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Can please someone help with finding the Taylor series expansion to this exponential? I am only looking for the first two terms, thanks!

$(1-\varepsilon)^{-(\kappa-1)/\kappa}$

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  • $\begingroup$ What's the $\&$ symbol? And what's the variable? $\endgroup$
    – Andrei
    Mar 25, 2020 at 20:01
  • $\begingroup$ @Andrei sorry there is no &, I just edited it, $\kappa$ is just a constant in this case. $\endgroup$
    – QUTADAH
    Mar 25, 2020 at 20:03
  • $\begingroup$ if $\kappa$ is a constant, then $\exp(-(\kappa-1)/\kappa)$ is also a constant $\endgroup$
    – Andrei
    Mar 25, 2020 at 20:06
  • $\begingroup$ @Andrei exactly. $\endgroup$
    – QUTADAH
    Mar 25, 2020 at 20:07

1 Answer 1

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For a constant $C$ and small $\varepsilon$, $$(1-\varepsilon)^C\approx 1-C\varepsilon+\frac {C(C-1)}2\varepsilon^2+...$$ Just plug in $$C=-\frac{\kappa-1}\kappa$$

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  • $\begingroup$ @Thanks alot! Do you have a reference how you came across this answer? $\endgroup$
    – QUTADAH
    Mar 25, 2020 at 20:23
  • $\begingroup$ Take the derivatives: $f(x)=(1+x)^p$. Then $f'(x)=p(1+x)^{p-1}$ and $f''(x)=p(p-1)(1+x)^{p-2}$ $\endgroup$
    – Andrei
    Mar 25, 2020 at 20:27
  • $\begingroup$ Thank you :) It worked. $\endgroup$
    – QUTADAH
    Mar 25, 2020 at 20:44

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