Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Using the definition $\sinh x = \dfrac{e^x-e^{-x}}{2},\;$ let's say we want to solve $\;y = \sinh x \;$ for $x$.

It's not hard to show that $\;\sinh x \;$ is bijective, so this should have exactly one solution:

$$\begin{align} y &= \frac{e^x-e^{-x}}{2} \\ \\ y &= e^x - e^{-x} \\ \\ 2ye^x &= (e^x)^2 - 1 \\ \end{align}$$ $$(e^x)^2 - 2y e^x -1 = 0$$ $$ e^x = y \pm \sqrt{1+y^2} $$

Now, the solution with "$-$" cannot be, since it's always negative and we can't take the log. Therefore, we get that $y = \sinh x \iff x = \ln(y+\sqrt{1+y^2})$.

What I don't understand is, why are there two solutions to the quadratic? You would think that since we start with the equation for a bijective function and apply reversible operations to it (multiplication by $2$, multiplication by $e^x$, subtracting $2ye^x$) we would get something that has only one solution, but somehow we get two and one has to be discarded. Why does that happen?

share|improve this question
    
You get one complex solution –  Amr Dec 26 '12 at 21:57
    
And, if you study the logarithm further, you get infinitely many complex solutions... –  GEdgar Dec 26 '12 at 22:36

1 Answer 1

You only get two solutions because you use the general method of solving quadratic equations over the all of the reals to solve $(e^x)^2 - 2y e^x -1 = 0$, after substituting $z$ for $e^x$. But $z$ does not range over all of the reals, so you should really either use a specific method to solve a quadratic equations over the positive reals, or use an ad hoc method to solve this equation without using any substitution. Of course a method to solve a quadratic equations over the positive reals will basically just consist of using those values of the quadratic formula that are positive. But one could state the general fact that with all variables ranging over the positive reals the equation $z^2=2bz+c$ has a unique solution $z=b+\sqrt{b^2+c}$, and using that fact with $z=e^x$ one would never see a second solution. What I want to say is: the spurious solution is more an artifact of the method used to find the solution than a characteristic of the original equation itself.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.