Calculating e^x-1 on classic HPs

[Thomas Klemm]

(01-11-2016 10:20 PM)Dieter Wrote:  What do you think?

Brilliant!

Let's assume: \(u=e^x+\varepsilon\) where \(\varepsilon \ll 1\)

Then: \(e^x-1=u-1-\varepsilon\)

But since: \(u=e^x+\varepsilon=e^x(1+e^{-x}\varepsilon)\)

\(\log(u)=\log(e^x)+\log(1+e^{-x}\varepsilon)\)

Now we can use: \(log(1+x)\approx x\) for small \(x\).

\(\log(u)\approx x+e^{-x}\varepsilon\)

Or: \(\varepsilon \approx (\log(u)-x)e^x\)

This leads to: \(e^x-1\approx u-1+(x-\log(u))u\)

Cheers
Thomas