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One sided differentiation prove

  1. Jan 24, 2009 #1
    i am given two differentiable function f and g .
    prove that for u(x)=max(f(x),g(x))
    and v(x)=min(f(x),g(x))

    there is one sided derivatives
    ??


    how to put mim ,max functions into the formula of derivative formula??

    [itex]f'(x)_+ = \lim_{h \to 0^+} \frac {f(x + h) - f(x)}h[/itex] (one-sided derivative...from the right)
    [itex]
    f'(x)_- = \lim_{h \to 0^-} \frac {f(x + h) - f(x)}h
    [/itex]
    (one-sided derivative... from the left)
     
  2. jcsd
  3. Jan 24, 2009 #2

    CompuChip

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    So let's fix a point x.
    I think you can show that if one function is larger at x then there is some one-sided interval for which it stays the larger one.
    So for example, if u(x) = f(x)) then there exists an [itex]\epsilon > 0[/itex] such that u(y) = f(y) for all [itex]y \in [x, x + \epsilon)[/itex]. So then [itex]u'(x)_+ = f'(x)[/itex].
     
  4. Jan 24, 2009 #3
    how to prove it?
     
  5. Jan 25, 2009 #4

    CompuChip

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    I am assuming you have done / are doing some analysis.
    What you could do, for example, is define h(x) := f(x) - g(x).
    Then if u(x) := max(f(x), g(x)) is equal to f(x) in some point x0, that is equivalent to h(x) being greater or equal to 0.

    Do you then see how your question reduces from
    "Show that there is a right one-sided derivative of u(x)"
    to
    "Show that if [itex]h(x_0) \ge 0[/itex] then there is some [itex]\epsilon > 0[/itex] such that [itex]h(x) \ge 0, \, \forall x \in [x_0, x_0 + \epsilon)[/itex]"
     
  6. Jan 25, 2009 #5
    i understand you explanation
    but i cant deside if it the needed proove

    deffentiable function is a function which on soome point its
    left side derivative equals its write side derivative.

    so
    if both functions are derivatable.
    then it doent matter what function we take

    there for sure will be a function where
    it has a point for which there is a derivative on both sides
    even though we need only one.
    so its an over kill
    am i correct?
     
  7. Jan 25, 2009 #6

    quasar987

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    There is something I would like to point out. It does not seem exactly true that if [itex]h(x_0) \ge 0[/itex] then there is some [itex]\epsilon > 0[/itex] such that [itex]h(x) \ge 0, \, \forall x \in [x_0, x_0 + \epsilon)[/itex].

    For instance, consider f(x)=-|x| and g(x)=-x². Then u(0)=f(0) but in no one sided interval of 0 do we have [itex]h(x) \geq 0[/itex].
     
  8. Jan 25, 2009 #7

    Hurkyl

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    That f isn't differentiable....
     
  9. Jan 25, 2009 #8

    quasar987

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    Good point, but the idea still stands. Take instead f(x)=-x² and g(x)=-x^4.
     
  10. Jan 26, 2009 #9

    CompuChip

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    For f(x) = -x2 and g(x) = -x4, the difference h(x) := f(x) - g(x) actually has the same sign throughout the entire interval ]-1, 1[.

    OK, so maybe you have to redefine h(x) as g(x) - f(x) to get it positive; that's irrelevant for the proof.
     
  11. Jan 26, 2009 #10

    CompuChip

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    Note that max(.., ..) is not in itself a differentiable function. For example,
    [tex]x \mapsto \max(x, -x)[/tex]
    is not differentiable at zero, because it's basically just [itex]x \mapsto |x|[/itex] - even though [itex]x \mapsto x, x \mapsto -x[/itex] are smooth.

    So if you plug two differentiable functions into max(.., ..) you don't necessarily get something differentiable. The point of the exercise is, that a one-sided derivative always exists though (in this case, [itex]u'(x)_\pm = \pm 1[/itex]).
     
  12. Jan 30, 2009 #11
    on what basis i can say that max has always one-sided derivative.

    in your example you had a formula
    [itex]
    u'(x)_\pm = \pm 1
    [/itex]

    but in my question its totally abstract
     
    Last edited: Jan 30, 2009
  13. Jan 30, 2009 #12
    this is the only equations i can construct

    [itex]
    \mathop {\mathop {\lim }\limits_{x \to x_0 - } \frac{{f(x_{} ) - f(x_0 )}}{{x - x_0 }} = \mathop {\lim }\limits_{x \to x_0 + } \frac{{f(x_{} ) - f(x_0 )}}{{x - x_0 }}}\limits_{} \\
    [/itex]
    [itex]
    \mathop {\mathop {\lim }\limits_{x \to x_0 - } \frac{{g(x_{} ) - g(x_0 )}}{{x - x_0 }} = \mathop {\lim }\limits_{x \to x_0 + } \frac{{g(x_{} ) - g(x_0 )}}{{x - x_0 }}}\limits_{} \\
    [/itex]
     
  14. Jan 30, 2009 #13
    "if u(x) = f(x)) then there exists an [itex]\epsilon > 0[/itex] such that u(y) = f(y) for all [itex]y \in [x, x + \epsilon)[/itex]. So then [itex]u'(x)_+ = f'(x)[/itex]."

    i dont know how to apply this this to my question

    i showed the differentiation equations
    i dont know how to continue.
     
  15. Jan 30, 2009 #14

    CompuChip

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    So I think you are missing the geometric picture that I have in mind and therefore cannot make sense of my algebra either :smile:

    Start by drawing two differentiable curves, which may or may not intersect in some point. Pick a point x0 on the x-axis. If the curves don't intersect there, then one will be bigger than the other one in x0 and a small interval around it. So if you consider the maximum of the two, u = max(f(x), g(x)), then u will just be equal to either f(x) or g(x) in some interval around x0, so the derivative of u will just be either f'(x) or g'(x), respectively. Since f and g are both differentiable, you actually have a ("two-sided") derivative.

    However, it is possible that the curves intersect in x0, and that u(x) changes from being equal to f(x) to being equal to g(x) - again: draw a picture! Then the derivative of u(x) will be equal to f'(x) on one side, and to g'(x) on the other side. Since f'(x0) isn't necessarily equal to g'(x0), you cannot really say that the ("two-sided") limit
    [tex]\lim_{x \to x_0} \frac{u(x) - u(x_0)}{|x - x_0|}[/tex]
    exists. However, the claim in your exercise is, that if you only take the limit from one side, that is, consider
    [tex]\lim_{x \downarrow x_0} \frac{u(x) - u(x_0)}{|x - x_0|}[/tex]
    or
    [tex]\lim_{x \uparrow x_0} \frac{u(x) - u(x_0)}{|x - x_0|}[/tex]
    then the limit does exist (and from the above story I hope you see that it will be equal to the derivative of the function along which you are approaching x0).

    Does the question and picture make sense to you now? I think it's important that you understand why it is true, before you try to show how it is true.
     
  16. Jan 30, 2009 #15
    so you basically say
    that maximum will pick either way some function
    and in every case around that x_0 it has to be differentiable because its defind
    to be differentiable on every x.

    correct?
     
  17. Jan 30, 2009 #16

    CompuChip

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    Yes

    Almost, except that it can happen that it picks different functions on different sides of x_0, so you can't take the limit on both sides as in ordinary derivative... but on each side separately it is defined to be equal to a differentiable function.
     
  18. Jan 30, 2009 #17
    so u(x) and v(x) around x_0 have to pick different functions
    because if one is bigger then the other is smaller.

    so if f(x)=u(x) then g(x)=v(x) and once again
    because f(x) and g(x) are differentiable then we have one sided derivative.

    i dont know how to write it thematically but here is a try

    [itex]
    \forall x_0 \in R
    [/itex]
    [itex]
    {\rm{ }}\exists \delta {\rm{ > 0 |x - x}}_0 | < \delta {\rm{ }}\mathop {\lim }\limits_{x \to x_0 } u(x) = \mathop {\lim }\limits_{x \to x_0 } v(x)
    [/itex]
     
    Last edited: Jan 31, 2009
  19. Jan 31, 2009 #18
    how to prove it??
     
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