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A problem involving Riemann Integrals

  1. Oct 22, 2011 #1
    I've been having some trouble with a maths problem and I hoped someone might be able to help.

    We don't seem to have been taught most of what we need to do this, I understand Riemann integrals but what we've been taught and what they're asking for is just different.

    2iuazgx.png

    I could do with a couple of hints or steps to get me in the right direction, it seems we've been taught different stuff to this, or stuff completely unrelated

    Any help at all would be greatly appreciated
     
  2. jcsd
  3. Oct 22, 2011 #2
    The function:
    [tex]
    \Psi(x, \epsilon) \equiv \left( \epsilon \, \vert f(x) \vert - \vert g(x) \vert \right)^2 \ge 0
    [/tex]
    is non-negative for all values of [itex]x[/itex] and [itex]\epsilon[/itex]. But, if you expand it as a polynomial in [itex]\epsilon[/itex], you will get a quadratic function:
    [tex]
    \psi(x, \epsilon) = \epsilon^2 \vert f(x) \vert^2 - 2 \epsilon \, \vert f(x) \vert \, \vert g(x) \vert + \vert g(x) \vert^2 \ge 0
    [/tex]
    But, for [itex]\epsilon > 0[/itex], we have the inequality
    [tex]
    \frac{\epsilon}{2} \, \vert f(x) \vert^2 + \frac{1}{2 \epsilon} \, \vert g(x) \vert^2 \ge \vert f(x) g(x) \vert, \; \forall x \in \mathbb{R}
    [/tex]
    Taking [itex]\epsilon \rightarrow 2\epsilon[/itex] would lead you to an inequality that is very useful for you
     
  4. Oct 22, 2011 #3
    Thank you so much, that actually looks like maths I can understand, not the unrelated gibberish our lecturers seem to enjoy telling us

    It's now almost 1 in the morning and I've been doing maths for the last 6 hours pretty much, I may be back tomorrow with a question about Holders Inequalities (same situation, my lecture notes dont match the questions in the slightest), but i'll have a better look when i've had some sleep

    Thanks again!
     
  5. Oct 23, 2011 #4
    Hey, I have another problem using Riemann Integrals:

    2k1sti.png

    In the case of part (ii), we haven't actually been taught what the symbol of the integral with the line through means or used it in any way

    Again, any help or just a nudge in the right direction would be much appreciated, this module has been great for giving us questions unrelated to what we've been taught
     
  6. Oct 23, 2011 #5
    Let's focus on the first problem. Did you solve it? If yes, post the solution here.
     
  7. Oct 23, 2011 #6
    Thats a good idea, here's what i've got (I'm no good with Latex so ill write it out longhand, say if anything is unclear)

    Given
    a
    ∫ (ε|f(x)| - |g(x)||2dx≥0 for all x, ε,
    b

    We can expand this to:
    a
    ∫ε2|f(x)|2-ε|f(x)||g(x)|-ε|f(x)||g(x)|+|g(x)|2dx
    b

    =
    a
    ∫ε2|f(x)|2-2ε|f(x)||g(x)|+|g(x)|2dx
    b

    So we know:
    a
    ∫ε2|f(x)|2-2ε|f(x)||g(x)|+|g(x)|2dx ≥ 0
    b

    then
    a
    ∫ε2|f(x)|2+|g(x)|2dx ≥
    b


    a
    ∫2ε|f(x)||g(x)|dx
    b

    Then dividing by 2ε:
    a
    ∫(ε/2)|f(x)|2+(1/2ε)|g(x)|2dx
    b

    a
    ∫|f(x)||g(x)|dx
    b

    We now need to substitute ε by 2ε:
    a
    ∫ε|f(x)|2+(1/4ε)|g(x)|2dx
    b

    =
    a
    ∫ε|f(x)|2dx +
    b

    a
    ∫(1/4ε)|g(x)|2dx ≥
    b

    a
    ∫|f(x)||g(x)|dx
    b

    =a
    ε∫|f(x)|2dx +
    _b

    _____a
    (1/4ε)∫|g(x)|2dx ≥
    _____b

    a
    ∫|f(x)g(x)|dx
    b

    or
    a
    ∫|f(x)g(x)|dx ≤
    b

    =a
    ε∫|f(x)|2dx +
    _b

    _____a
    (1/4ε)∫|g(x)|2dx
    _____b

    as required.

    Sorry that's such a mess, must teach myself how to represent the symbols using latex :P
     
    Last edited: Oct 23, 2011
  8. Oct 23, 2011 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That's because it is not a standard symbol. It is defined in this problem: it means "the integral of f from a to b divided by b- a" just as you are told.

     
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