chosen that the derivative exist for all directions. you can try derivative it using limitation.
$endgroup$
– Zang MingJie
Jan 18 at 12:02
$begingroup$
@Zang MingJie It is discontinuous along $x=0$ and also along $y=0$. Verify that the left and right derivatives there do not match.
$endgroup$
– lightxbulb
Jan 18 at 12:15
$begingroup$
@ZangMingJie having a partial derivative in all directions does not necessarily make your function differentiable.
$endgroup$
– Randall
Jan 18 at 12:18
$begingroup$
@lightxbulb $f vert_{x=0} = z - min(0, lvert y rvert)times frac{lvert y rvert}{y} = z$ can you explain why it is discontinuous.
$endgroup$
– Zang MingJie
Jan 18 at 12:33
$begingroup$
@Randall this maybe the answer I need, I want to know exactly the requirements for a function be differentiable. Any why my $f$ is not differentiable.
$endgroup$
– Zang MingJie
Jan 18 at 12:35
|
show 3 more comments
$begingroup$
You are taking a derivative where your function doesn't have one. $f(x) = |x|$ does not have a derivative at 0. However it is convex and has the subdifferential $[-1,1]$ at $0$.
Edit:
Your partial derivative wrt $x$ is not smooth. Consider $|y|>0$, then the function $g(x) = max(|x|,|y|)frac{|y|}{y}$ does not have a derivative at $x=0$. So even if your function has partial derivatives at $(0,0,0)$, the partial derivative wrt $x$ is not continuous.
answered Jan 18 at 11:56
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
My $f$ is already carefully chosen that the derivative exist for all directions. you can try derivative it using limitation.
$endgroup$
– Zang MingJie
Jan 18 at 12:02
$begingroup$
@Zang MingJie It is discontinuous along $x=0$ and also along $y=0$. Verify that the left and right derivatives there do not match.
$endgroup$
– lightxbulb
Jan 18 at 12:15
$begingroup$
@ZangMingJie having a partial derivative in all directions does not necessarily make your function differentiable.
$endgroup$
– Randall
Jan 18 at 12:18
$begingroup$
@lightxbulb $f vert_{x=0} = z - min(0, lvert y rvert)times frac{lvert y rvert}{y} = z$ can you explain why it is discontinuous.
$endgroup$
– Zang MingJie
Jan 18 at 12:33
$begingroup$
@Randall this maybe the answer I need, I want to know exactly the requirements for a function be differentiable. Any why my $f$ is not differentiable.
$endgroup$
– Zang MingJie
Jan 18 at 12:35
|
show 3 more comments
$begingroup$
You are taking a derivative where your function doesn't have one. $f(x) = |x|$ does not have a derivative at 0. However it is convex and has the subdifferential $[-1,1]$ at $0$.
Edit:
Your partial derivative wrt $x$ is not smooth. Consider $|y|>0$, then the function $g(x) = max(|x|,|y|)frac{|y|}{y}$ does not have a derivative at $x=0$. So even if your function has partial derivatives at $(0,0,0)$, the partial derivative wrt $x$ is not continuous.
answered Jan 18 at 11:56
lightxbulblightxbulb
945311
$endgroup$
You are taking a derivative where your function doesn't have one. $f(x) = |x|$ does not have a derivative at 0. However it is convex and has the subdifferential $[-1,1]$ at $0$.
Edit:
Your partial derivative wrt $x$ is not smooth. Consider $|y|>0$, then the function $g(x) = max(|x|,|y|)frac{|y|}{y}$ does not have a derivative at $x=0$. So even if your function has partial derivatives at $(0,0,0)$, the partial derivative wrt $x$ is not continuous.
answered Jan 18 at 11:56
lightxbulblightxbulb
945311
edited Jan 18 at 13:14
answered Jan 18 at 11:56
lightxbulblightxbulb
945311
answered Jan 18 at 11:56
lightxbulblightxbulb
945311
answered Jan 18 at 11:56
lightxbulblightxbulb
945311
945311
$begingroup$
My $f$ is already carefully chosen that the derivative exist for all directions. you can try derivative it using limitation.
$endgroup$
– Zang MingJie
Jan 18 at 12:02
$begingroup$
@Zang MingJie It is discontinuous along $x=0$ and also along $y=0$. Verify that the left and right derivatives there do not match.
$endgroup$
– lightxbulb
Jan 18 at 12:15
$begingroup$
@ZangMingJie having a partial derivative in all directions does not necessarily make your function differentiable.
$endgroup$
– Randall
Jan 18 at 12:18
$begingroup$
@lightxbulb $f vert_{x=0} = z - min(0, lvert y rvert)times frac{lvert y rvert}{y} = z$ can you explain why it is discontinuous.
$endgroup$
– Zang MingJie
Jan 18 at 12:33
$begingroup$
@Randall this maybe the answer I need, I want to know exactly the requirements for a function be differentiable. Any why my $f$ is not differentiable.
$endgroup$
– Zang MingJie
Jan 18 at 12:35
|
show 3 more comments
$begingroup$
My $f$ is already carefully chosen that the derivative exist for all directions. you can try derivative it using limitation.
$endgroup$
– Zang MingJie
Jan 18 at 12:02
$begingroup$
@Zang MingJie It is discontinuous along $x=0$ and also along $y=0$. Verify that the left and right derivatives there do not match.
$endgroup$
– lightxbulb
Jan 18 at 12:15
$begingroup$
@ZangMingJie having a partial derivative in all directions does not necessarily make your function differentiable.
$endgroup$
– Randall
Jan 18 at 12:18
$begingroup$
@lightxbulb $f vert_{x=0} = z - min(0, lvert y rvert)times frac{lvert y rvert}{y} = z$ can you explain why it is discontinuous.
$endgroup$
– Zang MingJie
Jan 18 at 12:33
$begingroup$
@Randall this maybe the answer I need, I want to know exactly the requirements for a function be differentiable. Any why my $f$ is not differentiable.
$endgroup$
– Zang MingJie
Jan 18 at 12:35
$begingroup$
My $f$ is already carefully chosen that the derivative exist for all directions. you can try derivative it using limitation.
$endgroup$
– Zang MingJie
Jan 18 at 12:02
$begingroup$
My $f$ is already carefully chosen that the derivative exist for all directions. you can try derivative it using limitation.
$endgroup$
– Zang MingJie
Jan 18 at 12:02
$begingroup$
@Zang MingJie It is discontinuous along $x=0$ and also along $y=0$. Verify that the left and right derivatives there do not match.
$endgroup$
– lightxbulb
Jan 18 at 12:15
$begingroup$
@Zang MingJie It is discontinuous along $x=0$ and also along $y=0$. Verify that the left and right derivatives there do not match.
$endgroup$
– lightxbulb
Jan 18 at 12:15
$begingroup$
@ZangMingJie having a partial derivative in all directions does not necessarily make your function differentiable.
$endgroup$
– Randall
Jan 18 at 12:18
$begingroup$
@ZangMingJie having a partial derivative in all directions does not necessarily make your function differentiable.
$endgroup$
– Randall
Jan 18 at 12:18
$begingroup$
@lightxbulb $f vert_{x=0} = z - min(0, lvert y rvert)times frac{lvert y rvert}{y} = z$ can you explain why it is discontinuous.
$endgroup$
– Zang MingJie
Jan 18 at 12:33
$begingroup$
@lightxbulb $f vert_{x=0} = z - min(0, lvert y rvert)times frac{lvert y rvert}{y} = z$ can you explain why it is discontinuous.
$endgroup$
– Zang MingJie
Jan 18 at 12:33
$begingroup$
@Randall this maybe the answer I need, I want to know exactly the requirements for a function be differentiable. Any why my $f$ is not differentiable.
$endgroup$
– Zang MingJie
Jan 18 at 12:35
$begingroup$
@Randall this maybe the answer I need, I want to know exactly the requirements for a function be differentiable. Any why my $f$ is not differentiable.
$endgroup$
– Zang MingJie
Jan 18 at 12:35
|
show 3 more comments
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