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I have read through several proofs of Bertrand's Theorem, including the one on Wikipedia. A typical proof can be found here (Santa Cruz Institute for Particle Physics). Almost all proofs using Newtonian mechanics, including the one on Wikipedia, includes a step to neglect the higher-order terms on the RHS of the equation $$ \frac{d^2 x}{d\theta}+(1-\frac{dJ}{dx}|_{x=0})x=O(x^2). $$ and subsequently, to deduce that $x=A\cos \beta t$ and that $\beta$ is rational.

I am quite a bit shocked since ignoring all terms in $O(x^2)$ is mathematically unacceptable. Of course, you can say that when $x$ is small, higher-order terms doesn't count, but such a statement is not mathematically rigorous; we must use $\epsilon$-$\delta$ language to describe that, which is very difficult.

Note: I am NOT saying the big O notation is not rigorous. I am saying that IGNORING or Negleting terms inside big O is not justified.

Can anyone suggest a more rigorous proof? Perhaps just fill the mathematical gaps for the document I linked above?

Edit: Even if the $O$ term is arbitrarily small, all we can say is that the exact solution and the approximate solution lies very close to each other. However, the author in the linked text is saying that if the approximate orbit is closed, then so is the exact orbit - this is NOT guaranteed by the fact that the approximate and exact solution lie very close to each other.

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  • $\begingroup$ There is absolutely nothing non-rigorous about $\mathcal O(x)$ notation, cf. big O notation for the rigorous definition. (Not saying that the proof was rigorous to begin with, but spelling out the $\epsilon$-$\delta$s will not make the proof any more rigorous than it already was). $\endgroup$ – AccidentalFourierTransform Nov 16 at 2:43
  • $\begingroup$ @AccidentalFourierTransform I am NOT saying big O is not rigorous. I am saying that IGNORING big O is not justified. $\endgroup$ – Ma Joad Nov 16 at 2:45
  • $\begingroup$ $\epsilon-\delta$ is unnecessary for clarifying the big O, but NECESSARY when assessing its effect on the solution of the DE. $\endgroup$ – Ma Joad Nov 16 at 2:49
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    $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. For the record, the link is Appendix A of Goldstein, 2nd edition. $\endgroup$ – Qmechanic Nov 16 at 14:45
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The big $O$ term is ignorable because we are showing that a necessary condition for closed orbits is that the potential harmonic or inverse square. We are led to investigate the stability of circular arbits under that perturbations are arbitrarily small so the $O$ term is arbitrarily small and can be ignored. If we going the other way --- trying to prove that having a harmanic or inverse square potential orbits is sufficient for closed orbits --- then we could not ignore the $O$ term. However, in this direction, we have the exact solutions and can see explicitly that the orbits are closed.

Note added in reponse to comment: No the Wikipedia author did not say that "if the approximate orbit is closed, then so is the exact orbit." He/she said the exact opposite. He/she says that if the always existing exact circular orbit is stable then i) $\beta^2>0$ in the stabilily equation. Then ii) he says that if the exact orbit is stably closed then so must the approximate orbit in the limit of very small perturbations, so $\beta$ is rational. He then shows that this excludes all potentials except power laws. Then again by a stability argument he shows that only three cases $\beta^2=0,1,4$ can possibly give stable closed orbits. All the time he is therefore excluding stably closed orbits for any other values. At this point in the proof he never shows or tries to show, that an exact orbit is closed.

It's easy to show that something is unstable. It is hard to show that something is stable over any finite range, and the author never tries to do that.
He only does shows that the orbits are stably closed for $\beta^2=1,4$ the in the (trivial) proof that follows from the exact solutions in those cases.

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  • $\begingroup$ Even if the $O$ term is arbitrarily small, all we can say is that the exact solution and the approximate solution lies very close to each other. However, the author in the linked text is saying that if the approximate orbit is closed, then so is the exact orbit - this is NOT guaranteed by the fact that the approximate and exact solution lie very close to each other. $\endgroup$ – Ma Joad Nov 16 at 22:11
  • $\begingroup$ I will respond by editing answer. $\endgroup$ – mike stone Nov 16 at 23:14
  • $\begingroup$ Thank you, but I still have not got it :( I admit that I made a mistake in my last comment, but why is this true: "if the exact orbit is stably closed then so must the approximate orbit in the limit of very small perturbations"? How can I prove this? $\endgroup$ – Ma Joad Nov 17 at 0:27

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