# Hyperbolic cosine looks like a parabola

1. Oct 18, 2012

### Mathguy15

Hello,

I wanted to know why the graph of the hyperbolic cosine function (1/2(e^x)+1/2(e^-x)) looks like a parabola. Is there any reason for this? I suppose the individual exponential functions both go to infinity in a symmetric way.... but I wanted a better reason :).

Thanks,
Mathguy

2. Oct 18, 2012

### phyzguy

They are only qualitatively similar. Cosh(x) looks like a parabola in the sense that it is concave upward and increases faster than linear, but in fact it is very different from a parabola. The height of a parabola (like y=x^2) increases as x^2 as you move away from 0. y=cosh(x) increases exponentially as x gets large. At x=100, the value x^2 is 10^4, while the value of cosh(x) is about 10^43, which is hugely larger.

3. Oct 18, 2012

### AlephZero

The Taylor series for $\cosh x$ is $1 + x^2/2! + x^4/4! + x^6/6! + \cdots$.

So the graph of $y = \cosh x$ does look similar to the parabola $y = 1 + x^2/2$ when $|x|$ is small, but not when $|x|$ is big.