Calculating e^x-1 on classic HPs
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01-12-2016, 01:41 AM
(This post was last modified: 01-12-2016 01:49 AM by Thomas Klemm.)
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RE: Calculating e^x-1 on classic HPs
(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 |
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