I am reading from two books a few definitions and a few theorems related to differentiability, differential, directional derivative, etc., and I got a bit confused so I want to clarify if I am understanding these things right.
Let's say we have a function $w = f(x,y,z)$ which is defined in some neighborhood $U$ of the point $(a,b,c)$. As usual, we denote
$\Delta{x} = x - a$
$\Delta{y} = y - b$
$\Delta{z} = z - c$
$\Delta{w} = f(a+\Delta{x}, b+\Delta{y}, c+\Delta{z}) - f(a,b,c) $
where $(x,y,z)$ is some other point from $U$.
This is our setup here.
The first statement is that when $f'_x(a,b,c), f'_y(a,b,c), f'_z(a,b,c)$ exist, that does not imply that the function $f$ is continuous in the point $(a,b,c)$. I saw some examples i.e. counterexamples, so OK, I think I understand that.
But then I am reading this statement/theorem. If $f'_x(a,b,c), f'_y(a,b,c), f'_z(a,b,c)$ exist and are continuous in the point (a,b,c), then $\Delta{w} = f(a+\Delta{x}, b+\Delta{y}, c+\Delta{z}) - f(a,b,c) $ can be written in this way
$\Delta{w} = f'_x(a,b,c) \Delta{x} + f'_y(a,b,c) \Delta{y} + f'_z(a,b,c) \Delta{z} + \epsilon_1 \Delta{x} + \epsilon_2 \Delta{y} + \epsilon_3 \Delta{z} $, where $\epsilon_1,\epsilon_2,\epsilon_3$ are functions of $a,b,c, \Delta{x}, \Delta{y} ,\Delta{z}$ which tend to $0$ as $(\Delta{x},\Delta{y},\Delta{z}) \to (0,0,0)$
From the conclusion of this theorem, I think it easily follows also that $f$ is continuous in the point $(a,b,c)$.
We just let $\Delta{x}, \Delta{y}, \Delta{z}$ go to zero and we see that $\Delta{w}$ also goes to zero. So it follows that $f$ is continuous in $(a,b,c)$. Right?
Comparing this to 1) I come to the conclusion that the difference in the two situations comes from the fact that in 2) we require also that the partial derivatives are also continuous in the point $(a,b,c)$, and this makes it possible to claim/prove that $f$ is also continuous. Is this indeed so?
- I am reading then another definition. We say that $w = f(x,y,z)$, whose partial derivatives exist in the point $(a,b,c)$, is differentiable in the point $(a,b,c)$, if $\Delta{w}$ can be written in the form
$$\Delta{w} = f'_x(a,b,c) \Delta{x} + f'_y(a,b,c) \Delta{y} + f'_z(a,b,c) \Delta{z} + \epsilon_1 \Delta{x} + \epsilon_2 \Delta{y} + \epsilon_3 \Delta{z} \ \ \ \ \ (*)$$ where $\epsilon_1,\epsilon_2,\epsilon_3$ are functions of $a,b,c, \Delta{x}, \Delta{y} ,\Delta{z}$ which tend to $0$ as $(\Delta{x},\Delta{y},\Delta{z}) \to (0,0,0)$
From this definition I am drawing the conclusion that the situation in 2) is just one particular case in which the function $f$ is differentiable. I mean, basically 2) is saying that if the partial derivatives exist and are continuous at $(a,b,c)$, then the function $f$ is differentiable at the point $(a,b,c)$ /differentiable in the sense of definition 3)/.
Is this so, i.e. am I understanding this correctly?
Also, in other words this definition is just saying that $\Delta{w} = dw + \epsilon_1 \Delta{x} + \epsilon_2 \Delta{y} + \epsilon_3 \Delta{z} $ i.e. the differential of w is "very good" approximation of $\Delta{w}$ (very good because these epsilons go to zero as $(\Delta{x}, \Delta{y}, \Delta{z}) \to 0$). Right? And when this is so, we say that $f$ is differentiable (differentiable as a whole, not partially) in the point $(a,b,c)$. Am I getting this right?
- Finally, there is another theorem which says that if $f$ is differentiable (in the sense of def. 3)) in the point $(a,b,c)$, then it also has directional derivatives at $(a,b,c)$ in the direction of any unit vector $v = (a_1, b_1, c_1)$. The proof the authors give for this one is kind of informal, I didn't like it much. But this statement I was able to prove formally myself by using $(*)$. It seems very easy to prove by using the definition 3) i.e. by using $(*)$.
So am I understanding all this stuff correctly? Am I drawing the correct conclusions for myself?
Actually in the book these defs and theorems were stated for two variables $f(x,y)$ but I stated them here for three variables $f(x,y,z)$ again to ensure that I am understanding all this correctly.
Just a partial (no pun intended) answer. A function $ w = f(x,y,z) $ is said to be differentiable at a point $ (a,b,c) $ is there exists a (bounded, but you can ignore this) linear mapping $ L(\Delta x,\Delta y,\Delta z) $ such that $$ f(a + \Delta x,b + \Delta y,c + \Delta z) - f(a,b,c) = L(\Delta x,\Delta y,\Delta z) + o(\Delta x,\Delta y,\Delta z) $$ for any increments $ \Delta x $, $ \Delta y $ and $ \Delta z $, where $$ \lim_{(\Delta x,\Delta y,\Delta z)\to 0}\frac{o(\Delta x,\Delta y,\Delta z)}{\lVert (\Delta x,\Delta y,\Delta z)\rVert} = 0\text{.} $$ You can check that the map $ L $, when it exists, it is unique. Usually, $ L $ is called the differential of $ f $ at $ (a,b,c) $, and is denoted $ \mathrm df(a,b,c) $.
If $ f $ is differentiable at $ (a,b,c) $, then it is continuous there, since $$ \lim_{(\Delta x,\Delta y,\Delta z)\to (0,0,0)}\left\lvert f(a + \Delta x,b + \Delta y,c + \Delta z) - f(a,b,c)\right\rvert = 0\text{,} $$ because $ \mathrm df(a,b,c) $ is linear and $ o $ acts as described above when its argument goes to zero.
Now you can easily show that $$ \mathrm df(a,b,c)(1,0,0) = {\frac{\partial f(x,y,z)}{\partial x}}\Bigg|_{(a,b,c)}\quad \mathrm df(a,b,c)(0,1,0) = {\frac{\partial f(x,y,z)}{\partial y}}\Bigg|_{(a,b,c)}\quad \mathrm df(a,b,c)(0,0,1) = {\frac{\partial f(x,y,z)}{\partial z}}\Bigg|_{(a,b,c)}\text{,} $$ so that you can write $$ f(a + \Delta x,b + \Delta y,c + \Delta z) - f(a,b,c) = \Delta x{\frac{\partial f(x,y,z)}{\partial x}}\Bigg|_{(a,b,c)} + \Delta y{\frac{\partial f(x,y,z)}{\partial y}}\Bigg|_{(a,b,c)} + \Delta z{\frac{\partial f(x,y,z)}{\partial z}}\Bigg|_{(a,b,c)} + o(\Delta x,\Delta y,\Delta z) $$ by linearity.
It's a (standard) theorem of calculus that a function $ w = f(x,y,z) $ is differentiable at $ (a,b,c) $ if and only if the partial derivatives $ \partial f/\partial x $, $ \partial f/\partial y $, $ \partial f/\partial z $ are defined in a neighbourhood of $ (a,b,c) $, and are continuous at $ (a,b,c) $.