What is the proof of the Taylor Theorem in n variables?

[Castilla]

Hello, guys. I am studying the Taylor Theorem for functions of n variables and in one book I've found a proof based on the lemma that I am copying here. I am having some trouble in following its proof so I seek your kind assistance.

The lemma rests on two items: the definition of a function of n variables differentiable in a point "a" and the Mean Value Theorem for functions of n variables.

I. A function f:U\rightarrow{R}, defined in an open set U \subset R^n, is said to be differentiable in a point (a_1,...,a_n) \in U when it fulfills these conditions:

1. There exist the partial derivatives \frac{\partial}{\partial x_1}f(a_1,...,a_n),..., \frac{\partial}{\partial x_n}f(a_1,...,a_n).

2. For every v = (v_1,...,v_n) such that a + v \in U we got
f(a+v) - f(a) = \sum_{i=1}^{n} v_i \frac{\partial}{\partial x_i}f(a) + r(v), where \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0.

II. The Mean Value Theorem.

Let the function f:U\rightarrow{R} be differentiable in the open set U \subset R^n, and the line [a, a+v] \subset U; then we can find a \theta \in (0,1) such that
f(a+v) - f(a) = \sum_{i=1}^{n} v_i \frac{\partial}{\partial x_i}f(a+ \theta v).

Now I state the

Lemma.- Let be the function r:B\rightarrow{R} of class C^2 in the open ball B \subset R^n of center (0,...,0). If for every i = 1,..., n we got r(0,...,0) = \frac{\partial}{\partial x_i}r(0,...,0) = \frac{\partial^2}{\partial x_j \partial x_i}r(0,...,0) = 0, then \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert^2}=0.

And here I copy literally the proof of the author:

"Proof.-

1. "Being r:B\rightarrow{R} a function of class C^1 (therefore differentiable) that gets null in the point (0,...,0) (and the same for its derivatives \frac{\partial}{\partial x_i}r), it follows from the definition of differentiable function that \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0". (My note: OK, this is fine).

2. "By the Mean Value Theorem, for each v = (v_1,..., v_n) \in B exists \theta \in (0,1) such that r(v) = \sum_{i=1}^{n} v_i \frac{\partial}{\partial x_i}r(\theta v). Therefore \frac{r(v)}{\Vert{v}\Vert^2}= \sum_{i=1}^{n} {\frac {1}{\Vert{v}\Vert}v_i \frac{\partial}{\partial x_i}r(\theta v)." (OK, this is fine also).

3. "Every partial derivative \frac{\partial}{\partial x_i}r
and its derivatives \frac{\partial^2}{\partial x_j \partial x_i}r, gets null in the point (0,...,0). Hence, from our initial observation (I suppose he refers to paragraph 1? ) it follows that (I do not understand this) \lim_{\Vert{v}\Vert\rightarrow 0} {\frac {1}{\Vert{v}\Vert}}{\frac{\partial}{\partial x_i}r(\theta v)} = 0 for all i = 1,...,n."

4. "Furthermore, each quocient \frac {v_i}{\Vert{v}\Vert}has absolute value \leqq 1. Therefore \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert^2}=0".

End of proof.

As I've said, the results of paragraphs 1 and 2 are OK. My trouble is the inference of paragraph 3. I know that each partial derivative \frac{\partial}{\partial x_i}r is on its own right a function differentiable in (0,...,0), so applying the definition we've seen before the lemma we got for every i = 1,...,n that \lim_{\Vert{v}\Vert\rightarrow 0} {\frac {1}{\Vert{v}\Vert} \frac{\partial}{\partial x_i}r(\theta v) = 0. But I don't catch up how this fact leads to the result of paragraph 3. Or maybe he gets that result in another way which escapes me.

Can I ask for your assistance?

P Castilla.

 

[TD]

Perhaps it's just me but the LaTeX doesn't appear at the bottom, from 4.

 

[Castilla]

In a moment I will copy again from 4 onwards.

P Castilla.

 

[Castilla]

OK, don't mind the last part of mi initial post. Here I repeat it from paragraph 4 onwards (making some correction).

4. "Furthermore, each quocient \frac {v_i}{\Vert{v}\Vert} has absolute value \leqq 1. Therefore
\lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert^2}=0.". End of the proof of the Lemma.

As I have said, the results of paragraphs 1 and 2 are OK. My problem is the inference of paragraph 3. How does he got it?

Let'see... I know that each partial derivative \frac{\partial}{\partial x_i}r is, on its own right, a function of n variables differentiable in (0,...,0), so if I apply the definition of "differentiable function" I got for every i=1,...,n that \lim_{\Vert{v}\Vert\rightarrow 0}{\frac{1}{\Vert{v}\Vert} \frac{\partial}{\partial x_i}r(v)}=0.

Does this statement, combined with \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0, lead to

\lim_{\Vert{v}\Vert\rightarrow 0}{\frac{1}{\Vert{v}\Vert}}{\frac{\partial}{\partial x_i}r(\theta v)}=0?? I don't catch how (see the \theta). Or maybe he gets it in other way which escapes me.

Can you help me?

P Castilla.

 

[AKG]

\lim_{\Vert{v}\Vert\rightarrow 0}{\frac{1}{\Vert{v}\Vert}}{\frac{\partial}{\partial x_i}r(\theta v)}=\lim_{\Vert{y/\theta}\Vert\rightarrow 0}{\frac{1}{\Vert{y/\theta}\Vert}}{\frac{\partial}{\partial x_i}r(y)} = \theta \lim_{\Vert{y/\theta}\Vert\rightarrow 0}{\frac{1}{\Vert{y}\Vert}}{\frac{\partial}{\partial x_i}r(y)}=\theta \lim_{\Vert{y}\Vert\rightarrow 0}{\frac{1}{\Vert{y}\Vert}}{\frac{\partial}{\partial x_i}r(y)}=0

 

[Castilla]

AKG:

Thanks for your time and answer. Yet let me bother with some aditionals:

1. You do not use lim r(v)/lvl = 0. What would be the author's reason to deduce that partial result if it was unnecesary?

2. May I request you to clarify the change of variable in your first and third equality?

(Apologies for my english).

Thanks in advance,

P Castilla.

 

[AKG]

The truth is I'm not entirely sure about all this. My first thought was to look at:

r(v) = \sum_{i=1}^{n} v_i \frac{\partial}{\partial x_i}r(\theta v)

and

\lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0

and do some substitution, but I wasn't able to make that work. In your last post, you asked whether:

\lim_{\Vert{v}\Vert\rightarrow 0}{\frac{1}{\Vert{v}\Vert} \frac{\partial}{\partial x_i}r(v)}=0

combined with

\lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0

leads to the desired result, and appeared to me that it did, and that's what I posted. I realize what I posted seems a little questionable (which is why you're asking me to explain the changes of variables in the first and third equalities) so I will try to justify it.

The first change of variables is simply v = \theta ^{-1}y. Note that \theta ^{-1} is a constant, and v and y are vectors. The next change of variables is not really a change of variables. Let K = \theta ^{-1} be a constant. Then I'm basically just arguing that:

\lim _{|Ky| \to 0} f(y) = \lim _{|y| \to 0} f(y)

If for every E > 0, there is a D > 0 such that |f(y) - L| < E for all |y| < D, then for every E > 0, there is a D' > 0 such that |f(y) - L| < E for all |Ky| < D', simply chosing D' = KD. So both limits are the same.

 

[Castilla]

AKG:

1. In the last equality of the post of 9.44 am you used \lim_{\Vert {v}\Vert\rightarrow 0}{\frac {1}{\Vert{v}\Vert} \frac{ \partial}{\partial x_i} r(v)} = 0. But I still do not see in which equality you used \lim_{\Vert{v}\Vert\rightarrow 0} \frac {r(v)}{\Vert{v}\Vert} = 0.

2. I am afraid \theta is not more constant than v.The TVM says that for every v we have a \theta_v. Hence\theta is a dependent variable... I don't know if this is harmless for your equalities.

Thanks for your patience.

Castilla.

 

[Castilla]

Hm, please don't forget this thread. Is only a question of Analysis 1.

 

[Castilla]

AKG:

1. In the last equality of the post of 9.44 am you used \lim_{\Vert {v}\Vert\rightarrow 0}{\frac {1}{\Vert{v}\Vert} \frac{ \partial}{\partial x_i} r(v)} = 0. But I still do not see in which equality you used \lim_{\Vert{v}\Vert\rightarrow 0} \frac {r(v)}{\Vert{v}\Vert} = 0.

2. I am afraid \theta is not more constant than v.The TVM says that for every v we have a \theta_v. Hence\theta is a dependent variable... I don't know if this is harmless for your equalities.

Thanks for your patience.

Castilla.

AKG:

Finally I have understood your equalities, so I take back "objection" Nr. 2. But the Nr. 1 keeps bothering me. Shall I suppose that the author included an unnecesary result in his proof of the lemma?

Castilla.