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I was told that Cosh is not a parabola. A hanging rope is Cosh. To me, that seems like a parabola. What is the difference?

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    $\begingroup$ The main difference is at infinity: $\cosh$ has an exponential growth, whereas a parabola has a quadratic growth. $\endgroup$ – Bernard Nov 21 at 22:05
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    $\begingroup$ Depends on what you mean by "parabola"! If you mean "function whose graph looks like a big U but it keeps going forever", then I daresay I would have a hard time arguing that $\cosh(x)$ is not one of these. But if you are more strict, and define a parabola only to be the graph of a quadratic function, then it is not too hard to prove that $\cosh(x)$ is not a quadratic function. $\endgroup$ – Prototank Nov 21 at 22:08
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They do look the same near $0$, which is why you ask. As others have pointed out, $\cosh$ grows much faster than a parabola.

I think the important difference is that when you write down the physics that tells you the rope is hanging in equilibrium the solution to that equation is the hyperbolic cosine, not a quadratic polynomial.

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Try googling "1 + x^2/2, cosh x" and look at the graphs. You can roll your mouse wheel and zoom in and out.

You're right -- near the origin, the graphs are nearly indistinguishable. That's because $1+x^2/2$ is the first two nonzero terms of the Taylor series for $\cosh x$, so we expect that.

However, far away from the origin, you will see that $\cosh x$ behaves like $\tfrac12e^{|x|}$, which grows much more rapidly than $\tfrac12 x^2$.

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By definition $\cosh(x) = \tfrac 12 (e^{+x} + e^{-x})$. As $x\to 0$ we have $\cosh(x) = 1 + \tfrac 12 x^2 +\mathcal O(x^4)$, so for small $x$ it looks like a parabola.

But $\cosh$ is actually more related to the hyperbola (hence hyperbolic cosine). Take the hyperbola $xy=1$ and rotate the coordinate system 45°, which we can do by change of coordiantes $x=u+v$ and $y=u-v$. Then $u^2 - v^2 =1$. cosh and sinh now give you a parametrization of the curve (recall that $\cosh^2 - \sinh^2 = 1$)

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The shape of a hanging rope is the same as the graph of the hyperbolic cosine: it's a catenary. And it grows much faster than a parabola. Actually, it grows exponentially fast, whereas the parabola corresponds to the graph of a quadratic function.

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It is similar to a parabola since

$$\cosh x=\frac{e^x+e^{-x}}{2}>0 \quad \cosh x=\cosh (-x)$$

is even, convex and

$$\lim_{x\to \infty }\cosh x = \infty \quad \lim_{x\to - \infty }\cosh x =- \infty$$

but they are really different functions, for example

$$y=ax^2 \implies y'=2ax \implies y''=2a \implies y'''=0$$

but

$$y=\cosh x \implies y'=\sinh x \implies y''=\cosh x \implies \ldots$$

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enter image description here

Graph of a parabola ($y=x^2$) and a hyperbolic cosine ($y=cosh(x)-1$).

The graph shows quite clearly how they are different. Parabolas grow faster at first but then slow down in comparison to the hyperbolic cosine.


You can create any number of U-shaped functions that don't quite look like each other. $y=x^4$ and $y=x^6$ are examples of a couple of others, as is $y=e^{x^2}$. In fact, you can take any probability distribution that's continuous and unbounded, and invert it and you'd get a U shape.

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    $\begingroup$ If you are going to adjust $\cosh$ to $\cosh(x)-1$, you might adjust the parabola to $\frac12x^2$ $\endgroup$ – Henry Nov 22 at 11:24
  • $\begingroup$ And what would that achieve? $\endgroup$ – Ingolifs Nov 22 at 19:34
  • $\begingroup$ It might show that your "Parabolas grow faster at first" is a result of scaling rather than the shapes of the curves $\endgroup$ – Henry Nov 23 at 8:13

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