How do I "convert" a hyperbolic function into a parabolic function? How can I find a parabolic function that mimics a hyperbolic one? How would I find the parabolic function for the hyperbolic function $y=5\cosh(\frac x5)$?
 A: For the hyperbolic cosine the Taylor series is $\cosh(x)=1+x^2/2!+x^4/4!+\cdots$. Hence $5\cosh x/5 = 5(1+(x/5)^2/2!+\cdots)$ and a close approximation to your function, for values close to $0$ is $y=5+{x^2 \over 10}$. 
A hanging cable or chain appears in the shape of a parabola but that is only an approximation. The ideal shape is hyperbolic or catenary.
A: It sounds like you want a parabola that traces a path similar to $\cosh$.  These are very different functions.  Far from the origin, parabolas can only rise as $kx^2$, while $\cosh$ is exponential, which rises much faster.  Over a particular range, you can find the best parabola that fits $\cosh(x)$ but it is a deception to claim that all even functions that go to $+\infty$ as $x \to \infty$ are similar in any other way.
A: For simplicity, let's take  parabola
 opening up with value $y=1$ at $x=0$.
Take the cosh function as
$y = b \cosh(x/b)$ and the parabola
as $y = a x^2+1$.
To get them as close as possible,
we will try to match the second derivative
at $x = 0$, since both have
zero slope there.
For the parabola,
$y''(0) = 2a$.
For the cosh,
$y' = \sinh(x/b)$
and $y'' = (1/b)\cosh(x/b)$,
so $y''(0) = 1/b$.
To have the parabola match this,
we must have $2a = 1/b$ or $a = 1/(2b)$.
Therefore the parabola that
most closely matches this hyperbola is
$y = x^2/(2b) + 1$.
In your case, $b = 5$ so the parabola is
$x^2/10+1$.
A: The circular trig functions relate to the complex numbers. The hyperbolic ones to the split-complex numbers. Presumably the best analogy for parabolic functions would be the dual numbers.
These are boring. $\exp(\epsilon x) = 1 + \epsilon x$, so we would define


*

*$\mathop{\text{cosp}}(x) = 1$

*$\mathop{\text{sinp}}(x) = x$

