# Inverse hyperbolic functions

I am attempting to learn inverse hyperbolic functions but I can't even follow the examples in the book. The example is $y= \sinh y = \frac{e^y - e^{-y}}2$ they they make it into $e^y -2x - e^{-y} = 0$ I don't really understand what is going on, are inverses equal to x? I just don't follow what he did there at all. There just isn't much explanation here, or at least enough for me to figure it out.

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That should read $x = \mathrm{sinh} y$, so that $y=\mathrm{sinh}^{-1} x$ (the inverse hyperbolic sine function) is what you're solving for. And what they did is basically multiply both sides of the equation by $2$ and then move everything onto one side. – anon Oct 7 '11 at 0:42
I stared at it for about 10 minutes and couldn't figure that out...either I am stupid or they should have just added that step. – user138246 Oct 7 '11 at 0:45
Jordan, you shouldn't have to apologize or call yourself stupid. Maybe you are doomed to be forever bad at math, or maybe you have a lot more potential and just can't find the right rhythm - I don't have an answer to that. But it seriously pains me to see you beating yourself up all the time. – anon Oct 7 '11 at 0:53
...and it's certainly unhealthy to be frequently, repeatedly calling yourself "stupid". Things like those, repeated enough, are self-fulfilling. You need to change that habit if you don't want to be forever stuck. – J. M. Oct 7 '11 at 1:03

The text of the question is not clear, and I am a few miles away from my copy of Stewart. But I imagine that the calculation goes something like this. Suppose that $$x=\sinh y= \frac{e^y-e^{-y}}{2}.$$ We want to solve for $y$ in terms of $x$. The first thing to do is to multiply both sides by $2$, so we won't have to carry fractions around. They are heavy.

So we get $2x=e^y-e^{-y}$. Now we can do one of several things. Maybe rewrite $e^{-y}$ as $1/e^y$. So now we have $$2x=e^y -\frac{1}{e^y}.$$ It will save typing, and be useful in other ways, to let $w=e^y$. We obtain $$2x=w -\frac{1}{w}.$$ Multiply both sides by $w$. We get $2xw=w^2-1$. Rearrange this equation a little. We get $$w^2-2xw-1=0.$$ This is a quadratic equation in $w$. The solutions are, by the Quadratic Formula, $$w=\frac{2x+\sqrt{4x^2+4}}{2}=x \pm\sqrt{x^2+1}.$$ Now we remember that $w=e^y$, so $$e^y=x \pm\sqrt{x^2+1}.$$ But note that $e^y$ is always positive, and $x-\sqrt{x^2+1}$ is negative. So the solution with the minus sign has to be rejected, and $$e^y=x + \sqrt{x^2+1}.$$ Take the natural logarithm ($\ln$) of both sides. We get $$y=\ln\left(x+\sqrt{x^2+1}\right).$$

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How do I take the natural logarithm of a number? – user138246 Oct 7 '11 at 1:17
@Jordan: The line in which Andre took the natural logarithm doesn't have a specific number, it has an expression. In order to represent taking the natural logarithm of it, you put $\ln()$ around it. This "undoes" the exponential function, so taking the natural logarithm of $e^y$ returns plain old $\text{ }y$. – anon Oct 7 '11 at 1:20
I don't know why I just can't wrap my head around that concept. So if I am taking loge of $e^y$ it is going to give me y because whatever is making e y will also make the ln y? Also what if I don't have x as a value, what if the value is 5 so I have 5 = sinh inverse? – user138246 Oct 7 '11 at 1:24
Remember, this is one of the standard facts I said in another post comes up over and over, and has to become automatic: $\ln(e^u)=u$. – André Nicolas Oct 7 '11 at 1:26
I don't follow the quadratic part at all. – user138246 Oct 7 '11 at 2:26

Try writing $y = \dfrac{e^x - e^{-x}}{2}$, so $y = \sinh x$, not $y= \sinh y$.

\begin{align} y & = \frac{e^x + e^{-x}}{2} \\ \\ 2y & = e^x + e^{-x} \\ \\ 0 & = e^x + e^{-x} - 2y \end{align} If you multiply both sides of this by $e^x$, you get $$0 = \left(e^x\right)^2 + 1 - 2ye^x,$$ so $$0 = u^2 + 1 - 2yu$$ and you've got a quadratic equation in $u$ $$au^2 + bu + c = 0,$$ where $a=1$, $b=-2y$, and $c=1$.

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I believe you have the $y$ and $x$ backwards. – anon Oct 7 '11 at 1:02