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.