Mixing
is function differentiable iff directional derivative is linear
$begingroup$
Original definition
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^m,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T_a(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
My definition
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^m,$
if there is a linear transformation $T$ such that for any unit vector $hat u$
$$
lim_{t to 0}
frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = T_a(hat u).
$$
Question: is there any difference between original definition and my definition
In the original definition, $mathbf h$ can approach to $mathbf 0$ by any trajectory; In my definition it can only be approached from certain direction, so my definition is weaker than the original definition. Substitute $mathbf h = t hat u$ to the original definition will get my definition.
If my definition is not true, can somebody provide a counter example ?
One may already noticed that
$$
partial_{hat u} f (a) = T_a(hat u)
$$
Since $T$ is linear, we can assume (here $nabla f$ is just a function, it happens to be equal to the gradient if exist):
$$
T_a(hat u) = hat u cdot nabla f(a)
$$
So my definition can be written as:
$$
partial_{hat u} f (a) = hat u cdot nabla f(a) Leftrightarrow ftext{ is differentiable at }a
$$
multivariable-calculus
asked Jan 19 at 20:57
Zang MingJieZang MingJie
1356
$endgroup$
add a comment |
$begingroup$
Original definition
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^m,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T_a(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
My definition
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^m,$
if there is a linear transformation $T$ such that for any unit vector $hat u$
$$
lim_{t to 0}
frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = T_a(hat u).
$$
Question: is there any difference between original definition and my definition
In the original definition, $mathbf h$ can approach to $mathbf 0$ by any trajectory; In my definition it can only be approached from certain direction, so my definition is weaker than the original definition. Substitute $mathbf h = t hat u$ to the original definition will get my definition.
If my definition is not true, can somebody provide a counter example ?
One may already noticed that
$$
partial_{hat u} f (a) = T_a(hat u)
$$
Since $T$ is linear, we can assume (here $nabla f$ is just a function, it happens to be equal to the gradient if exist):
$$
T_a(hat u) = hat u cdot nabla f(a)
$$
So my definition can be written as:
$$
partial_{hat u} f (a) = hat u cdot nabla f(a) Leftrightarrow ftext{ is differentiable at }a
$$
multivariable-calculus
asked Jan 19 at 20:57
Zang MingJieZang MingJie
1356
$endgroup$
add a comment |
$begingroup$
Original definition
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^m,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T_a(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
My definition
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^m,$
if there is a linear transformation $T$ such that for any unit vector $hat u$
$$
lim_{t to 0}
frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = T_a(hat u).
$$
Question: is there any difference between original definition and my definition
In the original definition, $mathbf h$ can approach to $mathbf 0$ by any trajectory; In my definition it can only be approached from certain direction, so my definition is weaker than the original definition. Substitute $mathbf h = t hat u$ to the original definition will get my definition.
If my definition is not true, can somebody provide a counter example ?
One may already noticed that
$$
partial_{hat u} f (a) = T_a(hat u)
$$
Since $T$ is linear, we can assume (here $nabla f$ is just a function, it happens to be equal to the gradient if exist):
$$
T_a(hat u) = hat u cdot nabla f(a)
$$
So my definition can be written as:
$$
partial_{hat u} f (a) = hat u cdot nabla f(a) Leftrightarrow ftext{ is differentiable at }a
$$
multivariable-calculus
asked Jan 19 at 20:57
Zang MingJieZang MingJie
1356
$endgroup$
Original definition
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^m,$
if there is a linear transformation $T$ such that
$$
lim_{lVert mathbf hrVert to 0}
frac{f(mathbf a+mathbf h)-f(mathbf a)-T_a(mathbf h)}{lVert mathbf hrVert} = mathbf 0.
$$
My definition
A function $f: A to mathbb{R}^n$, $A subseteq mathbb{R}^m$ is differentiable at a point $mathbf a in mathbb R^m,$
if there is a linear transformation $T$ such that for any unit vector $hat u$
$$
lim_{t to 0}
frac{f(mathbf a+t hat u)-f(mathbf a)}{t} = T_a(hat u).
$$
Question: is there any difference between original definition and my definition
In the original definition, $mathbf h$ can approach to $mathbf 0$ by any trajectory; In my definition it can only be approached from certain direction, so my definition is weaker than the original definition. Substitute $mathbf h = t hat u$ to the original definition will get my definition.
If my definition is not true, can somebody provide a counter example ?
One may already noticed that
$$
partial_{hat u} f (a) = T_a(hat u)
$$
Since $T$ is linear, we can assume (here $nabla f$ is just a function, it happens to be equal to the gradient if exist):
$$
T_a(hat u) = hat u cdot nabla f(a)
$$
So my definition can be written as:
$$
partial_{hat u} f (a) = hat u cdot nabla f(a) Leftrightarrow ftext{ is differentiable at }a
$$
multivariable-calculus
multivariable-calculus
asked Jan 19 at 20:57
Zang MingJieZang MingJie
1356
asked Jan 19 at 20:57
Zang MingJieZang MingJie
1356
edited Jan 19 at 21:12
Zang MingJie
asked Jan 19 at 20:57
Zang MingJieZang MingJie
1356
asked Jan 19 at 20:57
Zang MingJieZang MingJie
1356
asked Jan 19 at 20:57
Zang MingJieZang MingJie
1356
1356
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For a counterexample try
$$
f:mathbb{R}^2 longrightarrow mathbb{R}
$$
with
$$
f(x,y) = left{
begin{array}
11 &text{if } ::x=y^2:: text{ and }::(x,y)not=(0,0)\
0 & text{otherwise}
end{array}
right.
$$
I don't want to prove everything for you. Try to imagine how this function looks like, why it's not differentiable in 0 (by general definition) and why partial derivatives exist on all directions (following your definition).
EDIT:
Your intuition is very good. The difference between the original definition of differentiability, and the one you proposed is in the fact that the original definition allows $h$ to approach the point $a$ in any trajectory, while you allow it only on linear trajectories.
This counterexample is a function that is 0 everywhere, except on a small set $A={(x,y):|:x=y^2} setminus{(0,0)}$ where it is equal to 1. Looking at the function on this curve, it is obviously not continuous in 0. But it has directional derivatives in all directions in 0, since for every direction $hat u$, the point $that u$ is inside $mathbb{R}^2setminus A$ for $t$ small enough.
answered Jan 19 at 21:14
KoljaKolja
590310
$endgroup$
$begingroup$
You'll note that this function isn't even continuous in 0.
$endgroup$
– Kolja
Jan 19 at 21:18
$begingroup$
It doesn't satisfy my definition, my definition requires directional derivatives exist on all directions and they are linear.
$endgroup$
– Zang MingJie
Jan 19 at 21:20
$begingroup$
The directional derivatives are 0, aren't they? That's as linear as it gets.
$endgroup$
– Kolja
Jan 19 at 21:21
$begingroup$
Ahh... I got it.
$endgroup$
– Zang MingJie
Jan 19 at 21:24
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a counterexample try
$$
f:mathbb{R}^2 longrightarrow mathbb{R}
$$
with
$$
f(x,y) = left{
begin{array}
11 &text{if } ::x=y^2:: text{ and }::(x,y)not=(0,0)\
0 & text{otherwise}
end{array}
right.
$$
I don't want to prove everything for you. Try to imagine how this function looks like, why it's not differentiable in 0 (by general definition) and why partial derivatives exist on all directions (following your definition).
EDIT:
Your intuition is very good. The difference between the original definition of differentiability, and the one you proposed is in the fact that the original definition allows $h$ to approach the point $a$ in any trajectory, while you allow it only on linear trajectories.
This counterexample is a function that is 0 everywhere, except on a small set $A={(x,y):|:x=y^2} setminus{(0,0)}$ where it is equal to 1. Looking at the function on this curve, it is obviously not continuous in 0. But it has directional derivatives in all directions in 0, since for every direction $hat u$, the point $that u$ is inside $mathbb{R}^2setminus A$ for $t$ small enough.
answered Jan 19 at 21:14
KoljaKolja
590310
$endgroup$
$begingroup$
You'll note that this function isn't even continuous in 0.
$endgroup$
– Kolja
Jan 19 at 21:18
$begingroup$
It doesn't satisfy my definition, my definition requires directional derivatives exist on all directions and they are linear.
$endgroup$
– Zang MingJie
Jan 19 at 21:20
$begingroup$
The directional derivatives are 0, aren't they? That's as linear as it gets.
$endgroup$
– Kolja
Jan 19 at 21:21
$begingroup$
Ahh... I got it.
$endgroup$
– Zang MingJie
Jan 19 at 21:24
add a comment |
$begingroup$
For a counterexample try
$$
f:mathbb{R}^2 longrightarrow mathbb{R}
$$
with
$$
f(x,y) = left{
begin{array}
11 &text{if } ::x=y^2:: text{ and }::(x,y)not=(0,0)\
0 & text{otherwise}
end{array}
right.
$$
I don't want to prove everything for you. Try to imagine how this function looks like, why it's not differentiable in 0 (by general definition) and why partial derivatives exist on all directions (following your definition).
EDIT:
Your intuition is very good. The difference between the original definition of differentiability, and the one you proposed is in the fact that the original definition allows $h$ to approach the point $a$ in any trajectory, while you allow it only on linear trajectories.
This counterexample is a function that is 0 everywhere, except on a small set $A={(x,y):|:x=y^2} setminus{(0,0)}$ where it is equal to 1. Looking at the function on this curve, it is obviously not continuous in 0. But it has directional derivatives in all directions in 0, since for every direction $hat u$, the point $that u$ is inside $mathbb{R}^2setminus A$ for $t$ small enough.
answered Jan 19 at 21:14
KoljaKolja
590310
$endgroup$
$begingroup$
You'll note that this function isn't even continuous in 0.
$endgroup$
– Kolja
Jan 19 at 21:18
$begingroup$
It doesn't satisfy my definition, my definition requires directional derivatives exist on all directions and they are linear.
$endgroup$
– Zang MingJie
Jan 19 at 21:20
$begingroup$
The directional derivatives are 0, aren't they? That's as linear as it gets.
$endgroup$
– Kolja
Jan 19 at 21:21
$begingroup$
Ahh... I got it.
$endgroup$
– Zang MingJie
Jan 19 at 21:24
add a comment |
$begingroup$
For a counterexample try
$$
f:mathbb{R}^2 longrightarrow mathbb{R}
$$
with
$$
f(x,y) = left{
begin{array}
11 &text{if } ::x=y^2:: text{ and }::(x,y)not=(0,0)\
0 & text{otherwise}
end{array}
right.
$$
I don't want to prove everything for you. Try to imagine how this function looks like, why it's not differentiable in 0 (by general definition) and why partial derivatives exist on all directions (following your definition).
EDIT:
Your intuition is very good. The difference between the original definition of differentiability, and the one you proposed is in the fact that the original definition allows $h$ to approach the point $a$ in any trajectory, while you allow it only on linear trajectories.
This counterexample is a function that is 0 everywhere, except on a small set $A={(x,y):|:x=y^2} setminus{(0,0)}$ where it is equal to 1. Looking at the function on this curve, it is obviously not continuous in 0. But it has directional derivatives in all directions in 0, since for every direction $hat u$, the point $that u$ is inside $mathbb{R}^2setminus A$ for $t$ small enough.
answered Jan 19 at 21:14
KoljaKolja
590310
$endgroup$
For a counterexample try
$$
f:mathbb{R}^2 longrightarrow mathbb{R}
$$
with
$$
f(x,y) = left{
begin{array}
11 &text{if } ::x=y^2:: text{ and }::(x,y)not=(0,0)\
0 & text{otherwise}
end{array}
right.
$$
I don't want to prove everything for you. Try to imagine how this function looks like, why it's not differentiable in 0 (by general definition) and why partial derivatives exist on all directions (following your definition).
EDIT:
Your intuition is very good. The difference between the original definition of differentiability, and the one you proposed is in the fact that the original definition allows $h$ to approach the point $a$ in any trajectory, while you allow it only on linear trajectories.
This counterexample is a function that is 0 everywhere, except on a small set $A={(x,y):|:x=y^2} setminus{(0,0)}$ where it is equal to 1. Looking at the function on this curve, it is obviously not continuous in 0. But it has directional derivatives in all directions in 0, since for every direction $hat u$, the point $that u$ is inside $mathbb{R}^2setminus A$ for $t$ small enough.
answered Jan 19 at 21:14
KoljaKolja
590310
edited Jan 19 at 21:35
answered Jan 19 at 21:14
KoljaKolja
590310
answered Jan 19 at 21:14
KoljaKolja
590310
answered Jan 19 at 21:14
KoljaKolja
590310
590310
$begingroup$
You'll note that this function isn't even continuous in 0.
$endgroup$
– Kolja
Jan 19 at 21:18
$begingroup$
It doesn't satisfy my definition, my definition requires directional derivatives exist on all directions and they are linear.
$endgroup$
– Zang MingJie
Jan 19 at 21:20
$begingroup$
The directional derivatives are 0, aren't they? That's as linear as it gets.
$endgroup$
– Kolja
Jan 19 at 21:21
$begingroup$
Ahh... I got it.
$endgroup$
– Zang MingJie
Jan 19 at 21:24
add a comment |
$begingroup$
You'll note that this function isn't even continuous in 0.
$endgroup$
– Kolja
Jan 19 at 21:18
$begingroup$
It doesn't satisfy my definition, my definition requires directional derivatives exist on all directions and they are linear.
$endgroup$
– Zang MingJie
Jan 19 at 21:20
$begingroup$
The directional derivatives are 0, aren't they? That's as linear as it gets.
$endgroup$
– Kolja
Jan 19 at 21:21
$begingroup$
Ahh... I got it.
$endgroup$
– Zang MingJie
Jan 19 at 21:24
$begingroup$
You'll note that this function isn't even continuous in 0.
$endgroup$
– Kolja
Jan 19 at 21:18
$begingroup$
You'll note that this function isn't even continuous in 0.
$endgroup$
– Kolja
Jan 19 at 21:18
$begingroup$
It doesn't satisfy my definition, my definition requires directional derivatives exist on all directions and they are linear.
$endgroup$
– Zang MingJie
Jan 19 at 21:20
$begingroup$
It doesn't satisfy my definition, my definition requires directional derivatives exist on all directions and they are linear.
$endgroup$
– Zang MingJie
Jan 19 at 21:20
$begingroup$
The directional derivatives are 0, aren't they? That's as linear as it gets.
$endgroup$
– Kolja
Jan 19 at 21:21
$begingroup$
The directional derivatives are 0, aren't they? That's as linear as it gets.
$endgroup$
– Kolja
Jan 19 at 21:21
$begingroup$
Ahh... I got it.
$endgroup$
– Zang MingJie
Jan 19 at 21:24
$begingroup$
Ahh... I got it.
$endgroup$
– Zang MingJie
Jan 19 at 21:24
add a comment |
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