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Math Help - Problem 22

  1. #1
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    Problem 22

    1)Given a function defined on [a,b], we say it is null iff INT(a,b) f^2(x) dx=0.
    Note, f(x)=0 is null on any interval however the converse is not true.
    Show that if g(x) is any function defined on [a,b] and f(x) is a null function then,
    INT(a,b)g(x)f(x) dx=0

    2)Show that the diameter is the longest chord. (Not necessarily hard I want to see how many reasons you can come up with).

    3)Let a,b be positive integers. Given necessary and sufficient conditions so that,
    (ab)/(a+b)
    Is a positive integer.
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  2. #2
    Super Member malaygoel's Avatar
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    2)
    angle subtended by it is greatest at the centre
    half of diameter is greaterst(=radius, for rest chords, it is sqrt(r^2-a^2), where a is not zero)

    2)
    a=k
    b=k(k-a)
    where k-a+1 divides k

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    Malay
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  3. #3
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    (1)

    It seems false the way you have stated the problem. For consider (ix + n). It can easily be shown that

    Int(a,b) (ix + n)^2 dx = 0

    for

    [a,b] = [-sqrt(3n^2), sqrt(3n^2)]

    But with (g(x) = x) then

    Int(a,b) x(ix + n) dx = 2in^3 * sqrt(3)


    (2)

    Center the circle on the origin and use a simple max value formula:

    (d/dx) chord length = 0 or undefined

    (d/dx) 2 * sqrt(r^2 - x^2) = 0 or undefined

    -2x * (1 / sqrt(r^2 - x^2)) = 0 or undefined

    x = -r, 0, r

    Plug in the values

    2 * sqrt(r^2 - (-r)^2) = 0

    2 * sqrt(r^2 - 0) = 2*r

    2 * sqrt(r^2 - r^2) = 0

    So, the max value is obviously the line (x = 0) which passes through origin, a.k.a the circle's center.

    And finally:

    Chord that passes through center of circle = Diameter

    Done-and-done.
    Last edited by Chronos; April 10th 2007 at 08:23 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Chronos View Post
    (1)

    It seems false the way you have stated the problem. For consider (ix + n).
    I believe he's referring to real valued functions.

    -Dan
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    Oh. Well, if so, then the only valid f(x) would be (f(x) = 0). Because the square of any real function is going to be always positive, thus the only way to have zero area under the curve would be for the function to be a constant zero. And (0 * g(x) = 0).

    I assumed his question was more complicated than that.
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    Super Member Rebesques's Avatar
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    Quote Originally Posted by Chronos View Post

    I assumed his question was more complicated than that.

    That's because you assumed the functions involved are continuous! It's pretty trivial in that case

    If however they are assumed only integrable (as they should - ps. I love Hacker's problems, coz they are almost never "well posed", and that's exactly what makes a problem interesting ) than we can just apply Cauchy-Schwartz:



    Q(uite)E(asily)D(oneth).


    For 2, this came to my mind. Consider a tangent to the circle. Rotating this by the touching point, for an angle s ε (0,π) we obtain a chord of the circle, of length say l(s)>0. This function is defined on (0,π), is continuous, and can be extended to a function continuous on [0,π] by the (obvious) l(0)=l(π)=0. So it must attain a maximum in (0,π). We now see it is increasing for s<π/2 and decreasing for s>π/2, so the maximum is attained for s=π/2 - this chord is exactly the diameter.
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  7. #7
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    Quote Originally Posted by Rebesques View Post
    coz they are almost never "well posed",
    My PDE's are never well posed, however my problems are.
    and that's exactly what makes a problem interesting ) than we can just apply Cauchy-Schwartz:
    Exactly! My solution as well.
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