I have the following hyperbolic identity, which I solved, analytically:
$6~\text{sech}^2 x$ $= 4 + \tanh x$
The two solutions which I get are:
$x=\frac{1}{2} \ln3$ and $x=-\frac{1}{2} \ln5$
These roots, providing they are right, would satisfy the original equation. So to check I substituted these roots into the equation.
For $x=-\frac{1}{2} \ln 5$, I get the correct result $(\text{LHS}=\text{RHS})$.
But for the second root, I don't. It is pretty close but I get; $1.1×10^{-13}\approx 0$.
What I don't understand is where this error comes from, is it my calculator, or am I just doing something incorrect?
I apologise in advance if I haven't use $mathjax$ properly.
Cheers
$1.1 \times 10^{-13}$ is $0.00000000000011$. A 32-bit calculator can store numbers as big as $2^{32}$, which is only 10 digits in decimal. It's not surprising if there's a tiny bit of error when you're dealing with transcendental functions such as hyperbolic functions and logarithms. Use a more high-powered calculator and the error goes away.