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Math Help - Area Under an Inverted Catenary

  1. #1
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    Area Under an Inverted Catenary

    Greetings all, this is my first post on these forums.

    I'm attempting to construct an inverted catenary that will intersect the x-axis at y=0 and y=1 and have an area of 1 bound by the x-axis and the function.

    I started off with f(x)=cosh[1.5]-cosh[3x-1.5] and that satisfies the intersection points, but does not have an area of 1. So to get the area closer to one, I kept manipulating the numbers and integrating, until I came to f(x)=cosh[1.543405]-cosh[3.08681*x-1.543405]

    This gives an area of 1.0000 under the curve bound by the x-axis, but I'm sure it's not actually equalling 1, I've just reached the limit of my program's (winplot) decimal display.

    My goal is to find a more systematic approach, since repeatedly adjusting the numbers isn't exact. I wonder if there could be some sort of series representation for the function. I don't recognize the number 1.543405 as anything special, but perhaps it could be respresented exactly, rather than a decimal (eg. to say pi rather than 3.141592) and yield the proper results.

    Thanks in advance for any help!
    Cheers.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by yfronto View Post
    Greetings all, this is my first post on these forums.

    I'm attempting to construct an inverted catenary that will intersect the x-axis at y=0 and y=1 and have an area of 1 bound by the x-axis and the function.

    I started off with f(x)=cosh[1.5]-cosh[3x-1.5] and that satisfies the intersection points, but does not have an area of 1. So to get the area closer to one, I kept manipulating the numbers and integrating, until I came to f(x)=cosh[1.543405]-cosh[3.08681*x-1.543405]

    This gives an area of 1.0000 under the curve bound by the x-axis, but I'm sure it's not actually equalling 1, I've just reached the limit of my program's (winplot) decimal display.

    My goal is to find a more systematic approach, since repeatedly adjusting the numbers isn't exact. I wonder if there could be some sort of series representation for the function. I don't recognize the number 1.543405 as anything special, but perhaps it could be respresented exactly, rather than a decimal (eg. to say pi rather than 3.141592) and yield the proper results.

    Thanks in advance for any help!
    Cheers.
    Clearly, as you noted, f(x)=\cosh(a)-\cosh\left(2ax-a\right) will do. So we need to solve \int_0^{1}\left\{\cosh(a)-\cosh\left(2ax-a\right)\right\}dx=1\implies \cosh(a)-\frac{\sinh(a)}{a}=1 this has no closed form solution though. - Wolfram|Alpha. If you want to get close numerical solutions to this use the fact that \cosh(a)=\sum_{n=0}^{\infty}\frac{a^{2n}}{(2n)!},\  sinh(a)=\sum_{n=0}^{\infty}\frac{a^{2n+1}}{(2n+1)!  }. If you only take the first, say 4, terms of these series you can find an approximate closed form solution.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Clearly, as you noted, f(x)=\cosh(a)-\cosh\left(2ax-a\right) will do. So we need to solve \int_0^{1}\left\{\cosh(a)-\cosh\left(2ax-a\right)\right\}dx=1\implies \cosh(a)-\frac{\sinh(a)}{a}=1 this has no closed form solution though. - Wolfram|Alpha. If you want to get close numerical solutions to this use the fact that \cosh(a)=sum_{n=0}^{\infty}\frac{a^{2n}}{(2n)!},\s  inh(a)=\sum_{n=0}^{\infty}\frac{a^{2n+1}}{(2n+1)!}.
    That makes sense. I was just having trouble connecting the dots. Thanks!
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