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Functional Inequation

  1. Feb 9, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    The function f satisfies [itex] \dfrac{f(x)}{f(y)} \leq 2^{(x-y)^2} x,y \in D [/itex] where D denotes domain set of the function, then f(x) can be

    I have a set of options as well but I'm not posting it now. I will post it if required, later.

    3. The attempt at a solution
    I have dealt with functional equations but this seems more daunting as it is rather an inequation. First, to simplify it, I take natural logarithm of both sides. Then

    [itex] log f(x) - logf(y) \leq (x-y)^2 log2 [/itex]

    My next thought is to differentiate the expression wrt x. But I'm not sure whether it would be helpful as I don't want to do it uselessly.
     
  2. jcsd
  3. Feb 9, 2014 #2

    pasmith

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    "Can be", rather than "is", suggests that checking which of those options satisfies this condition is the way forward.
     
  4. Feb 9, 2014 #3

    utkarshakash

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    So how should I check the given options?
     
  5. Feb 9, 2014 #4

    pasmith

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    Calculate [itex]f(x)/f(y)[/itex] for each case, and check whether the result is less than or equal to [itex]2^{(x-y)^2}[/itex].
     
  6. Feb 9, 2014 #5

    Mark44

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    One other thing is that, since (x - y)2 ≥ 0 for any real x and y, it must be true that f(x)/f(y) ≥ 20 = 1. You didn't show what the options are, but if there are any that are less than 1, you can eliminate them from further consideration.

    Edit: Never mind on the above. I was looking at the wrong direction of the inequality in post #1.

    BTW, we don't call it an "inequation" - we call it an inequality.
     
    Last edited: Feb 9, 2014
  7. Feb 9, 2014 #6

    pasmith

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    I'm not sure how that follows from [itex]\dfrac{f(x)}{f(y)} \leq 2^{(x-y)^2}[/itex].
     
  8. Feb 9, 2014 #7

    Mark44

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    Good point. I must have gotten my inequality sign turned around.
     
  9. Feb 9, 2014 #8

    utkarshakash

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    For example, one option is

    [itex] \int_0^x 2t^3 dt [/itex]

    This function is x^4/2. You are saying to simply plug x in one and y in another. Doing that gives

    [itex] \left( \dfrac{x}{y} \right) ^4 [/itex]

    Now how do I check whether this is less than [itex]2^{(x-y)^2} [/itex] or not? Should I start substituting some random values?
     
  10. Feb 9, 2014 #9

    haruspex

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    You don't say what D is. Is it specified separately for each f option?
    Since the right-hand side of the inequation becomes weak when x and y are far apart, and the inequation is trivially true when x = y, I would concentrate on y and x differing by a small amount.
     
  11. Feb 9, 2014 #10

    Ray Vickson

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    You can look at the function
    [tex] F(x,y) = \left(\frac{x}{y}\right)^4 - 2^{(x-y)^2},[/tex]
    and try to maximize it, to see if its maximum is ≤ 0.
     
    Last edited: Feb 9, 2014
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