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Dy/dx = 0

  1. Aug 16, 2008 #1
    If dy/dx=0,
    for all x in the function y= f(x)'s domain,then
    how can we say... only by using limits, that the function is a constant one?
     
  2. jcsd
  3. Aug 16, 2008 #2

    morphism

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    Only by using limits? What's wrong with the way this is usually proved, i.e. via the mean value theorem?
     
  4. Aug 16, 2008 #3
    the mean value theorem here....im not sure...cuz we might get a limit of the function staying constant..but i dont see how the function is EXACTLY constant, being proved by only using the first principle...

    thanks
     
  5. Aug 16, 2008 #4
    As far as a limit definition:

    dy/dx = lim[ F(x+h) - F(x)]/h

    In order for that to be true, then:

    F(x+h) - F(x) = 0

    =>

    F(x+h) = F(x)

    Which means that you have an equation which does not vary with respect to x, otherwise there would be something left over from that subtraction. The only equation which does not vary with respect to x is a function that is constant, relative to x.
     
  6. Aug 16, 2008 #5
    cant F(x+h)-F(x) be <<h ...I dont think it can as h is a dynamic variable,which is infinitesimally small...so am i correct
    Please correct me if im wrong
    thanks
     
  7. Aug 17, 2008 #6

    nicksauce

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    If (A-B)/C = 0, and C is not 0, then it has to follow that A=B.
     
  8. Aug 17, 2008 #7
    But this is a limit

    i mean lim(a--->b)a-b/c=0

    it duznt mean a=b!
     
  9. Aug 17, 2008 #8
    No, it's lim(c -> 0) of (a-b)/c
     
  10. Aug 17, 2008 #9
    ok...i got my mistake...but cant [tex]a-b[/tex] be something like [tex]c^2[/tex]

    then the limit still is 0 but we cant say a=b...can we?
     
  11. Aug 17, 2008 #10

    HallsofIvy

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    In order that this be true, f must be differentiable at every point in the interval. Your limit calculations are not using that.
     
  12. Aug 17, 2008 #11

    morphism

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    This is not true. Take F(x)=x^2 for instance, and apply the limit definition at x=0:

    [tex]F'(0) = \lim_{h \to 0} \frac{F(0+h) - F(0)}{h} = \lim_{h \to 0} \frac{h^2}{h} = 0.[/tex]

    But F(0+h)-F(0)=h^2 is not zero for all h.

    This is why you need to use a deeper result like the mean value theorem. It tells you that if x is not equal to y, then there is a c such that

    [tex]\frac{F(x) - F(y)}{x - y} = F'(c) = 0 \implies F(x) = F(y).[/tex]

    [Note: I'm assuming that F is differentiable everywhere so that the hypotheses of the MVT are satisfied.]
     
  13. Aug 17, 2008 #12
    Is it correct to simply say ..that in a function we cant have an infinitesimal so the limit of the derivative is not zero unitl it IS constant
    or like we have a function [tex]f(x)=a(x)-b(x)[/tex]
    if a`(x)=b`(x) for all x

    then a(x)=b(x) necessarily using the above argument(infinitesimals)
     
    Last edited: Aug 17, 2008
  14. Aug 17, 2008 #13

    morphism

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    That doesn't really make any sense.
     
  15. Aug 17, 2008 #14
  16. Aug 18, 2008 #15
    So how would we exactly prove the required result using only limits and logic?

    thanks
     
  17. Aug 18, 2008 #16

    HallsofIvy

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    Use limits (and logic) to prove the mean value theorem! One point that has been made repeatedly here is that you can't use limits at a single point because saying that y= 0 identically (on some interval) is a property of that interval, not a single point.
     
  18. Aug 18, 2008 #17
    ok...i get your point:smile:

    but what if we have a function like...
    [tex]
    f(x+y)=f(x)+y^3[/tex]
    and it is provided that the function is continuous and differentiable
    so this gives us
    [tex]
    f(x+h)=f(x)+h^3[/tex]

    or
    [tex]
    \frac {f(x+h)-f(x)}{h}=h^2[/tex]

    [tex]
    \lim_{h \to 0} \frac {f(x+h)-f(x)}{h}=0[/tex]

    but i dont think the function is constant....
    is there anything wrong in the function?
     
  19. Aug 18, 2008 #18

    HallsofIvy

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    What reason do you have to believe such a function exists? If so, taking x= 0, f(y)= y3. But then
    [tex]f(x+y)= (x+y)^3= x^3+ 3x^2y+ 3xy^2+ y^3= f(x)+ 3x^2y+ 3xy^2+ y^3\ne f(x)+ y^3[/tex]
    a contradiction.

     
  20. Aug 18, 2008 #19
    I agree.. that the function wont exist...

    but,using the MVT arent we still just proving the tendency not the EXACT equality?
    i know this is irritating

    but

    thanks anyways
     
  21. Aug 18, 2008 #20

    morphism

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