Best approximation of a catenary by a parabola

A parabola and a catenary can look very similar but are not the same. The graph of

y = x^2

is a parabola and the graph of

y = cosh(x) = (e^x + e^-x)/2

is a catenary. You've probably seen parabolas in a math class; you've seen a catenary if you've seen the St. Louis arch.

Depending on the range and scale, parabolas and catenaries can be too similar to distinguish visually, though over a wide range enough range the exponential growth of the catenary becomes apparent.

For example, for x between -1 and 1, it's possible to scale a parabola to match a catenary so well that the graphs practically overlap. The blue curve is a catenary and the orange curve is a parabola.

The graph above looks orange because the latter essentially overwrites the former. The relative error in approximating the catenary by the parabola is about 0.6%.

But when x ranges over -10 to 10, the best parabola fit is not good at all. The catenary is much flatter in the middle and much steeper in the sides. On this wider scale the hyperbolic cosine function is essentially e^|x|.

Here's an intermediate case, -3 < x < 3, where the parabola fits the catenary pretty well, though one can easily see that the curves are not the same.

Now for some details. How are we defining best" when we say best fit, and how do we calculate the parameters for this fit?

I'm using a least-squares fit, minimizing the L^2 norm of the error, over the interval [M, M]. That is, I'm approximating

cosh(x)

with

c + kx^2

and finding c and k that minimize the integral

The optimal values of c and k vary with M. As M increases, c decreases and k increases.

It works out that the optimal value of c is

and the optimal value of k is

Here's a log-scale plot of the L^2 norm of the error, the square root of the integral above, for the optimal parameters as a function of M.

More on catenaries

• Surface of revolution with minimal area
• When length equals area