# Precise form of the mean motion theorem

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.

• @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. – Alexandre Eremenko Aug 18 '17 at 19:39
• 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 $\exp$.) – David E Speyer Dec 17 '18 at 16:11
• @David E Speyer: Yes, thanks. I corrected. – Alexandre Eremenko Dec 17 '18 at 16:46
• Something with the last sentence is wrong because $|a_1| > \sum_{k=1}^{n}|a_k|$ makes no sense. – Sam Hopkins Dec 18 '18 at 0:53
• @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! – Alexandre Eremenko Dec 19 '18 at 3:39