# Why NOT tanh(3x) for Hyperbolic solution of a cubic

Discussion in 'Math Research' started by padraighos, Apr 23, 2006.

Hello from Ireland,

Since Tanh(x) would seem to have the same range of values as Cos(x) ie
-1 to +1,
I can not understand why ALL cubics of the form:-

x^3 + px +q = 0

can not be solved using hyperbolic functions.

While Hyperbolic Sinh(3x) & Cosh(3x) lend themselves for solutions
where the magnitude of Q is > 1, I tried expanding Tanh(3x) to see if
it could likewise be used instead of the

Trignometric Cos(3x) to solve such a cubic as has to depend on Cos(3x)
for solution.

I am having difficulty in expanding Tanh(3x) in a way that would lend
itself towards that end.

It would need to be something like this that i found on the internet:-
TANH(3x) = (3*TANH(x) + TANH^3(x)) / (1 + 3*TANH^2(x))
but alas i dont know where to go from here with it

Can anyone help or even show me why this might not be even possible.

Thank You

2. ### Oscar Lanzi IIIGuest

A real problem is that the tank function over the reals is one-to-one.
This means one root for tanh 3x over the reals corresponds to just one
root for tanh x. So the tanh function does not have the right structure
to generage three real roots.

The sinh, cosh, and cos functions are all really the same function,
rotated and translated in the complex plane. Thus extension of the
hyperbolic functions to include the circular cosine (or sine, which
would work just as well for three-real-root cubics) is quite natural.

--OL

Oscar Lanzi III, Apr 29, 2006