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Solving a Putnam Equation

  1. Dec 3, 2004 #1
    I'll describe a problem from William Lowell Putnam Mathematical Competition (54th, problem A1):

    There is an image of the graph y = sin x where x ranges from 0 to pi. The graph of y = k where 0 < k < 1 is also drawn there.

    Clearly y = k intersects y = sin x at two places, call them (x1, k) and (x2, k).

    Let us define

    A1 = area bounded above by y = k, below by y = sin x, left by x = 0, and right by x = x1

    A2 = area bounded above by y = sin x, below by y = k (the left endpoint is at x = x1 and the right endpoint is at x = x2)

    Now we must find k (NOTE by me: the problem asks us to find, not approximate) such that that A1 = A2.

    I reduced this problem to the equation:

    [tex]
    \sin{x_1} = \frac{2}{\pi - 2 + 2x_1}
    [/tex]

    Where k = sin x1. (try to derive this equation :) )

    I can approximate x1, and thus k, using a graphing calculator (and I think it's possible to approximate it using newton's method) but I can't see how I can find the exact value of k. Is it even possible? If so, how?

    Thank you.
     
  2. jcsd
  3. Dec 3, 2004 #2

    AKG

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    You know that sin is symmetric about [itex]\pi /2[/itex]. Define t such that sin(t) = k. So, we want to find:

    [tex]A_1 = A_2[/tex]

    [tex]2\int _0 ^t k - \sin (x)\, dx = \int _t ^{\pi - t} \sin (x) - k\, dx[/tex]

    [tex]2\left (kx + \cos (x)\right )_0 ^t = \left (-\cos (x) - kx\right ) _t ^{\pi - t}[/tex]

    [tex]2(kt + \cos (t) - 1) = -\cos (\pi - t) - k(\pi - t) + \cos (t) + kt[/tex]

    From here, it's just some simple algebra to solve for k. You'll need to use the fact that sin(t) = k, and [itex]\cos (a + b) = \cos (a) \cos (b) - \sin (a) \sin (b)[/itex].
     
  4. Dec 3, 2004 #3

    StatusX

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    the simplest equation I could get was:

    [tex] \cos(x_2) + x_2 \sin(x_2) = 1 [/tex]

    This gives an approximate answer for x2 of 2.33112237. I don't think this has a closed form solution.
     
  5. Dec 3, 2004 #4

    Hurkyl

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    You sure you have the right problem? That doesn't look like A1 of the 54th exam...
     
  6. Dec 3, 2004 #5

    AKG

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    Oops, I didn't even read the problem correctly. My approach should be modified to start with:

    [tex]\int _0 ^t k - \sin (x)\, dx = \int _t ^{\pi - t} \sin (x) - k\, dx[/tex]

    which makes the problem much harder.
     
  7. Dec 3, 2004 #6

    AKG

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    I have a solution, but something is wrong with it, but what?:

    [tex]\int _0 ^t k - \sin (x)\, dx = \int _t ^{\pi - t} \sin (x) - k\, dx[/tex]

    [tex](kx + \cos (x))_0 ^t = (-\cos (x) - kx)_t ^{\pi - t}[/tex]

    [tex]kt + \cos (t) - 1 = -\cos (\pi - t) - k(\pi - t) + \cos (t) + kt[/tex]

    [tex]-1 = -(\cos (\pi )\cos (t) + \sin (\pi )\sin (t)) - k\pi + kt[/tex]

    [tex]-1 = \cos (t) - k\pi + kt[/tex]

    [tex]\cos (t) = k(\pi - t) - 1[/tex]

    [tex]1 - \sin ^2 (t) = k^2(\pi - t)^2 - 2k(\pi - t) + 1[/tex]

    Recall that sin(t) = k, so:

    [tex]- \sin ^2(t) = \sin ^2(t)(\pi - t)^2 - 2\sin (t)(\pi - t)[/tex]

    [tex]\sin ^2 (t) ((\pi - t)^2 + 1) = 2\sin (t)(\pi - t)[/tex]

    [tex]\sin (t) ((\pi - t)^2 + 1) = 2(\pi - t)[/tex]

    [tex]\sin (t) = \frac{2(\pi - t)}{(\pi - t)^2 + 1}[/tex]

    Now, note that:

    [tex]\sin (\pi - t) = \sin (\pi )\cos (t) - \cos (\pi )\sin (t)[/tex]

    [tex]\sin (\pi - t) = 0(\cos (t)) - (-1)\sin (t) = \sin (t)[/tex]

    Therefore:

    [tex]\sin (\pi - t) = \frac{2(\pi - t)}{(\pi - t)^2 + 1}[/tex]

    Let y = [itex]\pi - t[/itex]:

    [tex]\sin (y) = \frac{2y}{y^2 + 1}[/tex]

    Differentiate with respect to y:

    [tex]\cos (y) = \frac{2y^2 + 2 - 4y^2}{(y^2 + 1)^2}[/tex]

    [tex]\cos (y) = \frac{2(1 - y^2)}{(y^2 + 1)^2}[/tex]

    Now, we know that [itex]\sin ^2 (y) + \cos ^2 (y) = 1[/itex], so:

    [tex]1 = \left (\frac{2(1 - y^2)}{(y^2 + 1)^2}\right )^2 + \left ( \frac{2y}{y^2 + 1}\right )^2[/tex]

    [tex]1 = \frac{4 - 8y^2 + 4y^4}{(y^2 + 1)^4} + \frac{4y^2}{(y^2 + 1)^2}[/tex]

    [tex]1 = \frac{4 - 8y^2 + 4y^4}{(y^2 + 1)^4} + \frac{4y^2(y^2 + 1)^2}{(y^2 + 1)^4}[/tex]

    [tex](y^2 + 1)^4 = 4 - 8y^2 + 4y^4 + 4y^6 + 8y^4 + 4y^2[/tex]

    [tex]y^8 + 4y^6 + 6y^4 + 4y^2 + 1 = 4y^6 + 12y^4 - 4y^2 + 4[/tex]

    [tex]y^8 - 6y^4 + 8y^2 - 3 = 0[/tex]

    [tex](y^2 + 3)(y^2 - 1)^3 = 0[/tex]

    [tex]y = \pm 1[/tex]

    [tex]\pi - t = \pm 1[/tex]

    [tex]t = \pi \pm 1[/tex]

    Of course, [itex]t < \pi[/itex], so we have:

    [tex]t = \pi - 1[/tex]

    Now, substitute this into the earlier equation:

    [tex]\cos (t) = k(\pi - t) - 1[/tex]

    Recall that k = sin(t), so:

    [tex]\cos (\pi - 1) = \sin (\pi - 1) (\pi - (\pi - 1)) - 1[/tex]

    [tex]\cos (\pi )\cos (1) + \sin (\pi )\sin (1) = (\sin(\pi )\cos (1) - \cos (\pi )\sin (1)) - 1[/tex]

    [tex]-\cos (1) = \sin (1) - 1[/tex]

    [tex]\cos (1) + \sin (1) = 1[/tex]

    Which is wrong. So where did I mess up?
     
    Last edited: Dec 3, 2004
  8. Dec 4, 2004 #7
    Wrong Equation

    Sorry guys I got my equation wrong. Here's the correct one:

    We use the fact that

    [itex]A_1 = x_1\sin{x_1} - \int_0^{x_1} \sin x dx[/itex]

    and

    [itex]A_2 = \int_{x_1}^{\pi-x_1}\sin x dx - (\pi - 2x_1)\sin{x_1}[/itex]

    By setting A1 = A2 and rearranging, we get:

    [itex](\pi - x_1)\sin x_1 = \int_0^{\pi-x_1}\sin x dx[/itex]

    so

    [itex](\pi - x_1)\sin x_1 = \left[-\cos x\right]_0^{\pi-x_1}[/itex]
    [itex](\pi - x_1)\sin x_1 = -\cos(\pi-x_1) + 1[/itex]

    using the identity cos(pi - x) = -cos x and rearranging we get

    [itex](\pi - x_1)\sin x_1 -\cos x_1 - 1 = 0[/itex]

    In which I don't see how to find the exact value of x1.

    The problem is from Howard Anton's Calculus 7th edition, Supplementary exercise no. 19. It says that "This exercise is based on Problem A1 of the Fifty-Fourth Annual William Lowell Putnam Mathematical Competition". Well, the author said "based on" so it could be different (not exactly the same) :). A mistake on my part to misinterpret it.
     
  9. Dec 4, 2004 #8
    My last equation gave [itex]x_1[/itex] approximately 0.8105 so k is approximately 0.7246. How did you derive your equation, StatusX?
     
  10. Dec 4, 2004 #9

    StatusX

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    Since y is a constant, the derivative of both sides is 0. What you're doing is finding values of y where the slope of both functions are equal, but it would be a pure coincidence if they also happened to intersect at these points. Try to solve x = x^2 this way. You get 1 = 2x, and x=1/2.
     
  11. Dec 4, 2004 #10

    StatusX

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    our answers agree. x2 = [tex]\pi[/tex] - x1


    the way i derived it was just by solving:

    [tex] \int_0^{x_2} \sin(t) - \sin(x_2) dt = 0[/tex]
     
    Last edited: Dec 4, 2004
  12. Dec 4, 2004 #11
    Aaaah.... now THAT was simple & elegant :)
     
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