That is, the inverse of (e^x - e^-x) / 2. The answer is x + sqrt(x^2 + 1). If you start with x = (e^y - e^-y ) / 2, what intermediary steps lead to the answer?
You can clear the denominator:
2x = e^y - e^-y;
you can re-write e^-y as 1/e^y:
2x = e^y - 1/e^y;
you can then clear that denominator:
2xe^y = e^2y - 1;
I'm basically out of ideas at this point. Hmm term on the right is a difference of two squares. I would appreciate any comments!
When Plato go to , he thought of that as a quadratic equation, of the form with a= 1, , b= -2x, and c= -1. The quadratic formula says that the solutions to are of the form . Setting a= 1, b= -2x, c= -1 that gives .
But notice that so that and, or course, is never negative so we can drop the "-" sign to get exactly what Plato gave.
(It can be simplified a little more: and so that .)