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From Morin's Classical Mechanics, on the chapter of Small Oscillations in Lagrangian Mechanics, he does this approximation on the last equality, I don't understand what happened there.

I get the first approximation, where we neglect the terms of order higher δ^2.

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The 1st one is a direct equivalence because it is a definition.

The 2nd one is a Taylor's expansion to first order.

The 3rd one is a factorisation of common terms.

The 4th one is a geometric series expansion of the denominator, chopped off at first order.

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  • $\begingroup$ Ah, right. Thx! $\endgroup$
    – Lyu
    Commented Apr 20, 2023 at 14:34
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    $\begingroup$ While the second equality can be called a "Taylor expansion to first order" of the denominator, I would rather call it "expanding the binomial and truncating to first order" ... The final term, btw, can easily obtained as the Taylor expansion of the initial expression to the first order. $\endgroup$
    – Sebastian Riese
    Commented Apr 20, 2023 at 15:36
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    $\begingroup$ Yes, I was also annoyed at how long the argument is, when in fact it could have been gotten in a lot fewer steps, by factorising the r0 cubed out, and then one expansion. $\endgroup$
    – naturallyInconsistent
    Commented Apr 20, 2023 at 15:38
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Alternatively use the expansion of $\left(1+x\right)^{-3}$ at $x = 0$, where $x = \delta/r_0$: $$ \begin{eqnarray} r^{-3} &=& \left(r_0 + \delta\right)^{-3} \\ &=& r_0^{-3} \left(1 + \delta/r_0\right)^{-3} \\ &=& \frac{1}{r_0^3} \left[1 - 3\left(\frac{\delta}{r_0}\right) + 6\left(\frac{\delta}{r_0}\right)^2 - 10\left(\frac{\delta}{r_0}\right)^3 + ...\right] \end{eqnarray} $$

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