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  1. MHB Craftsman
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    #1
    Does anyone perhaps have a good way for me to get a lasting 'intuition' about what inverse hyperbolics are? I look at, for example, the well known sin x; it is periodic.

    Then, it seems, sinh x is a reflection of sin x about the line y=x.
    (I found an example at )
    It ends up not very dissimilar from sin x , but with a limited range - it is not periodic?

    Then arcsin x is again a reflection of sinh x about y=x. It looks closer to what sinx was , also not periodic?(example at )

    But what do hyperbolic and inverse hyperbolic functions do - apart from causing me to see double after a while ....Sin is a wave, I can look at ripples in a pond etc. The others?

  2. Indicium Physicus
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    #2
    None of the hyperbolic functions, if taken over the reals, are periodic. You have the basic definitions:
    \begin{align*}
    \sinh(x)&=\frac{e^x-x^{-x}}{2} \\
    \cosh(x)&=\frac{e^x+e^{-x}}{2} \\
    \tanh(x)&=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{e^x+e^{-x}}.
    \end{align*}
    The reason people use them at all is mostly because they have some nice identities that you can use sometimes to reduce exponential expressions. For example, suppose you have a second-order linear ODE with constant coefficients, and the roots of the characteristic equation are distinct and real. Then your solutions are exponentials. But you could just as well use hyperbolic trig functions instead. What do you gain? Sometimes a lot. I used them in just this scenario, which was embedded in a larger problem of using Laplace Transforms to solve the diffusion equation. You take the LT in the time domain, and the result is a second-order ODE in the spatial variable. It turns out that if you use hyperbolic trig functions, then doing the inverse LT is a bit easier, because of the simplifications that the hyperbolic trig identities gave you.

    Another note: the hyperbolic secant function is also of particular interest, since it is the shape of solitons in fiber optic cables. It's very close to the normal curve in shape.

    Now, getting to your question: all inverses of single-variable functions, including the inverse hyperbolics, can be found graphically by reflecting over the line $y=x$. This is the graphical analogy of switching the $x$ and $y$ variables, and solving for the new $y$. The result is the function inverse.

    You wrote
    Quote Quote:
    Then, it seems, $\sinh(x)$ is a reflection of $\sin(x)$ about the line $y=x$.
    Not quite. $\arcsin(x)$ or $\sin^{-1}(x)$ is a reflection of $\sin(x)$ about the line $y=x$ - that is the true function inverse. The hyperbolics and the regular trig functions are related via complex numbers:
    \begin{align*}
    \sinh(x)&=- i \sin(ix) \\
    \cosh(x)&=\cos(ix).
    \end{align*}

    Does that help?

  3. MHB Craftsman
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    #3 Thread Author
    That is reassuring thanks Ackbach, I had already noticed the wealth of identities they are involved in, which is part of what made me concerned that I was ignorant of some higher purpose(s); thinking of them as mainly mathematical conveniences makes them less fearsome :-) . Also interesting about the hyperbolic secant function, thanks.

    For sinh(x) = -isin(ix) etc., graphically is that something like a mapping from the real to the complex plane?

  4. Indicium Physicus
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    #4
    Quote Originally Posted by ognik View Post
    That is reassuring thanks Ackbach, I had already noticed the wealth of identities they are involved in, which is part of what made me concerned that I was ignorant of some higher purpose(s); thinking of them as mainly mathematical conveniences makes them less fearsome :-) . Also interesting about the hyperbolic secant function, thanks.

    For sinh(x) = -isin(ix) etc., graphically is that something like a mapping from the real to the complex plane?
    Well, $x$ is real on both sides. To see why this identity is the way it is, I recommend using the Euler identity $e^{i x}=\cos(x)+i \, \sin(x)$. I'm not sure how you would think of it graphically. Maybe other, wiser heads here on MHB could help out with that one. It's definitely not a mapping from the reals to the complexes. But I don't know what it actually is.

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