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Thread: Fake proof that cos(t) is not bounded

  1. #1
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    Fake proof that cos(t) is not bounded

    Using Laplace transforms I created a proof that cos(t) is not bounded above. I cannot find where the error is even though the proof doesn't use very complicated theorems. I am not so familiar with the limitations of Laplace transforms so I suspect that the error is at the start or the end where I use Laplace.

    If cos(t) is not bounded above then for any c>1 there exists some t such that cos(t)=c

    Taking the Laplace transform of the equation

    \mathcal{L} \left\{ cos(t) \right\} = \mathcal{L} \left\{ c \right\}

    \frac{s}{s^2+1} = \frac{c}{s}

    Assuming s is not equal to zero

    \frac{s^2}{s^2+1} = c

    There is no solution for real s, instead consider a purely complex value s=xi, where x \in R \backslash 0

    \frac{-x^2}{-x^2+1} = c

    Plotting \frac{x^2}{x^2-1} gives the following graph:

    Fake proof that cos(t) is not bounded-wolframplot.png

    As can be seen from the graph, the function is continuous for x>1

    Now using the following limits:

    \lim_{x \to 1^+}\frac{x^2}{x^2-1} =\infty

    and
    \lim_{x \to \infty}\frac{x^2}{x^2-1} =1

    Since c \in (1,\infty) then by the intermediate value theorem there exists some x_0 \in (1,\infty) such that \frac{x^2}{x^2-1} =c

    Therefore a solution in the Laplace domain s_0=x_0i exists and this implies that a solution in the time domain exists for some t>0

    Is anyone able to spot the error in my reasoning? I didn't properly prove that the function was continuous when applying the intermediate value theorem but I don't think this is the blunder.
    Attached Thumbnails Attached Thumbnails Fake proof that cos(t) is not bounded-wolframplot.png  
    Last edited by Shakarri; Jan 26th 2016 at 03:27 AM.
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    Re: Fake proof that cos(t) is not bounded

    Quote Originally Posted by Shakarri View Post
    There is no solution for real s...
    \frac{s^2}{s^2 + 1} = c has the solution s = \sqrt{\frac{c}{1 - c}} for 0 < c < 1. (Of course you can extend this to -1 < c < 0.) Thus -1 < c < 1 are the only real possibilities. So your initial statement "If cos(t) is not bounded above then for any c>1 there exists some t such that cos(t) = c" can't be true for real c and you have your contradiction.

    -Dan
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  3. #3
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    Re: Fake proof that cos(t) is not bounded

    Thanks Dan, I always thought that since s can be complex that solutions in the Laplace domain could be complex and still correspond to a positive real solution in the time domain.
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