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Show that f'(x) grad f(x) is positive

  1. Aug 16, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I'm preparing a maths exam so I run through a lot of problems, and I am unsure whether I tackled this one correctly or not:

    Prove that for every differentiable function ##f: \mathbb{R}^3 \to \mathbb{R}## and for every point ##x \in \mathbb{R}^3## the following statement is true:

    ##\lim\limits_{t \to 0} \frac{f(x+t \nabla f(x)) - f(x)}{t} \geq 0##

    2. Relevant equations

    I have the following definitions:

    ##f'(x) = \frac{\partial (f_1, ..., f_m)}{\partial (x_1, ..., x_n)} (x)##

    Directional derivative: ##f'(x)v = \lim\limits_{t \to 0} \frac{1}{t} (f(x + tv) - f(x))##

    3. The attempt at a solution

    So what I did was pretty straightforward:

    ##\lim\limits_{t \to 0} \frac{f(x + t \nabla f(x)) - f(x)}{t} = f'(x) \nabla f(x)##
    ##= [\partial_1 f(x) \partial_2 f(x) \partial_3 f(x)] \cdot [\partial_1 f(x), \partial_2 f(x), \partial_3 f(x)]##
    ##= (\partial_1 f(x))^2 + (\partial_2 f(x))^2 + (\partial_3 f(x))^2 \geq 0##

    Is that legit? Not sure about my handling from ##f'(x)##. Should I have used the total differential definition? Would I obtain the same result? I'm still unclear about total derivatives :)


    Thanks a lot in advance for your answers.


    Julien.
     
  2. jcsd
  3. Aug 16, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    For a vector ##\vec{u} \in R^3## we have
    $$\lim_{t \to 0} \frac{f(\vec{x} + t \vec{u}) - f(\vec{x})}{t} = \langle \vec{u}, \vec{\nabla} f(\vec{x}) \rangle,$$
    where ##\langle \vec{p}, \vec{q} \rangle## denotes the inner product of the vectors ##\vec{p}, \vec{q}##. You are taking ##\vec{u} = \vec{\nabla} f(\vec{x})##.
     
    Last edited: Aug 16, 2016
  4. Aug 16, 2016 #3

    fresh_42

    Staff: Mentor

    It looks right to me. I would have written the limit as ##(\nabla_{(\nabla f)(x)}f)(x) \overset{(*)}= (\nabla f (x))\cdot (\nabla f (x)) = |\nabla f (x)|^2##. The total differentiability of ##f## is hidden in ##(*)## which allows you to write the directional derivative in direction ##v## (##=(\nabla f)(x)## in this case) as ##D_v f(x) = \nabla f(x) \cdot v##. Other than directional differentiability and partial differentiability (which are simply special directional derivatives), the total differentiability allows us to choose any direction. In this case ##v=(\nabla f)(x)## is a linear combination of partial derivatives.
     
  5. Aug 17, 2016 #4
    @Ray Vickson Hi and thanks for your answer! I like the way you write the definition with the inner product. At the end it gives the same result that I wrote, right?

    @fresh_42 Hello and thank you for your answer as well. :) Of course I realised the equality holds only because the problem states that the function is differentiable. I am still unsure about the meaning of the total derivative. So far I understand it like that: the partial differentials hold for analytical use of a function, while the total differential becomes useful when wanting to approximate the slope at a given point. Would you support that statement? If not, could you maybe give me an example of partial and total differentiation so that I can visualise it? My interpretation is based on the following observation (for the case ##f: \mathbb{R}^n \to \mathbb{R}##):

    ##f' (\vec{x}) = \frac{\partial (f_1, ... , f_m)}{\partial (x_1, ... , x_n)} (x) = \mbox{vector}##
    ##f' (\vec{x}) = \sum \frac{\partial f}{\partial x_i} dx_i = \mbox{scalar}##

    First of all, is that correct? And if so, would you say that "the total differentiability allows us to choose any direction" when we plug in arbitrary ##dx_i## for a given point ##\vec{x}##?


    Thanks a lot again to both of you for your answers, I really appreciate your explanations!


    Julien.
     
  6. Aug 17, 2016 #5

    fresh_42

    Staff: Mentor

    Not really. Partial derivatives also approximate the slope at a given point. Just in the direction of the coordinates. A total differential approximates a function ##f## as a whole by a linear function ##L## (the total differential), i.e. you may write ##f(\vec{x}+\vec{v}) = f(\vec{x}) + L(\vec{v}) +r(\vec{v})## such that ##\lim_{\vec{v} \rightarrow 0} \frac{r(\vec{v}}{||\vec{v}||} = 0##. This means ##f## can be approximated by ##L## such that the error terms in ##r## run faster towards zero than ##L## does, or as it often occurs (e.g. in Taylor expansions), ##r## has terms in ##\vec{v}## of higher orders than ##1##.

    I have difficulties with naming everything ##x##. You should try to get used to distinguish when you are speaking of a function in the variable ##x##, coordinates ##x_i## or ##x^i## and eventual evaluation points ##x_0=p,## e.g. where the derivative is evaluated at. I'm not sure if I got you right. E.g. you may denote a gradient better by $$grad(f) = \nabla f = \sum_{i=1}^{n}\frac{\partial f}{\partial x_i} \vec{e}_i$$ or more physical $$\nabla f = \sum_{i=1}^{n}\frac{\partial f}{\partial x_i} dx_i$$ and $$\nabla_p f = \sum_{i=1}^{n}\left(\frac{\partial f}{\partial x_i}\right)_{x=p} dx_i$$

    In addition, how can ##f'(\vec{x})## simultaneously be a vector and a scalar? Only by evaluation of ##\nabla_p f## at a certain point, which should not be denoted by the variable name.

    An example for the difference between partial and total differentiation is the following:
    $$f(\vec{x}) = \begin{cases} \frac{2x_1x_2}{x_1^2+x_2^2} & \text{if } (x_1,x_2) \neq (0,0) \\ \\ 0 & \text{if } (x_1,x_2) = (0,0) \end{cases}$$
    ##f## is partially differentiable at ##(0,0)## but not continuous at ##(0,0)##. And you might want to calculate what happens along the line ##x_1 = x_2##. The latter is an example of a direction that is not along a coordinate.
     
  7. Aug 18, 2016 #6
    @fresh_42 Thank you again for your answer, it is very complete and I am slowly starting to understand. I will study the matter for some more time on my own and probably come back here for more questions later on :)

    Julien.
     
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