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Math Help - Composition of an Inverse Hyperbolic Function

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    Composition of an Inverse Hyperbolic Function

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    \displaystyle \sinh({\tanh^{-1}{x}}) = \frac{x}{\sqrt{1-x}\sqrt{1+x}}
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Sakai View Post
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    \displaystyle \sinh({\tanh^{-1}{x}}) = \frac{x}{\sqrt{1-x}\sqrt{1+x}}
    Let y=\tanh x. Let's first find its inverse.

    Now, x=\tanh y\implies x=\dfrac{e^y-e^{-y}}{e^y+e^{-y}}=\dfrac{e^{2y}-1}{e^{2y}+1}

    Thus, xe^{2y}+x=e^{2y}-1\implies (1-x)e^{2y}=1+x\implies e^{2y}=\dfrac{1+x}{1-x}.

    Thus, in order for us to take \ln of both sides, we require that |x|<1 (do you see why?). Therefore,

    2y=\ln\left(\dfrac{1+x}{1-x}\right)\implies y=\tanh^{-1}x=\frac{1}{2}\ln\left(\dfrac{1+x}{1-x}\right)=\ln\sqrt{\dfrac{1+x}{1-x}}.

    Now,

    \begin{aligned}\sinh\left(\tanh^{-1}x\right)&=\dfrac{e^{\tanh^{-1}x}-e^{-\tanh^{-1}x}}{2}\\ &=\frac{1}{2}\left(\sqrt{\dfrac{1+x}{1-x}}-\sqrt{\dfrac{1-x}{1+x}}\right)\\ &= \ldots\end{aligned}

    Can you take it from here?
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    Quote Originally Posted by Chris L T521 View Post
    Thus, in order for us to take of both sides, we require that |x|<1 (do you see why?).
    Let me try.

    If x\ge1, then 1-x\le0, thus \ln(1-x)\not\in\mathbb{R}.
    If x\le-1, then 1+x\le0, thus \ln(1+x)\not\in\mathbb{R}.
    If -1< x < 1 , then (1-x) > 0 and (1+x) > 0, thus \ln(1-x)\in\mathbb{R} and \ln(1+x)\in\mathbb{R}.

    So if we take the logarithm of \displaystyle \frac{1+x}{1-x}, it would only make sense for |x|<1.

    I hope that's right.
    \begin{aligned}\sinh\left(\tanh^{-1}x\right)&=\dfrac{e^{\tanh^{-1}x}-e^{-\tanh^{-1}x}}{2}\\ &=\frac{1}{2}\left(\sqrt{\dfrac{1+x}{1-x}}-\sqrt{\dfrac{1-x}{1+x}}\right)\\ &= \ldots\end{aligned}

    Can you take it from here?
    Sure.

    \displaystyle \frac{1}{2}\left(\sqrt{\dfrac{1+x}{1-x}}-\sqrt{\dfrac{1-x}{1+x}}\right) = \frac{\sqrt{1+x}\sqrt{1+x}-\sqrt{1-x}\sqrt{1-x}}{2\sqrt{1-x}\sqrt{1+x}} \displaystyle = \frac{1+x-(1-x)}{2\sqrt{1-x}\sqrt{1+x}} = \frac{2x}{2\sqrt{1-x}\sqrt{1+x}} = \frac{x}{\sqrt{1-<br />
x}\sqrt{1+x}}, as required.

    Thanks. I will now try to tackle more of these.
    Last edited by Sakai; August 21st 2010 at 08:11 PM.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Sakai View Post
    Let me try.

    If x\ge1, then 1-x\le0, thus \ln(1-x)\not\in\mathbb{R}.
    If x\le-1, then 1+x\le0, thus \ln(1+x)\not\in\mathbb{R}.
    If -1< x < 1 , then (1-x) > 0 and (1+x) > 0, thus \ln(1-x)\in\mathbb{R} and \ln(1+x)\in\mathbb{R}.

    So if we take the logarithm of \displaystyle \frac{1+x}{1-x}, it would only make sense for |x|<1.

    I hope that's right.
    Great job!!

    Quote Originally Posted by Sakai View Post
    \displaystyle \frac{1}{2}\left(\sqrt{\dfrac{1+x}{1-x}}-\sqrt{\dfrac{1-x}{1+x}}\right) = \frac{\sqrt{1+x}\sqrt{1+x}-\sqrt{1-x}\sqrt{1-x}}{2\sqrt{1-x}\sqrt{1+x}} \displaystyle = \frac{1+x-(1-x)}{2\sqrt{1-x}\sqrt{1+x}} = \frac{2x}{2\sqrt{1-x}\sqrt{1+x}} = \frac{x}{\sqrt{1-<br />
x}\sqrt{1+x}}, as required.

    Thanks. I will now try to tackle more of these.
    Well done. I hope this is helpful when you tackle more problems like this.
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  5. #5
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    Another method is to use the basic identity cosh^2(x)- sinh^2(x)= 1, which can be derived from the definition of hyperbolic functions in terms of exponentials as Chris L T521 did or from the defintion using x as arclength on a hyperbola.

    Dividing both sides of that equation by sinh^2(x) we have coth^2(x)- 1= \frac{1}{sinh(x)} or \frac{1}{tanh^2(x)}- 1= \frac{1}{sinh^2(x)}

    \frac{1- tanh^2(x)}{tanh^2(x)}- \frac{1}{sinh^2(x)}
    sinh^2(x)= \frac{tanh^2(x)}{1- tanh^2(x)}
    sinh(x)= \sqrt{\frac{tanh^2(x)}{1- tanh^2(x)}}

    so
    sinh(tanh^{-1}(x))= \sqrt{\frac{tanh^2(tanh^{-1}(x))}{1- tanh^2(tanh^{-1}(x))}}
    sinh(tanh^{-1}(x))= \sqrt{\frac{x^2}{1- x^2}}= \frac{x}{\sqrt{(1- x)(1+ x)}}

    with, of course, the conditions on x that Chris L T521 gave.
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