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Thread: Hyperbolic Functions

  1. #1
    Junior Member Lonehwolf's Avatar
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    Hyperbolic Functions

    Well, I was a lousy ass during the lessons so I ended up quite rundown by a truck when I checked out the questions.. private lessons turned out to help a great deal, but i'm stuck with this one question that i really can't move past it (i skipped it for the time being but would like getting it solved for once :P)

    Solve, for real values of x, the following equation:

    $\displaystyle 4 tanh^2 x - sech x = 1$

    Trying to open it up using the sinh x and cosh x rules got me a little bit lost in it

    $\displaystyle sinh x = \frac{ e^x - e^{-x}}{2}$

    and $\displaystyle cosh x = \frac{e^x + e^{-x}}{2}$

    Solution to that one would be awesome

    One more question my brain cells just won't go past is this one:

    Solve, for real values of x, the following equation:
    $\displaystyle sinh^2 x - 5 cosh x + 5 = 0$

    The powers just confuse me, trying to use the identities led me to

    $\displaystyle (cosh 2x -1) - 10 cosh x + 9 = 0$

    (...messing a bit around with it...)

    $\displaystyle e^{2x} + e^{-2x} - 10e^x + 10e^{-x} + 18 = 0$

    And from then onwards, my brain shuts down. I can't get past that =/
    Last edited by Lonehwolf; Nov 19th 2008 at 11:34 AM.
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  2. #2
    Junior Member Lonehwolf's Avatar
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    Just noticed I should have used the sin / cos identities for killing the power first... but I'd still get stuck anyways >_<
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  3. #3
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    Hello, Lonehwolf!

    I used the identity: .$\displaystyle \text{sech}^2x + \text{tanh}^2x \:=\:1$


    Solve, for real $\displaystyle x\!:\;\;4\:\!\text{tanh}^2x - \text{sech }\!x \:= \:1$

    We have: .$\displaystyle 4(1 - \text{sech}^2x) - \text{sech }x \:=\:1 \quad\Rightarrow\quad 4\:\!\text{sech}^2x + \text{sech }x - 3 \:=\:0$

    Factor: .$\displaystyle (\text{sech }\!x + 1)(4\:\!\text{sech }\!x -3)\:=\:0$


    Then: .$\displaystyle \text{sech }\!x + 1 \:=\:0 \quad\Rightarrow\quad \text{sech }\!x \:=\:-1 $ . . . Impossible: $\displaystyle \text{sech }\!x$ is always positive.

    And: .$\displaystyle 4\:\!\text{sech }\!x - 3 \:=\:0 \quad\Rightarrow\quad \text{sech }\!x \:=\:\frac{3}{4} \quad\Rightarrow\quad \cosh x \:=\:\frac{4}{3}$


    We have: .$\displaystyle \frac{e^x + e^{-x}}{2} \:=\:\frac{4}{3} \quad\Rightarrow\quad 3e^x + 3e^{-x} \:=\:8 \quad\Rightarrow\quad 3e^{2x} -8e^x + 3 \:=\:0$

    Quadratic Formula: .$\displaystyle e^x \;=\;\frac{4\pm\sqrt{7}}{3} $

    Therefore: .$\displaystyle x \;=\;\ln\left(\frac{4\pm\sqrt{7}}{3}\right) \;\approx\; \begin{Bmatrix}\;0.795365461 \\ \text{-}0.795365461 \end{Bmatrix} $

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  4. #4
    Junior Member Lonehwolf's Avatar
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    You're something =o

    That makes definitely a lot of sense to me, so 1 problem outta 2 is solved

    Regarding the second problem, I still can't get it to budge Any help much appreciated. I'll give it another try once i finish the last exercise.

    $\displaystyle sinh^2 x - 5 cosh x + 5 = 0$
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  5. #5
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    Hello again, Lonehwolf!

    Solve: .$\displaystyle \sinh^2\!x - 5\cosh x + 5 \:=\: 0$
    Identity: .$\displaystyle \cosh^2\!x - \sinh^2\!x \:=\:1$


    We have: .$\displaystyle \sinh^2\!x - 5\cosh x + 5 \:=\:0$

    Then: .$\displaystyle \overbrace{(\cosh^2\!x - 1)} - 5\cosh x + 5 \:=\:0 \quad\Rightarrow\quad \cosh^2\!x - 5\cosh x + 4 \:=\:0$

    Factor: .$\displaystyle (\cosh x - 1)(\cosh x - 4)\:=\:0$

    Hence: .$\displaystyle \begin{Bmatrix}\cosh x \:=\:1 & \Rightarrow & \dfrac{e^x+e^{-x}}{2} \:=\:1 \\ \\[-3mm]
    \cosh x \:=\:4 & \Rightarrow & \dfrac{e^x+e^{-x}}{2} \:=\:4 \end{Bmatrix} \quad\hdots\quad\text{etc.}$


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