How was $\cosh (x) = \frac{e^x + e^{-x}}{2} $ computed?

Question

I recently came across this fact in my differential equations book, and haven't been able to figure out how this was computed.

That is the definition of $\cosh$. – Henricus V. Mar 24 '16 at 03:38 — Henricus V., Mar 24 '16 at 03:38
Was there some derivation or intuition that led there? – adamcatto Mar 24 '16 at 03:40 — adamcatto, Mar 24 '16 at 03:40
If you want to prove lhs =rhs then use $e^x=cos(x)+isin(x)$ – Archis Welankar Mar 24 '16 at 03:47 — Archis Welankar, Mar 24 '16 at 03:47
Thank you for the link, it's very informative – adamcatto Mar 24 '16 at 03:53 — adamcatto, Mar 24 '16 at 03:53

score 4 · Accepted Answer · answered Mar 24 '16 at 03:44

Your question presumes that $\cosh$ was a pre-existing function, and someone happened to show that it is equal to $\frac{e^x+e^{-x}}{2}$.

It's more like $\frac{e^x+e^{-x}}{2}$ was recognized to be an important function, and so someone decided to give it a name. Because of its properties and certain similarities to $\cos$, it was named $\cosh$.

So there is nothing to show. $\cosh(x)=\frac{e^x+e^{-x}}{2}$ because that is the definition of $\cosh(x)$.

score 0 · Answer 2 · edited Mar 24 '16 at 05:24

0

$\cos\left(iz\right)=\dfrac {e^{i\cdot i\cdot z}+e^{-i\cdot i\cdot z}}{2}=\dfrac{e^{-z}+e^{z}}{2}\quad\implies \quad \cosh z=\cos\left(iz\right)$

edited Mar 24 '16 at 05:24

Vlad

6,710

answered Mar 24 '16 at 03:48

Rayees Ahmad

1,325

This seems to wander along the edge of circular reasoning. Perhaps the Question, in its lack of clarity, invites this, but your answer could be improved by making clear what is a definition and what is a "calculation". – hardmath Mar 24 '16 at 04:31

How was $\cosh (x) = \frac{e^x + e^{-x}}{2} $ computed?

2 Answers2