Consider an exponential polynomial $$f(t)=\sum_{k=1}^na_k\exp(i\lambda_kt),$$ where $a_k$ are complex and $\lambda_k, t$ real. The usual form of the Mean Motion Theorem says that the limit $$\lim_{t\to+\infty}\frac{\arg f(t)}{t}$$ exists. (If $f$ has real zeros one defines $\arg f$ by bypassing them along small half-circles in the upper half-plane).

All books that I know mention that this was conjectured by Lagrange, and proved by P. Bohl for $n=3$ and by B. Jessen and H. Tornehave (1945) in general, after earlier incomplete proofs by H. Weyl and P. Hartman. However in the paper of Bohl, a much more subtle question is actually studied, namely whether we have $$\arg f(t)=ct+O(1).$$ He shows that for $n=3$ this is sometimes the case, sometimes not, and gives an exact condition in terms of $a_k,\lambda_k$. My question is:

Has anyone ever continued this line of inquiery? Can $$\arg f(t)=ct+o(t)$$ be improved: a) in general, b) under some additional conditions on $a_k,\lambda_k$?

Of course one such condition is known since Lagrange: if $|a_1|>\sum_{k=2}^n|a_k|$, then the error term is $O(1)$.

Ref. P. Bohl, "Über ein in der Theorie der säkularen Störungen vorkommendes Problem", J. reine angew Math. 135 (1909) 189-283. There is a Russian translation in P. Bohl, Collected Works, Riga, Znanie, 1974.

  • $\begingroup$ @მამუკა ჯიბლაძე: Thanks for the editing. I am unable to remember how to type foreign language accents and umlauts in html. And Math Jack only handles the formulas. $\endgroup$ – Alexandre Eremenko Aug 18 '17 at 19:39
  • $\begingroup$ Should the formula be $\sum a_k \exp(i \lambda_k t)$? (This differs from the stated formula by an $i$ in the argument of $\exp$.) $\endgroup$ – David E Speyer Dec 17 '18 at 16:11
  • $\begingroup$ @David E Speyer: Yes, thanks. I corrected. $\endgroup$ – Alexandre Eremenko Dec 17 '18 at 16:46
  • $\begingroup$ Something with the last sentence is wrong because $|a_1| > \sum_{k=1}^{n}|a_k|$ makes no sense. $\endgroup$ – Sam Hopkins Dec 18 '18 at 0:53
  • 1
    $\begingroup$ @Icv: Buhl's result shows that for $n=3$ it can be improved in cases, different from Lagrange's case. Look in Bohl's paper! $\endgroup$ – Alexandre Eremenko Dec 19 '18 at 3:39

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