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Homework Help: Differentiable function with g(0) = 0 and etc

  1. Apr 3, 2010 #1
    1. The problem statement, all variables and given/known data

    I apologize for not knowing how to use Latex, so I will type the problem as it is read...


    Prove that for all x greater than or equal to 0, we have

    the integral from 0 to x for [g(x)]^3 dx which is less than or equal to

    (the integral from 0 to x for g(x) dx)^2.

    Suppose that g is a differentiable function with g(0) = 0 and 0 is less than g prime which is less than or equal to 1.

    Also, the brackets around the g(x) are supposed to be there, so it is to the power 3, as in C^3. The parentheses around the second part of the problem are there to show it is to the power 2.


    Sorry for the non-Latex format.


    2. Relevant equations

    I do not know where to start, but think it involves MVT (the Mean Value Theorem) and/or Rolle's Theorem.


    3. The attempt at a solution

    I tried working with a square root, but was unsure. My skills are rusty.
    Thanks for any help.
     
  2. jcsd
  3. Apr 4, 2010 #2

    Mark44

    Staff: Mentor

    With LaTeX, I think this is what you're trying to say. Double-click any of the integrals I have to see how I wrote the LaTeX script.

    Suppose g(0) = 0 and that g is differentiable, with 0 < g'(x) <= 1.
    Show that, for all x > 0,
    [tex]\int_0^x (g(t))^3 dt <= \int_0^x (g(t))^2 dt [/tex]

    Note that definite integrals that have a variable in the limit of integration almost always use a different variable (a so-called dummy variable) as the variable of integration.

    I think you're going to have trouble proving this statement, since it doesn't seem to be true. As a counterexample, consider g(t) = t/2. This function satisfies g(0) = 0, is differentiable, and satisfies 0 < g'(t) <= 1.

    [tex]\int_0^x (t/2)^3 dt = \frac{x^4}{32}[/tex]

    [tex]\int_0^x (t/2)^2 dt = \frac{x^3}{12}[/tex]

    For x suitably large, x4/32 > x3/12, so this function contradicts the statement you're supposed to prove.

    Have you left out anything?
     
  4. Apr 4, 2010 #3

    Office_Shredder

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    That's because the inequality is supposed to be

    [tex]
    \int_0^x (g(t))^3 dt \leq (\int_0^x g(t)dt)^2
    [/tex]

    Noticing that both functions are 0 at 0, you might consider showing that the derivative of one is larger than the derivative of the other
     
    Last edited: Apr 4, 2010
  5. Apr 4, 2010 #4
    Thank you Office_Shredder and Mark44.

    Office_Shredder has it written correctly, the part about the parentheses is correct, but it is g(x) dx, not with t's. Though that shouldn't make a difference, right??

    Mark44, thank you as well, your Latex writing of the problem is correct, although it is for all x[tex]\geq[/tex] 0.

    Your math work is helpful, but since the second part of the problem is different, it changes things.

    So, can I just do ...

    [tex]
    \int_0^x (g(t))^3 dt \leq (\int_0^x g(t)dt)^2
    [/tex]

    which is then,
    [tex]\frac{x^4}{32}\leq(\frac{x^2}{4})^2
    [/tex]

    which is...

    [tex]\frac{x^4}{32}\leq\frac{x^4}{16}
    [/tex]

    I have my example now, but how do I go about writing a proof for it?
    Also, thanks again with the Latex help, as well as even looking at my problem.
     
  6. Apr 4, 2010 #5

    Office_Shredder

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    You can answer the problem by taking derivatives. If you have two functions f(x) and h(x) such that f(0)=h(0), if f'(x)<h'(x) for all x>0, then f(x) < h(x) (if you're not sure why consider a proof by contradiction and the mean value theorem)

    So we have two functions (the integrals) and want to show one is smaller than the other, knowing that they are both equal at zero. Try taking the derivative of both and showing one is smaller than the other
     
  7. Apr 4, 2010 #6
    Office_Shredder....

    So, if I use a simple function of just x, then the derivative of it is...

    3x^2 [tex]\leq[/tex] (1)^2
    which is not true, if my values for x [tex]\geq 0[/tex] .


    Is that what you mean (did I do that right)? Is that the contradiction you were referring to, since that is true when x [tex]
    \leq 0
    [/tex] ?
     
  8. Apr 4, 2010 #7

    Mark44

    Staff: Mentor

    No, you can't substitute a particular function for g(x), such as g(x) = x. All you know about g is that it is differentiable and 0 < g'(x) <= 1. When I used a specific function, I was giving a counterexample (which wasn't actually a counterexample because I hadn't interpreted the problem correctly). To show that a statement is false, any specific function that is a counterexample will do, but to show that the statement is true, you cannot use specific functions.
     
  9. Apr 4, 2010 #8

    Office_Shredder

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    I'm confused what functions you took the derivative of. If your function is x, you're asked to compare [tex] \frac{x^4}{4}[/tex] (the integral of t3) and [tex]\frac{x^4}{4}[/tex] (the square of the integral of x) which certainly pass the derivative test I proposed.

    Do you know what the derivatives of [tex] \int_0^x g(x)^3 dx[/tex] and [tex]( \int_0^x g(x)dx)^2[/tex] are?
     
  10. Apr 4, 2010 #9
    If you mean the derivatives of the integrals==> the Fundamental Theorem of Calc., then is not just g(x)^3 and (g(x))^2?
     
  11. Apr 4, 2010 #10

    Office_Shredder

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    The first one is right but the second derivative isn't. You're squaring the integral so you have to use the chain rule
     
  12. Apr 4, 2010 #11
    Then, just 2g(x)?
     
  13. Apr 5, 2010 #12

    Mark44

    Staff: Mentor

    No, that's not it either.

    You have [tex]F(x) = \left(\int_0^x g(t) dt \right)^2[/tex]

    What's F'(x)? It's not 2g(x).

    That's sort of like saying that if H(x) = (f(x))2, then H'(x) = 2f(x)
     
  14. Apr 5, 2010 #13
    I am getting confused a bit.
    Let's see, the derivative of the integral of g(x) dx is just g(x), right?

    So, then, with the square on it is it, g2(x)g'(x)
    ???
     
  15. Apr 5, 2010 #14

    Mark44

    Staff: Mentor

    [tex]F'(x) = 2\left(\int_0^x g(t) dt \right) \frac{d}{dx} \left(\int_0^x g(t) dt \right)[/tex]

    Isn't that how the chain rule works?
     
  16. Apr 5, 2010 #15
    Oops. I misplaced my 2. Thanks.

    So, I was thinking about the application of the MVT...
    Any ideas, hints if I am in the right direction, because I did a similar problem that made me work in the opposite direction of this one, using MVT.

    It read: let a,b be differentiable on [tex]\Re[/tex] and suppose that a(0) = b(0) and a'(x)[tex]\leq[/tex] b'(x) for all x [tex]\geq[/tex] 0. Show that a(x) [tex]\leq[/tex] g(x) for all x[tex]\geq[/tex] 0.

    The solution: let x[tex]\geq[/tex] 0. The function b(x) - a(x) is continuous on [0,x] and differentiable on (0,x). By MVT, [tex]\exists[/tex] c [tex]\in[/tex] (0,x) st (b(x)) - a(x)) - (b(0) - a(0)) = (b'(c) - a'(c)) *(x-0) [tex]\geq[/tex] 0, since both terms on the RHS are positive.
    Since b(x) - a(x) [tex]\geq[/tex] 0, a(x) [tex]\leq[/tex] b(x) for all x[tex]\geq[/tex] 0.
     
  17. Apr 5, 2010 #16

    Office_Shredder

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    So once you take the derivatives the question is why is

    [tex]g(x)^2 \leq 2 \int_0^x g(x)dx[/tex]

    Again, you have two functions that are 0 at 0. Once you take the derivative again it becomes obvious which one is larger than the other, then just work your way back up with conclusions
     
  18. Apr 12, 2010 #17
    I found this problem elsewhere, so I was wondering how to prove a specific part of it to be positive, because my teacher wanted a better explanation of it.

    http://www.mathlinks.ro/viewtopic.php?t=325915

    My questions were posted in the forum there as well.

    Also, I realize that I misread a portion of someone's post when I asked the question about g'(x) and h(x).
    So, how do I know that h(x) and h'(x) are positive, and/or the whole problem is positive then?
    Thanks to all.
     
  19. Apr 12, 2010 #18
    I figure now to just subtract the left side from the right, setting the equation to f(x).


    Then, take the first derivative of it, which is g(x) [2[tex]\int[/tex]0^x g(t) dt - (g(x)2)]

    To prove f' increases, use the fact that g'[tex]\geq[/tex] 0, demonstrating g(x) increases.

    So, it can be said that h(x)=f'(x)= 2[tex]\int 0^x [/tex] g(t) dt - (g(x)2) increases.


    Taking h'(x) = 2g(x) - 2g'(x)g(x) = 2g(x)(1-g'(x)) >= 0, because of the problem's conditions.

    My teacher wants a better explanation of this last part, and I don't know what else to do.
    I essentially need to prove that every part is positive.
    I offered this as an explanation...
    If g'(0)=0, then f'(0)=0; thus, f'(x)>=0 implies f"(0)>=0.
    (My teacher said I should be able to get away with not having to take the second derivative.)
    Also, since g'(x)>=0 ==> g(x)>=g(0)=0


    Thanks for looking at it.
     
  20. Apr 13, 2010 #19

    Mark44

    Staff: Mentor

    This thread is getting long and seems to be getting afield from the OP, so it's probably reasonable to recap things.

    Given: g(0) = 0, 0 < g' <= 1
    Show that
    [tex]\int_0^x g^3(t)dt \leq \left( \int_0^x g(t)dt \right)^2[/tex]

    Since g is an increasing function (g' > 0), it turns out that we can look the derivatives of the two functions above (the two integrals, which are functions of x). If we can show that the derivative of the integral on the left is <= the derivative of the integral on the right, that will prove the inequality we're trying to prove.

    Equivalently we can show that the difference of the derivatives of the two integrals is <= 0, so lets' look at that difference.

    [tex]d/dx \int_0^x g^3(t)dt - d/dx \left( \int_0^x g(t)dt \right)^2[/tex]
    [tex]= g^3(x) - 2 \int_0^x g(t)dt~d/dx \left(\int_0^x g(t)dt\right) [/tex]
    [tex]= g^3(x) - 2g(x) \int_0^x g(t)dt [/tex]
    [tex]= g(x)[g^2(x) - 2 \int_0^x g(t)dt] [/tex]

    We know that g(x) >= 0, so to show that the expression above is <= 0, we need to show that the expression in parentheses is <= 0. Using Office_Shredder's suggestion, take the derivative with respect to x of
    [tex](g^2(x) - 2 \int_0^x g(t)dt) [/tex]

    and show that this quantity is <= 0.
     
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