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Taylor theorem in n variables

  1. Aug 30, 2005 #1
    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 [tex] f:U\rightarrow{R}, [/tex] defined in an open set [tex] U \subset R^n,[/tex] is said to be differentiable in a point [tex] (a_1,...,a_n) \in U[/tex] when it fulfills these conditions:

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

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

    II. The Mean Value Theorem.

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

    Now I state the

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

    And here I copy literally the proof of the author:

    "Proof.-

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

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

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

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

    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 [tex] \frac{\partial}{\partial x_i}r [/tex] is on its own right a function differentiable in [tex] (0,...,0) [/tex], so applying the definition we've seen before the lemma we got for every [tex] i = 1,...,n [/tex] that [tex]\lim_{\Vert{v}\Vert\rightarrow 0} {\frac {1}{\Vert{v}\Vert} \frac{\partial}{\partial x_i}r(\theta v) = 0 [/tex]. 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.
     
  2. jcsd
  3. Aug 31, 2005 #2

    TD

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    Perhaps it's just me but the LaTeX doesn't appear at the bottom, from 4.
     
  4. Aug 31, 2005 #3
    In a moment I will copy again from 4 onwards.

    P Castilla.
     
  5. Aug 31, 2005 #4
    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 [tex] \frac {v_i}{\Vert{v}\Vert}[/tex] has absolute value [tex] \leqq 1.[/tex] Therefore
    [tex] \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert^2}=0.[/tex]". 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 [tex] \frac{\partial}{\partial x_i}r [/tex] is, on its own right, a function of n variables differentiable in [tex] (0,...,0) [/tex], so if I apply the definition of "differentiable function" I got for every [tex] i=1,...,n [/tex] that [tex] \lim_{\Vert{v}\Vert\rightarrow 0}{\frac{1}{\Vert{v}\Vert} \frac{\partial}{\partial x_i}r(v)}=0 [/tex].

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

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

    Can you help me?

    P Castilla.
     
  6. Aug 31, 2005 #5

    AKG

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    [tex]\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[/tex]
     
  7. Aug 31, 2005 #6
    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.
     
  8. Aug 31, 2005 #7

    AKG

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

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

    and

    [tex] \lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0 [/tex]

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

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

    combined with

    [tex]\lim_{\Vert{v}\Vert\rightarrow 0} \frac{r(v)}{\Vert{v}\Vert}=0[/tex]

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

    [tex]\lim _{|Ky| \to 0} f(y) = \lim _{|y| \to 0} f(y)[/tex]

    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.
     
  9. Aug 31, 2005 #8
    AKG:

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

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

    Thanks for your patience.

    Castilla.
     
  10. Sep 1, 2005 #9
    Hm, please don't forget this thread. Is only a question of Analysis 1. :frown:
     
  11. Sep 2, 2005 #10
    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.
     
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