Question

Someone explain me, please.

Answer 1

don't get what 0|x-a| mean

Answer 2

My prof said that o stands for "order", but what it mean for o|x -a| in \(\mathbb R\) and for o||x-a|| in \(\mathbb R^n\)

Answer 3

oh the third one should be the second one.:) I am sorry.

Answer 4

I honestly don't know, I have seen this before with computer programming I think called "Big O notation" or something very similar and I believe it is just an error term.

Answer 5

But it is used a lot in the lecture. Sometimes, it is =0, sometimes, it is not. :)
That's why I want to know what it is.

Answer 6

wow this is waaaay too hard for me sorry D:

Answer 7

im pretty sure that's a 0

Answer 8

you mean zero?

Answer 9

just adding 0 is a common trick

Answer 10

yes

Answer 11

why do we have to do it?

Answer 12

the fact that they divide the |x-a| and have it set equal to 0

Answer 13

so, it is there to "decorate" only? hahaha... I am sorry for my words but do not know how to say.

Answer 14

what are you proving?

Answer 15

I am proving D(f+g) = Df +Dg

Answer 16

example of adding 0

|a|=|a-b+b|<=|a-b|+|b| implies |a|-|b|<=|a-b|

Answer 17

you are proving that the differential operator is a linear operation?

Answer 18

Nope, that is my lecture. My problem is proving D(f+g) = Df + Dg, use definition of D

Answer 19

and I am out at that point (o||x -a||), I don't get what it mean

Answer 20

OK so do you see how that part that that says linearization of f at x=a

do you see how that if you put in a you get f(a) = f(a)?

Answer 21

yes, I see, because the middle term is Df (x-a) when x --> a , it is =0
the last term is the same, then f(x) = f(a)

Answer 22

my bus stop is in 1 more stop. Ill get back on in 10 when I get home.

Answer 23

ok, I am waiting.

Answer 24

I need you on my Ring theory also, please.

Answer 25

Sorry it took a while, I did some research so my answer would be more 'complete'.
Well, from what I see, what you got here is basically the beginning of multi-variable calculus.

I'll try to show the intuition using 'simple functions'.
With a single-variable scalar functions (simple functions) the derivative will describe the 'slope of the tangent to the function at the given point'.
This tangent line can be used as a 'linear approximation' to the function at the tangency point.
The idea is that the tangent line is very close to the function in some region around that point since the function is differentiable there and therefore `doesn't change instantly` (is `smooth`) there.
Basically that is the idea behind Taylor series.
You can go here: http://en.wikipedia.org/wiki/Taylor%27s%20theorem and see that they start with `linear approximation` and try to get more accurate using higher degree polynomials.

That said, if we have a function \(f(x)\) we can say that the tangent line equation at point \(a\) will be:
$$
t(x) = f(a) + f'(x) \cdot (x-a)
$$Which is the `linear approximation` (or the `linearization`) of \(f(x)\).

Since \(t(x)\) is an approximation, it is (probably) not the same as \(f(x)\). In order to define \(f(x)\) using the linear approximation we have to take the 'approximation error' into account.
I'll define the error as \(E(x)\) which corresponds to the `Remainder` in the Taylor theorem. Means \(E(x)\) will give the difference between our approximation and the original function at a given point, and hence the Error of the approximation.
You can see at wiki that the remainder can also be defined using `little-o notation` (which is your \(o(|x-a|)\)).

In the first picture you have an example of such `linearization` of the function \(f(x)\).
We can define \(E(x)\) in many ways. To fit it more to your example we'll make \(E(x)\) such that \(E(0)\) would be at the tangency point. so:
$$
E(x) = f(x + a) - t(x + a) \\
E(x-a) = f(x) - t(x)
$$Then:
$$
f(x) = t(x) + E(x-a)
$$And since we know \(t(x)\) we can substitute:
$$
f(x) = f(a) + f'(a)(x) +E(x-a)
$$Which is pretty much the formula you have in picture #1.

Now for the limit beneath it.
Again, remember that it is just for intuition and therefore I work with 'simple' functions.
If we have a tangent line to \(f(x)\) at point \(a\) then, as we know, the slope of the line \(m\) is the derivative at the tangency point \(a\):
$$
m = f'(a)
$$We can use the definition of the derivative (for our simple function):
$$
f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}
$$And say:
$$
m = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}
$$Notice that in this limit \(x\) approaches \(a\), but it can approach it in different directions.
For real numbers that would mean either \(a\) is smaller or bigger than \(a\). If \(a\) and \(x\) are points in 2d space or are complex numbers for example, then such limits can evaluate in infinite number of different directions.
If this limit evaluates to the same value (we get same 'm') no matter from what 'direction' \(x\) approaches \(a\) then we found a single slope for the tangent line and the function is therefore differentiable at \(a\).

Manipulating the limit:
$$
\lim_{x \to a}\frac{f(x)-f(a)}{x-a} - m = 0 \\
\lim_{x \to a}\frac{f(x)-f(a)}{x-a} - \frac{m(x-a)}{x-a} = 0 \\
\lim_{x \to a}\frac{f(x)-f(a) - m(x-a)}{x-a} = 0 \\
$$
And since we know \(m\) is meant to be \(f'(x)\) we can substitute:
$$
\lim_{x \to a}\frac{f(x)-f(a) - f'(a)(x-a)}{x-a}= \lim_{x \to a} \frac{E(x-a)}{x-a} = 0 \\
$$Which is pretty much your limit, without absolute values (which we can add and wouldn't affect the limit in this case).
Though I think the substitution is somewhat useless and ironic since we're trying to see if there is a value for \(f'(x)\).

At the third image and then second image you have a generalized definition of differentiablity for multi-variable vector functions.
The multi-variables are expressed by \(a\) and \(x\) being points in perhaps more than one dimensional space and therefore contain multiple components (for example x,y,z).
Notice that in multi-variable functions the tangent is not necessarily a line. A tangent for a 2 variable scalar function would be a plane for example.
\(Df(a)\) is basically the Jacobian matrix ( http://en.wikipedia.org/wiki/Jacobian%20matrix%20and%20determinant), which is a matrix holding partial derivatives of \(f(x)\). This is basically a generalized form of the derivative of \(f(x)\).
And the limit becomes
$$
\lim_{x \to a} \frac{||f(x) - f(a) - Df(a)(x-a)||}{||x-a||} = 0
$$When working with multiple-variable vector functions it is not so simple (for me) to show how to get to the limit as with the 'simple' functions. (but my brain is pretty much melting at this point, so maybe tomorrow idk)

My resources:
My main resource (very nice, I recommend reading it):
http://mathinsight.org/differentiability_multivariable_introduction (Notice there are 'Next' and 'Previous' buttons on the right. This is an introduction page to the subject and the next chapters cover more of it).
And a bit of those:
http://en.wikipedia.org/wiki/Linear_approximation
http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
http://en.wikipedia.org/wiki/Linearization

Answer 26

Thank you very much. :)