Hyperbolic cosine looks like a parabola

[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

 

[phyzguy]

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

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.

 

[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.