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Proof Help: Directional Derivatives

  1. Oct 11, 2009 #1
    Prove the following is not true:

    Let f : R^2->R^2 be a nonlinear function. For any vectors a,v in R^2;

    f(a+v)-f(a)=[Df(a)]v


    In terms of my attempt, I've been trying to find some combination of a and v that ensure this fails, but so far the best I've come up with is to start with:

    [Df(a)]v=lim{h->0} (f(a+hv)-f(a))/h

    And then trying to show that f(a+v)-f(a) is the same thing only as h->1; again, no luck. I really just need a hint or two as to which way to approach the proof.
     
  2. jcsd
  3. Oct 11, 2009 #2

    LCKurtz

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    To show an equation is false you don't have to prove a big theorem. You just have to find an example where it doesn't work. Have you actually tried any examples? Take something simple with a squared term, an easy point and an easy unit vector. Almost anything you try should give a counterexample. Just show the equation you are given doesn't work using the formula for directional derivative.
     
  4. Oct 11, 2009 #3
    Read the text; finding a counter example for one function f simply proves that it doesn't hold for that ONE function f; the problem asks to show this is true for ANY nonlinear function.
     
  5. Oct 11, 2009 #4

    LCKurtz

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    Read the text yourself. It asks you to prove a theorem is not true.
     
  6. Oct 11, 2009 #5

    Office_Shredder

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    You can use that f is non-linear. What does that mean? Either
    1) There exist x, y so that f(x+y) is NOT f(x)+f(y)

    2) There exists x, c a constant so that f(cx) is NOT cf(x).

    Cleverly pick your point a to be one at which f is 'not linear' so to speak (i.e. you can use a to demonstrate f is non-linear by one of these).
     
  7. Oct 11, 2009 #6

    Landau

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    This is basic logic. You need to disprove a statement of the form "for all nonlinear functions f and vectors a,v: ...". This means finding one particular nonlinear function f and particular vectors a and v such that ... does not hold, is enough.
     
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