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Comparing two functions

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to show that the function f(x)=[itex]\frac{1}{x^{2}}[/itex] is greater than g(x)=[itex]\frac{1}{x}[/itex]-ln(1+[itex]\frac{1}{x}[/itex]) for all x>0.


    2. Relevant equations



    3. The attempt at a solution
    This is really simple, but I'm having some trouble showing it rigorously. I thought about trying to compare the derivatives, but it's not that simple. Are there any good ways to do this?
     
  2. jcsd
  3. Oct 1, 2012 #2

    SammyS

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    Well... then the function (f-g)(x) must be positive ...

    the function (f/g)(x) must be greater than 1, or be negative ...

    the function (g/f)(x) must be less than 1, or be negative.
     
  4. Oct 1, 2012 #3
    Right. That was the other method I tried. I just can't figure out a way to do it without some "hand-waving."
     
  5. Oct 2, 2012 #4

    jbunniii

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    Not sure if this simplifies anything, but if [itex]0 < x < 1[/itex], then [itex]1/x^2 > 1/x[/itex] so clearly [itex]f(x) > g(x)[/itex]. So you only have to worry about [itex]x > 1[/itex].
     
  6. Oct 2, 2012 #5

    HallsofIvy

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    Using that, a fourth method: it is easy to show this is true for x= 1 so if you can prove that f'(x)> g'(x) for all x>1 you are done.
     
  7. Oct 2, 2012 #6

    SammyS

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    How about comparing f(1/x) and g(1/x) ?
     
  8. Oct 13, 2012 #7
    Hmmm, OK, how does this help me?
     
  9. Oct 13, 2012 #8

    Ray Vickson

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    Well, derivatives are a bit simpler, etc.

    RGV
     
  10. Oct 14, 2012 #9
    I've been able to show that f(1/x)>g(1/x) for all x>1. From this, how exactly can I conclude that f(x)>g(x)?
     
  11. Oct 14, 2012 #10

    SammyS

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    That shows that f(x)>g(x) for 0 < x < 1, which isn't a big help.
     
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